Solid State Guitar Amp Forum | DIY Guitar Amplifiers

 Probably a simple fix but when I turn the effects knob (eff)  to the right the volume cuts out, starting at about 12 o'clock and then goes completey dead when turned all the way to the right. I used Deoxit and got rid of all the scratchiness but same problem. I removed the pot and measured 22K as it should be. As I turn the pot to the right it's as if the signal is being shorted to ground and then at the far right: dead, volume completely cuts out. ???

What are you using in the loop?  Try a plain patch cord.

It kills the sound because you have it turned all the way over to the loop.   That is your FX loop balance control.  All the way to that end means the only thing going on to the power amp is whatever is in the FX return jack.  Unless there is something in the loop, this will always happen.  If you do not use the loop, then leave the control at the other end.

Thanks. So obvious now. Amp is working as it should. I've never used outboard effects so I've learned something new. :)

This amp is dirty! What do people recommend for tolex/faceplate?

Aggressive cleaners can attack plastic if they have solvents or scratch labels off if abrasive, so classic cleaner (at least for starters)  is some cloth which you wet under the tap and then squeeze so it's humid but not dripping (you don't want water seeping inside) where then you add a drop of dishwasing type detergent.

Wipe everything in many ways, pull knobs if necessary, use a similarly treated Q Tip to clean hard to reach places, on large very dirt encrusted Tolex surfaces an old toothbrush and lots of patience will help.

Then wipe everything off, the small amount of detergent will leave behind a slight foam.

Then wash cloth under the tap, squeeeze again and sweep old water+detergent off.

Repeat a couple times.

Wipe it dry with a new dry clean cloth, let it dry naturally, if you touch it you will notice it later.

It will be clean but dull, then you can apply some spray furniture polish or whatever you use on car seats.

Most people skip the cleaning process and straight apply spray polish, net result is a still very dirty, gunky amp.

Can't suggest brands because every Country has its favorites but you know what I'm talking about.

Remember cleaning knobs, strap handles, etc. with  water with a drop detergent (don't overdo it)  and an old toothbrush.

Quote from: J M FaheyRemember cleaning knobs, strap handles, etc. with  water with a drop detergent (don't overdo it)  and an old toothbrush.

Dish washing liquid, warm water, several brushes and lots of toilet paper/tissue.

:dbtu: :dbtu:


Also, metho is pretty safe for cleaning off other gunk like coffee or wine, but don't use it on finger grime (front panel) because it turns it into a nasty sticky sludge.

Great help and answers. Thanks guys!

With this amp and others is there a way to have the circuit board out of the amp with the appropriate wires still connected so that I can test components with the amp in it's "on" state? Is it a matter of making sure the pots that need to be grounded are grounded?

Quote from: Hawkis there a way to have the circuit board out of the amp with the appropriate wires still connected so that I can test components with the amp in it's "on" state?

Every one is different.

I've worked on a huge range of equipment, and how easy it is to faultfind and repair covers a vast range from bloody wonderful to bloody appalling.

There is a special case of Murphy's Law which states that "an intermittent fault will only occur when the last of 24 retaining bolts are tightened down".  Well I once had exactly that with an industrial controller in a big flameproof box.


The very best:

A Philips car cassette player in for service.  Open lid and find a stack of three large PCB's.  Single multipole connector on one corner - unplug.  Now the whole lot swings up out of the mechanism and flat over the back, then the rest unfolds to the right, and the final layer folds forward and the connector can again be plugged in and the whole thing run while spread out flat on the bench.  Cunning design.


One of the very worst (and there are very many more contestants for this award, the world if full of "cost-effective" stinkers*):

Early Yamaha 8-channel digital mixing desk with flying faders.  Now generally speaking Yamaha stuff is pretty good, I've got a fair bit and it all works and is built well.  But there are exceptions and this mixer was an extreme one.

The flying faders where now slipping a lot, but investigation suggested that it was a design problem that couldn't be fixed.

One of the inputs was dead which turned out to be a dead A/D converter module which was unobtainable.

The proximate reason for it coming in was that it had started to forget its presets, and I suspected a dying backup battery.

When I opened the mixer up I found several large PCB's in layers about an inch apart, and the lot laced through with many un-insulated tinned copper wire links scattered all over the boards and running from top to bottom.  I did locate the backup battery but it was soldered in to the centre of one of the middle layer boards - simply impossible to access without seriously demolishing the PCB assembly.

{* saw a very impressive looking iPod dock/amp/speakers junked after less than a year.  Inside it was boards and modules just indifferently glued to the inside of the fiberboard case, as crude and rough and sloppy as one could imagine.)


Every time I open up some item of gear I can tell what sort of people designed it and put it together, and what sort of attitudes they have to the planet and their fellow humans.

With guitar amps, Fenders the component boards are so wired down you have no hope of swing one up, but other amps, I'm thinking Overreem, all the connections to the boards come off one end - a few screws and presto, you can work on the board hot, standing on end.

We had two rack amps in the PA, a New Zealand Perreaux, and an American CueTek.  The most significant difference between them was that the Perreaux had a removable lid and bottom, leaving the amp totally accessible, while the CueTek was poured into a U-channel case and you couldn't get at anything.

A lot of gear is designed to be manufactured.  It is not designed to be disassembled.  At all.  Ever.  Much less repaired.

A large part of my life has been figuring out how to get assemblies open, particularly stuff that wasn't intended to be opened again.

QuoteWith guitar amps, Fenders the component boards are so wired down you have no hope of swing one up, but other amps, I'm thinking Overreem, all the connections to the boards come off one end - a few screws and presto, you can work on the board hot, standing on end.

We had two rack amps in the PA, a New Zealand Perreaux, and an American CueTek.  The most significant difference between them was that the Perreaux had a removable lid and bottom, leaving the amp totally accessible, while the CueTek was poured into a U-channel case and you couldn't get at anything.

A lot of gear is designed to be manufactured.  It is not designed to be disassembled.  At all.  Ever.  Much less repaired.

A large part of my life has been figuring out how to get assemblies open, particularly stuff that wasn't intended to be opened again.

Thanks Roly, your years of experience pretty much says it all. Especially "designed to be manufactured not repaired". I'll soon discover what I can and can't do with amps, and hopefully I'll have some stories to pass down. :tu:

So this amp is working and using a voltage divider equation there should be 1.92 volts across R118 (should there?). But when I measure with my voltmeter I get .007vdc. I put a 1khz signal through the amp then I played my guitar and measured the voltage across the resistor and measured it referenced to ground but nowhere near this 1.92v. What am I missing here? I think I'm missing the obvious but what?

R117=820ohms, R116=470ohms

820+470/470=2.74X.7=1.92V

R118 is a 0,33 ohm resistor in the emitter lead of TR8.  How did you decide it ought to have 1.92v across it?  That would be about 6A flowing through it.

What do you consider to be a voltage divider here, and what is it across?

R117,116 do form a voltage divider, but they do not put a voltage on R118, rather they divide the voltage that drops across R118 for use by the base of TR12.

Whatever voltage drops across R118 is also across the parallel R117,116 pair.  820+470 is 1290 total, so at the base of TR12, we expect 470/1290 x (V of R118).

This is a classic limiter circuit.  The voltage across R118 varies with the current through it, the more current, the higher the voltage.  If that current, hence voltage, gets high enough, the divided voltage at TR12 base is enough to turn the transistor on, and that then causes the collector to drag down the signal at the base of TR8.

Enzo thanks so much for the response. Below I quote several times from a post on another forum that didn't involve myself. This person gave a fairly thorough analysis of this amp to someone trying to figure it out. When I was referring to a voltage divider I thought I knew what I was talking about as in this same post he referred to R74 and R73 as a voltage divider and set about making calculations 1800 + 680/680 X .7=2.55 volts. He did this to calculate Collector Emitter voltage across TR9


"For TR8 the resistors R116, 117, 118 and transistor TR12 and D5 make a current limiter.
For TR10 the resistors R96, 97, 115 and transistor TR11 and D4 make a current limitter."

So, below, is where my intial confusion came because I couldn't measure the voltage he refers to: 1.92V
"The calculations for the current are similar to the calculations at TR9.
It will result in 1.92 volts accross R118 , this will give a limit of 1.92 / 0.33 = 5.82 A
For R96 it will result in 2.14 Volts , this will give a limit of 2.14 / 0.33 = 6.48 A."

I somewhat understand what you're saying about  pulling down the signal on TR8 if the voltage gets to high. By signal are you referring to current, therefore this config. acts as a current limiter? 

One day the penny will drop, but it sure is taking awhile! :-[

The limiter action serves to limit the current through R118 and thus TR8.  If the current gets past a certain point, then the base or TR12 is turned on so the transistor conducts.  That prevents any additional current going into the base of TR8.

Off hand I don't know why there would be a difference between the voltages at R118 and R96.  I don't know what entire discussion took place with your 1.92 volts, but I suspect it was confused.  I would not expect 1.92v there, but I might suspect that 1.92v there could be the threshold for turning on TR12.  let me try that.

If I start with 1.92v across R118, that also means 1.92v across R117,116.  That is 820 over 470.  The voltage at their junction would be 470/1290 times 1.92.  I get 0.7v, which is about what ought to turn on Tr12.  I would expect the base voltage of that transistor to be closer to 0.5v, but 0.7v is the classic figure.  Thus I have to imagine they were discussing the voltage across r118 required to initiate the limiting function.  It is not a voltage you would expect to find there unless the amp was maxed out.  It would not be a steady DC voltage though.


R73,74 is a completely different circuit.

Quote from: HawkI'll soon discover what I can and can't do with amps

One of the great joys of my life is opening things that were never intended to be opened, generally while repairing something that was never intended to be repaired.

I fancy that I can hear, somewhere far across a distant sea, some dude in a suit yelling in pain "You're killing me here!  THROW IT OUT!  BUY A NEW ONE!  MAKE ME RICH!".  And I thumb my nose - I'm on the side of the planet here; repair, reuse, recycle.   :dbtu:

Quote from: Hawkthen I played my guitar and measured the voltage across the resistor and measured it referenced to ground

The voltage across the emitter resistor is a proxy for the current passing through it and the transistor.  The voltage to ground is effectively the output voltage, whichever side of the emitter resistor you measure it.

We basically have two situations, static or idle where all the voltages and currents are steady and a multimeter is the most suitable measuring instrument; and dynamic or being driven with signal when the CRO is really a more suitable instrument than a multimeter.


Here are two popular forms of output stage current limiting;
(http://www.renardson-audio.com/prot.gif)
(Note that the output stage consists only of Q2 and the 0.2r emitter resistor, everything else is protection).

Like Enzo, to me that discussion looks like determining at what current the protection would cut in.

To find the ballpark: assuming a 50W amp with +/-35V rails into an 8 ohm load.  Now we know all these numbers are rubbery and subject to a lot of "yeah, but...", however to a first approximation we can estimate the (lossless) peak output current to be;

I = Vrail/Rload

35/8 = 4.375 amps peak.  This is the minimum must-not-operate current because it is within specs.

A typical emitter resistor is 0.22 ohms, so the resulting peak voltage across the emitter resistor will be;

E = Ipk * Re

4.375 * 0.22 = 0.96 volts peak (or not quite a volt peak).

Now as Enzo said, the "given" value for the Base-Emitter cut-in voltage for a silicone transistor is "0.6V" (+/-0.1V in reality).  If we simply connected the protection transistor Base to the emitter resistor it would start to conduct (and clamp the drive into the power transistor) too early, limiting output peaks that are within ratings.  We actually want it to limit for some current above the allowable peak of 4.375 amps, and this might represent a voltage almost double, say 1.0 to 1.2 volts, and this is where the (roughly 2:1) divider across the emitter resistor comes in.

The question being addressed in the other forum is, "at what current does this circuit actually limit?" and the currents quoted look to be just about what I would expect;

QuoteIt will result in 1.92 volts across R118 , this will give a limit of 1.92 / 0.33 = 5.82 A
For R96 it will result in 2.14 Volts , this will give a limit of 2.14 / 0.33 = 6.48 A.

So the +ve and -ve peak current limits are a bit different, but outside the normal working peak current of 4.375 A, and still well within the capabilities of modern transistors.  What has been prevented is that very high current spike, tens of amps when the output gets shorted under full drive, that kills the output transistors.

So this is worked backwards, starting with the Vbe ("0.6V" nominal) of the protection transistor, multiplied by the resistive divider, giving a trigger voltage across the emitter resistor, and finally the current for that trigger voltage.


The only time there should be any activity in these protection circuits is when the amp is being driven into a load of lower resistance than it was designed to tolerate, e.g. a 4 ohm cab on an 8 ohm minimum.


{Sidebar: on the cct I have the OP transistors are shown with double emitter arrows.  This means that these devices are Darlington transistors;

(http://upload.wikimedia.org/wikipedia/commons/thumb/3/34/Darlington.svg/220px-Darlington.svg.png)

... which have inbuilt driver transistors, and thus their Vbe is double normal, around "1.2V" nominal.}

Hey Enzo and Roly,

Wow, thank you Enzo and Roly for your time again. I feel like I should be paying you guys! I went over all your equations and tried them on my own and I understand them. A few more pennies dropping for sure. You can look at this stuff for months and years on your own and not have a clue what it is you're really seeing. So thanks for the virtual apprenticeship :tu:

Thanks also for the info about the Darlington Transistors with the built-in driver transistors, I wouldn't have known that.

QuoteThe voltage across the emitter resistor is a proxy for the current passing through it and the transistor.  The voltage to ground is effectively the output voltage, whichever side of the emitter resistor you measure it.

When you say a proxy do you mean I can treat as a gauge for  current flow? With multimeter check for high voltage, therefore high current, low voltage low current?



What is the purpose of D4,D5, what is their function in the current limiting circuit? To take only the pos/neg half of the signal? D5--neg. half, D6, pos. half? Both turning on at.7v?


What is the purpose of TR9, R73,R74, TR4, TR7, C49, C47? (I might as well work backwards to fully understand this power amp). Thanks again.

Also, a buddy is bringing over an old tube radio he wants me to fix for him. Can anyone suggest a good forum to ask questions?

In this very section of the forum, up top are some "sticky" threads, the ones that don't move down over time.  One of those is titled Book ABout Solid State Amps.  Go download it and read it.

By measuring the voltage across any resistor, you can determine the current flowing through it.  That is Ohm's Law, one of the most fundamental concepts in electronics.

Tr9 and the rest of your list are the bias circuit.  They provide what I call a voltage space between the bases of the opposing output transistors.

Quote from: HawkI feel like I should be paying you guys!

Insert Card Here...
(http://www.thebuzzmedia.com/wp-content/uploads/2010/01/atm-card-skimmer-slot-insert-hidden.jpg)

Quote from: HawkWhen you say a proxy do you mean I can treat as a gauge for  current flow? With multimeter check for high voltage, therefore high current, low voltage low current?

Exactly.  Current, voltage and resistance are locked together via Ohms Law.

Your older moving coil or d'Arsonval type meter actually responds to current, so voltage must be implied by the large resistance in series with the meter.

Modern digital meters typically only measure voltage, so current must be implied from a small resistance in parallel with the meter.

Similarly with resistance measurement a moving coil meter puts a fixed voltage across the terminals and measures the resulting current, while a DMM applies a constant current to the terminals and measures the resulting voltage.

"E equals I multiplied by R" said Georg Simon Ohm - and you can bet your life on it.

---

Quote from: HawkWhat is the purpose of D4,D5, what is their function in the current limiting circuit? To take only the pos/neg half of the signal? D5--neg. half, D6, pos. half? Both turning on at.7v?

Ohdearohdearohdear.  On the (fuzzy) circuit I'm looking at <8080-pwr.gif> D4 and D5 are drawn the wrong way around!  Both are supposed to be pointing downwards, not upwards - as drawn the protection won't work.

They are there to protect the associated clamp transistor from being reversed biased on the "other" half cycle.

Quote from: HawkWhat is the purpose of TR9, R73,R74, TR4, TR7, C49, C47?

See;
http://www.ssguitar.com/index.php?topic=3592.msg28780#msg28780 (http://www.ssguitar.com/index.php?topic=3592.msg28780#msg28780)

---

Quote from: HawkAlso, a buddy is bringing over an old tube radio he wants me to fix for him. Can anyone suggest a good forum to ask questions?

Do not twiddle any of the many presets - these require special equipment to set.

Basic superheterodyne radio receiver and waveforms.
(http://archive.hnsa.org/doc/missile/img/figapp21.jpg)

Most electronics faults are common, simple, and fairly easy to fix, however there are some considerable differences between an audio amplifier and a radio receiver (I assume AM only), and a bunch of different concepts and techniques that are required.  If the problem turns out to be simple, e.g. a valve not lighting up, or in the audio section, then you're away, but ahead of the audio section there are a number of other stages;typically an Intermediate Frequency amplifier with tuned LC loads working at 455kHz, before that a Radio Frequency mixer stage, a Local Oscillator running around a megahertz, and perhaps a RF preamp stage (as above).

Radios made for the consumer market have a long history and are not helped by the commercial competition which saw very many designs with all sorts of odd circuit arrangements with the aim of getting maximum performance with minimum cost - and some of the worst cost cutting would make your eyes water.  There were also Patent wars where manufacturers did things differently because they had to, or pay.


By all means have a look (run a mile if it's an AC/DC set, they are killers), check that the chassis is grounded for your safety, then just the usual stuff, valves lighting up?  HT voltage?  Any signs of life in the audio?  Can we see signals anywhere (for which a CRO is very handy indeed)?

Same old story, signal flows from antenna connection to speaker, how far is it getting, and why not?

But having said that, don't be surprised if you very suddenly find yourself well out of your depth.

My first job in electronics was production line servicing AM radios, so I have repaired literally thousands.  Most service techs get to be expert in one or two particular fields, get to know radios, TV's, VCR's, cellphones, laptops, or microwaves and other kitchen appliances, &c&c, and often not a lot outside their area.  I have covered a lot of fields; design, production, industrial, bio-med, off-grid power, theater/film, AM/FM/UHF-TV broadcast, &c&c, and tried to be an "electronics generalist" so the list of things I've worked on (and normally fixed) is huge.

I'm quite happy to hand-hold you through a radio repair, but you need to clearly understand that this opens a whole new bag of concepts (and frankly, I don't think is possible via forum or email if the problem is in the IF or RF stages).

Now we warn folks that building electronics projects can be highly additive, but we don't often mention that if you want to be any good at electronics you had better be prepared for a lifetime of reading, study, and a bit of research (even if you are only ever going to do guitar amps).

You want a flying start? - then NEETS, here;
http://www.fcctests.com/neets/neets.htm (http://www.fcctests.com/neets/neets.htm)

A good grounding in Module 1 is vital, but after that you can just browse Modules as the need arises ('tho you may well find that Module 17, Radio-Frequency Communications Principles, first needs an understanding of Module 12, Modulation Principles, and perhaps others).


Yagotta stay curious 'bout stuff.

I recently bought a couple old nice looking tube table radios for $5.  I'll either fix them or turn them into cute guitar amps.  But if I make them work, frankly, what I expect to find is bad caps.  I doubt the IF and RF stages will be out of tune, at least unless someone was in there messing with them.   If it is a tube radio, it is at least 40 years old, so very likely the electrolytics are dried out.  Depending upon age, chances are also good the coupling caps are leaky.  ANy of the old wax covered ones I can just about guarantee are history.  Disc ceramics are probably OK.  Film caps may or not be, one determines that easiest by checking for DC on the downhill side of them.

For many years I restored old tube jukebox amplifiers, and that was the majority of the work I did on them, replacing most of the caps.

Here is a clean looking copy of the Marshall schematics:

Thank you Enzo, that is so much easier on my poor old eyes.

{And here we note that D4 and D5 are drawn the right way around - but now they have forgotten the values for R97 and R117!  (sigh)   "No circuit is without errors".}

When it comes to repairing anything I have observed that commonsense and a good set of eyeballs can take you far - i.e. paying attention helps a lot.

Quotebut now they have forgotten the values for R97 and R117!  (sigh)   "No circuit is without errors".}

You sure?  ;)

What about:



hint: the same board is used in 2 different amps, the 8080 combo (80W/8 ohms)  and the 8100 head (100W/4 ohms) .

Not only they need different rail voltages, but also different short protection.

And my paying more attention would have helped.   :-[

Okay, bum rap, that was off the screen in the x2 magnification I was looking at it.  Still, they did manage to get the diodes the right way around this time.

Roly, I kept trying to insert my card into that image but no luck! I'll keep trying! haha!

Roly, thanks for posting that radio info and the schematic and the information you've gleaned from working on radios.  Also, the current limiting info and link are great as well as the Neets site. Great info indeed!!

Enzo, I have downloaded that book you suggested and have read about 70 pages, and there are sections I have to keep re-reading but it is a great book and I will finish and re-read for sure.

So Beyond the Output Transistors we have resistors and caps between the OT's and the speaker jacks, plus some caps. Are these covered in the solid state amp book? Or can you give me a basic understanding of their importance? Thanks.

Also, What does O/P stand for? I believe it's the middle rail separating the +VE/-VE.

How long has this forum been in operation? I'm tempted to print off all this info but rather than do that and save some trees I'd rather return to it electronically a month/year/decade down the road. Is there a way to bookmark this post as it will eventually descend further down the post line and be tougher to locate. Can it be saved electronically? Thanks.

Where do you see "O/P"?

In one context it may mean output, while in the context of a forum thread it may mean original poster.

I'm referring to the O/P at the bottom of Roly's current limiting diagrams at top of page. Thanks

Quote from: Hawk on March 19, 2015, 09:03:03 PM
I'm referring to the O/P at the bottom of Roly's current limiting diagrams at top of page. Thanks

In this context "O/P" = output, the amplifier "half-rail".


Working backwards.

Quote from: HawkHow long has this forum been in operation? I'm tempted to print off all this info but rather than do that and save some trees I'd rather return to it electronically a month/year/decade down the road. Is there a way to bookmark this post as it will eventually descend further down the post line and be tougher to locate. Can it be saved electronically? Thanks.

If there is one thing that forums generally lack it's indexing, being able to find what you want in ways other than Google.  Proper indexing of topics has to have a human input from somebody who knows a bit about the craft, sundry wrinkles of terminology that are, or are not, conceptually linked.

The mechanics are there to be used.
Each forum, tread and post has a heading, and that heading contains the URL of the item (right click copy link location, paste, e.g. your post)
http://www.ssguitar.com/index.php?topic=3707.msg28813#msg28813 (http://www.ssguitar.com/index.php?topic=3707.msg28813#msg28813)

The page number contains the thread page;
http://www.ssguitar.com/index.php?topic=3707.0 (http://www.ssguitar.com/index.php?topic=3707.0)

The heading bar;

QuoteSolid State Guitar Amp Forum | DIY Guitar Amplifiers >
    Solid State Amplifiers >
    Amplifier Discussion >
    Marshall Valvestate 8100

Also contains URL links.

As you can see above I make free use of hotlinking content.  To me that is the whole idea of the internet.

In Cyberspace the "distance" is the retreval time.

Every single post on every forum is simply an available item, and all you have to do is point to it - from anywhere.

Any page on the net may consist almost entirely of links to external content.  In fact I ran for a short time an edited index to a forum.  I didn't want to stop people going to the forum, but a group of people wanted to present an edited version ordered by topics.

You can make up HTML pages for your personal use, or topic sub-folders in your browser bookmarks.

Quote from: HawkSo Beyond the Output Transistors we have resistors and caps between the OT's and the speaker jacks, plus some caps. Are these covered in the solid state amp book? Or can you give me a basic understanding of their importance? Thanks.

(http://webpages.charter.net/dawill/tmoranwms/Circuits_2008/2N6254_Power_Amp.gif)

Okay, mid-right we have the power transistors driving the half-rail via a couple of very low value "emitter" resistors.

Next we have a low value resistor in series with a cap to ground.  This is the Zobel stability network.  It is to prevent parasitic HF or VHF oscillation modes in the output pair.

Then we have a small inductor and resistor in parallel.  This is to isolate the output stage from any capacitive loads (such as crossovers or piezo tweeters) that may cause instability.  The resistor is to damp the self-resonance of the inductor (and is often also the form for winding the coil).

According to Doug Self (http://www.douglas-self.com/) the values of these components is not critical.

  Roly, perhaps you could comment on the origin of the term "half rail".
To someone who has only ever seen split supply designs, it may not make sense without context.   :)

Thanks g1, I was hoping it was implicit, but you're right, it should be nailed down.


In the amp circuit above the output or "half-rail" is the bit between the speaker output terminal (22uH inductor) and the 22k Negative FeedBack (NFB) resistor.


We also have two supply "rails", +38V and -38V with respect to ground.  ("rail" simply means that it is a distributor (of power or signal) and has quite a few connections; "buss" is sometimes also used in a similar context, audio mixers, power distribution).

In between we have the output "rail" connected to the output pair emitter resistors, the speaker, and the Negative FeedBack (22k above).  This output rail is also known as the "half-rail" because when the amp is working correctly it adopts a voltage half way between the supply and ground, or half way between the split supplies (+/-100mV).

This is actually a very important test because (apart from not lighting up your speakers - which is why we keep hammering "disconnect the speaker") if the half rail is right it is a strong indication that the whole output stage is working correctly - it has to be to achieve this "rebalanced" condition (the OP stage is basically a giant power op-amp that rebalances itself to the midpoint because of a lot of DC NFB - if something in this NFB loop isn't working properly then the amp is highly unlikely to "centre" or rebalance the half-rail).

On first encounter the half-rail voltage can also give you a strong indication as to where to look first.  If, say, the upper OP transistor is shorted then the half-rail will be close to the +ve supply, similarly for the lower transistor, and if it is at some other odd voltage it suggests that both OP transistor have blown open.  Like a pulse the half-rail voltage can tell you a lot about the patient with only one measurement.

In the early days of transistor amps it was fairly common to use a single supply;

(http://www.ozvalveamps.org/repairs/solidstateamprepair/singlerailhp.jpg) (http://www.ozvalveamps.org/repairs/solidstateamprepair/capcoupling.jpg)

...where the output would bias to half the supply voltage, hence "half (supply) rail", and needs to be DC blocked by a large output capacitor.  You still find this arrangement in smaller combo amps, particularly older ones of Asian origin.  (The output cap is more commonly between the OP stage and the speaker, with one side of the speaker grounded.)

(http://www.ozvalveamps.org/coronet/84_12_sb-r.jpg)


The trend a bit later was to use a split supply;

(http://www.ozvalveamps.org/repairs/solidstateamprepair/dualrailpsuhp.jpg) (http://www.ozvalveamps.org/repairs/solidstateamprepair/directcoupling.jpg)

...which eliminated the need for a specific output DC blocking cap because the "half rail" was now half way between +ve and -ve, i.e. at ground.  This is by far the most common arrangement these days, and particularly in higher powered amps.


Actually there isn't a lot to choose between the two arrangements, but Doug Self does make an interesting observation that in the latter the speaker is DC coupled to the output stage, and if the OP stage should fail you can get speaker combustion.

Some people (generally Hi-Fi buffs) don't like having a big electrolytic in series with the speaker with all those audio amps passing through (because, ya know, capacitors - they tend to have a thang about capacitors) , but you can redraw both arrangements to show that for audio frequencies it's a distinction without a difference.  In the split rail design the output currents also flow through the power supply filter caps anyway, it just isn't as obvious.

Ground
If you examine the circuit fragments above you will notice that the ground is in different places.  Which bit of the circuit is actually grounded to the chassis is really a bit arbitrary.  You may be using a metal speaker jack that has one side connected to the metal case.  With the single supply it then makes sense to call the -ve end of the supply "ground".

It is not unknown to come across 1970's Asian PNP single supply designs where the supply +ve is grounded (and the negative supply coded red - caution!).

If you are using a split rail supply design then the more logical place is the mid-point of the filter caps, but it is also possible to ground the half-rail output and let the supplies flap up and down (as one manufacturer does - to me this is the tail wagging the dog, but I'm sure they have their reasons).

I just grabbed this circuit at random to illustrate a couple of points about the output section;

(http://webpages.charter.net/dawill/tmoranwms/Circuits_2008/2N6254_Power_Amp.gif)

... but on closer examination it contains a couple of real design bloopers.  :o >:(

One error means this circuit would not work at all, and even if corrected the other error would mean it would then not work at all well.

No prizes for the techs, but can you spot them Hawk?


{proving once again that you can't just take any circuit on the internet at face value.}

Roly, I'll give your challenge a try. What through me off at first was seeing the NFB (22K resistor) running down the middle of the schem. I'm used to seeing them run on the outside of the schem them go to the input of an op amp or tone stack. So then I'm thinking where is the half rail? Without a half rail the system won't work or balance out (if we're to see it as an OpAmp). Doesn't the +38/-38 need a reference point?
Then I'm thinking, there is no current limiting here (but maybe there doesn't have to be if this is an older design). And should there not be diodes to separate the pos/neg swing of the signal before it heads to the speaker? Or would then both neg/pos rails be sending the full wave form and possibly it would be out of phase by 180 degrees. Hope I've made some sense here and I'm not a mile off base. :-\


P.S. Thanks Roly, G1, Enzio, Fahey for all this great info. I'm grateful for it and will read and re-read.

Quote from: HawkI'm used to seeing them run on the outside of the schem them go to the input of an op amp

Above I compared one of these OutPut stages to "a power op-amp", which is what it essentially is, and like an op-amp it has inverting ("-", NFB) and non-inverting ("+") inputs, they just aren't normally marked as such.

This example uses an actual op-amp as its first stage;
(http://www.next.gr/uploads/64/18-Watt-Amplifier-using-TLE2141C.jpg)
(and runs the NFB round the bottom, just for you  ;) )


Feedback in s.s. amps is a bit of a trick because it is both DC and AC.

(Ignoring the values) this is the outline of the majority of s.s. power amps;
(http://www.openobject.org/objectsinflux/wp-content/uploads/2008/07/op-amp_circuit.jpg)

Notice that the bottom leg of the feedback divider is connected to ground via a capacitor.

For DC signals this cap is effectively open circuit, so this means the amp is a voltage follower with 100% NFB, unity gain.  The DC point is set by the 10k to ground on the "+" input, so it basically "follows" ground for DC.

For AC signals where the reactance (or AC resistance), Xc, of the cap is low then the ratio of the feedback resistors sets the AC gain.

Revisiting our highly dubious circuit, the feedback components are the 22k on the half rail (generically called Rf for "feedback"), the 1k resistor (generically called Rs for "shunt" i.e. "across"), in series with the 10uF to ground.

The attenuation of Rf and Rs at AC is;

Rs / Rf+Rs

1 / (1+22) = 0.04347826 times

dB = 20 log₁₀(V1/V2)

20 * log10(0.04347826) = -27.23455689dB

This means that the power amp has a forward gain of +27.2dB between input and output.

The supply rail is +38V, but let us assume that when we take into account all the E-B drops that we actually get 35Vpk.

There are a couple of ways to work this out, but if we take the maximum peak output voltage and multiply it by the NFB network loss, we should have the peak input voltage for full output;

35 * 0.04347826 = 1.52173910

1.5Vpk or about 1V_RMS - which is exactly what I would expect.

Quote from: Hawkand possibly it would be out of phase by 180 degrees.

Close enough to earn a small prize;
(http://www.photographyblogger.net/wp-content/uploads/2013/04/Lollipops-14.jpg)
(I do hope you aren't diabetic or anything)

You're right, by drafting convention in s.s. amps the NFB normally runs back down the middle, while in valve amps the OP transformer gets in the way, so it has to go around, typically under.


Both of the problems with this circuit relate to the arrangement of the input transistors.  The minor work-but-not-well, problem actually struck me first.  But as they say you never find just one roach in a kitchen, so while having a deeper look it only later struck me that there is something quite fundamentally wrong with this circuit.

Let's start at the input and inject a notional small negative-going signal and work out what will happen.

This will cause the first transistor to conduct more, to turn on, so the join of the two emitters will also go negative, also turning on the second transistor.  This in turn draws more collector current so it will also turn on the third transistor (top-center, sometime called the Voltage Amplification Stage, VAS).

This in turn will cause it to pull its collector more positive, and since this is also the bias point it will turn on the clutch of upper transistors, and turn off the lower clutch, so the output half rail will be pulled positive.

Now this positive-going signal is taken back to the base of the second transistor, but it has just had its emitter pulled negative, so this positive-going feedback signal will actually try and turn on the second transistor harder, and the VAS, and the output clutch.

In other words the feedback is not countering or rebalancing the input signal, rather it is reinforcing it - i.e. the negative feedback isn't, it's positive!  {oh Gor blimey!  :o ::) :duh }


{I've already written quite a bit in other threads about the need to keep your wits about you when looking at circuits on the net.  I followed this one up and it has been up for years, and not a mention anywhere that it cannot possibly work.  But he ain't so bad; there are other sites that are riddled with such nonsense - redcircuits and runoffgroove come to mind, and a certain zombie "Blackface" FET preamp full of trimpots that just won't stay dead.}


Okay, the lesser problem.  As you know we want these output stages which are direct-coupled to the loudspeaker to have minimal DC offset, when idle the voltage on the half-rail needs to be within a hundred millivolts of zero or standing excessive current will flow through the loudspeaker.  Zero is the ideal, but for several real-world reasons we are generally happy of it's only tens of millivolts.

As you may be now guessing, the prime determinant of output offset voltage is the input arrangement, and may illuminate why the Long Tail Pair is so popular as the first stage.

Now, just pretending for a moment that this circuit doesn't have the NFB blooper and in fact has proper NFB.  What would the output DC offset be in this amp?


{hint: the base of the first transistor is tied to ground via the 10k, so applying very basic transistor knowledge, the voltage on its emitter must be...  and therefore...?}

Still laughing  :lmao:  at the "Coronet" cheesy 60's Japanese amplifier.

Which would have been very hard to sell in Argentina  :o

Why?

Coronet is the brand of a famous (at least when I was a kid)  condom, and became almost a synonym for one

(http://i.ebayimg.com/images/g/SpAAAOSwq7JT-SvD/s-l300.jpg)

so a typical half drunk (or equivalent)  Musicians dialogue would have been :

- "Is it true that you use a Coronet?"

- "Yup"

- "Not surprised everybody says you are a d*ck  :lmao: "

QuoteThe DC point is set by the 10k to ground on the "+" input, so it basically "follows" ground for DC.

Could you possibly re-explain this, not quite sure how that works.

QuoteFor AC signals where the reactance (or AC resistance), Xc, of the cap is low then the ratio of the feedback resistors sets the AC gain

. 
Are you referring to the 10K and 4K7?

QuoteFeedback in s.s. amps is a bit of a trick because it is both DC and AC.

So what is blocking DC from entering the speaker? I see C6 which goes to ground, does that mean that it is easier for DC to go to ground then the speaker? I expected to see a DC blocking coupling cap between OT's and speaker...hmmm

QuoteThe attenuation of Rf and Rs at AC is;
Rs / Rf+Rs
1 / (1+22) = 0.04347826 times
dB = 20 log10(V1/V2)
20 * log10(0.04347826) = -27.23455689dB

Roly, looks like you are using a voltage divider equation, but not sure how you get V1,/V2 also equal to 0.04347826? Could you detail this a little more if possible?

Quote
This in turn will cause it to pull its collector more positive, and since this is also the bias point it will turn on the clutch of upper transistors, and turn off the lower clutch, so the output half rail will be pulled positive.
Now this positive-going signal is taken back to the base of the second transistor, but it has just had its emitter pulled negative, so this positive-going feedback signal will actually try and turn on the second transistor harder, and the VAS, and the output clutch.

Thanks Roly for the details, what you're saying about the feedback circuit reinforcing makes sense. But what exactly is happening when you say:

Quoteand since this is also the bias point it will turn on the clutch of upper transistors, and turn off the lower clutch

How is the lower clutch turned off? Is it all about pos relative to negative, depending on the situation? Upper Clutch being more positive, lower clutch less positive and closer to a value less than turn on voltage? Thanks. :tu:

QuoteNow, just pretending for a moment that this circuit doesn't have the NFB blooper and in fact has proper NFB.  What would the output DC offset be in this amp?
{hint: the base of the first transistor is tied to ground via the 10k, so applying very basic transistor knowledge, the voltage on its emitter must be...  and therefore...?}

Here goes... :-[...I know that we want a very small DC offset or we'll get DC current travelling to the speakers and that will fry the speaker. Not sure how to calculate that, even with the hint provided. So, I can honestly say that I need to go back to transistor basics as I'm honestly not sure how to calculate the emitter voltage, wondering if it should follow the voltage somehow at the base of the transisor minus .6/.7v??  :-[ Even with this great info my brain can't wrap my head around this...this transistor stuff grips me but kind of torments me as well, it's slippery!!

For a long time here Durex was both a brand of condom and a brand of sticky tape.

{in a previous life I built a bunch of testers for Ansell Rubber, a bit starchy, "This is the Prophylactic Division, no condom jokes here please (sniff)" - did you know they can inflate up to a metre?  Amazing.  They also made weather balloons that looked like a condom for an elephant.  Did a lot of instrumentation for the rubber&plastics industry, all gone now, but.  "Youth employment" means "hospitality", i.e. waiting on tables; nobody actually makes anything any more.  Another old client, the car industry and component manufacturers, all closing down next year.  The only saving grace about the bunch of steaming neo-cons in Canberra ATM is that they turned out to also be extremely incompetent and couldn't organise an orgy in a brothel.  They would blow us all up ... if only they could just figure how to get the pin out...  :grr }

---

Okay Hawk, as often, the last first.

The voltage on the first transistor Base is ground because 10k resistor.  (There is a small offset due to the very small Base current (microamps) producing a small drop across the 10k, and that is in large part where the residual offset comes from in many designs).

If the Base is at ground, then the Emitter must be at ~+0.6V (basic silicon diode theory), then we have another E-B junction in the second transistor, and another ~+0.6V.  So the Base of the second transistor, where the DC and AC feedback goes, has an offset of 0.6+0.6 = 1.2 VOLTS to ground!  In an 8 ohm speaker that would result in a standing current of;

I = E/R
1.2/8 = 0.1 or 100mA.  Not ruinous, but not too damn flash either.   >:(


For DC the power amp is a voltage follower, that is it has 100% NFB giving unity gain (x1.00), and whatever DC voltage is presented at the input, the output will take the same voltage (albeit with a huge current capacity; you could replace the speaker with a DC motor and you would have a big servo.).

(http://upload.wikimedia.org/wikipedia/commons/thumb/6/68/Unity_gain_buffer_amp.svg/400px-Unity_gain_buffer_amp.svg.png)

In this case the DC level is set by the fact that the 10k input resistor goes from Base to ground - so that is the DC reference voltage into the amp.  In a perfect amp this will result in exactly zero volts on the output when the amp is idle.

Quote from: HawkAre you referring to the 10K and 4K7?

(http://webpages.charter.net/dawill/tmoranwms/Circuits_2008/2N6254_Power_Amp.gif)

No.  Rf is the 22k going from the half-rail to the Base of the second transistor, Rs is the 1k going to the 10uF cap.

Yes the voltage divider formula applies.

Rs / Rf + Rs

"The Rs-th part of the Rf plus Rs whole".

The voltage divider question is, "given a voltage V across two known resistors in series, what is the voltage at their join?"

Feedback and op-amp gain setting is the same thing transposed, so the question is, "what ratio of resistors do I need to set a given gain?"

Here we start with the idea that the input will be a nominal 1 volt level, and that to produce full output the amp will need some voltage gain to get full output swing between the supply rails (whatever those may be).  In this case we need about 23:1 (22+1:1).

We are effectively working it backwards.  Rather than knowing the total voltage and resistors, and finding the join voltage, we start with the join voltage and top voltage and work out the resistor ratio as the unknown.

Quote from: HawkSo what is blocking DC from entering the speaker?

In directly connected designs, nothing.  The power amp working correctly.  That's why an amp fault could potentially burn out a speaker.  This is why the DC balance is important.

Quote from: Hawk

QuoteThe attenuation of Rf and Rs at AC is;
    Rs / Rf+Rs
    1 / (1+22) = 0.04347826 times
    dB = 20 log10(V1/V2)
    20 * log10(0.04347826) = -27.23455689dB

Roly, looks like you are using a voltage divider equation, but not sure how you get V1,/V2 also equal to 0.04347826? Could you detail this a little more if possible?

Yes that is a version/transposition of the voltage divider formula.

dB's are a ratio, X compared to Y, gain as output over input, V1/V2 can also be Vin/Vout, or the ratio of interest, in this case 0.04347826 times.

Concept: The reverse loss of the NFB divider sets the gain of the forward amplifier to compensate.

The op-amp always acts to bring its two inputs into alignment.  Whatever signal goes in on "+" the op-amp will do whatever is required on the output to make the "-" input match.  In this case if we apply 1 volt then the output has to go to whatever voltage is required by the 22k/1k NFB divider to produce exactly 1 volt on the join, the Base of the second input transistor.

Quote from: HawkHow is the lower clutch turned off? Is it all about pos relative to negative, depending on the situation? Upper Clutch being more positive, lower clutch less positive and closer to a value less than turn on voltage?

At idle both upper and lower output transistor "clutches" are just conducting thanks to the idle bias setting.  The half-rail is at ground potential, and the bias network is pushing the Bases just far enough apart to let through a dribble of current.

The you come along with your gee-tar and whang a mighty chord.

When this signal goes up (i.e. the bias network is pulled positive), the upper clutch is turned on fairly hard, and at the same time the lower clutch has its dribble of Base drive removed and turns right off, the half-rail heads north and current starts to flow to the speaker.  On the other half cycle the same thing happens in reverse, the lower clutch conducts and the upper turns off.

Not to brag or anything, but the store shelves here all have condoms with my name on them:
(http://pics2.ds-static.com/prodimg/12799/300.jpg)

QuoteAt idle both upper and lower output transistor "clutches" are just conducting thanks to the idle bias setting.  The half-rail is at ground potential, and the bias network is pushing the Bases just far enough apart to let through a dribble of current.

On my valvestate, TR6 B-E voltage reads .535 Volts. When I play my guitar it doesn't change. I've always known about the the turn on voltage in a transistor, and you mention that while idling there is a dribble of current going through. Is that because we're so close to .6 v so there is some current as it's about to turn on? Also, now that I'm trying to get what's in my head into my hands and the pcb, I thought for some reason that when I played the guitar the resistance of the Transistor would drop (it would turn on) and therefore so would the voltage, but I suppose if the current increases and you multiply that by the resistance you get the same voltage, no can't be right...hmmm, wondering why the .535 stays the same, I expected to see it change.
So, now that I've written this I'm thinking about your OP amp explanation and output trying to keep up with the Input for unity gain and I'm wondering if this voltage is supposed to stay the same although the current increases (enough to drive a servo motoer). But then I would also ask how does this static EB voltage stay the same when we have full output swing between the rails, don't we have one side pulling the other side down and vica versa, therefore a collective voltage drops on either the pos or neg side of the rail... :-[ P.S. Excellent responses so far and greatly appreciated!)

Enzo, as an electronics champ I'm glad you've taken great pains to be "shielded'. :tu:

Remember the signal goes positive and negative from rest.  The more positive it goes, the more it turns on the positive outputs by driving their bases.  The bases of the negative side are ALSO going more positive at the same time.  But that just reverse biases those bases so the negative outputs don't conduct.  And opposite for negative signal excursions.

The voltage across the EB junction may stay the same - remember it is a "diode", not a resistor - changing current through that junction controls current through EC.  SO while you play, look at the voltage on the collector of that transistor.

A positive going input to the op amp inverts to negative at its output.  That pulls down the emitter of TR6, increasing EB current.  That pulls more current through R84 via the collector.  That in turn pulls more current through the EB of TR7, pulling up on its collector.  Up and down ar easier to write than positive and negative.  Pulling up the collector of TR7 also pulls up the base of output TR8.  The emeitter of TR8 follows, and thus the speaker output.  The output is thus in phase with the input to the power amp.

Thanks Enzo, your OpAmp explanation plus the idea of pulling up and down helped me look at this entire circuit differently and opened up my eyes to it. Of course, a few more things as I drill down...

Quotethe voltage across the EB junction may stay the same - remember it is a "diode", not a resistor - changing current through that junction controls current through EC.  SO while you play, look at the voltage on the collector of that transistor.

I tried that: referenced to ground, on collector of TR6 I get 36.7vdc at idle. At full volume, guitar playing, 35.8vdc. So, from this info, it looks like the pos. signal excursion has pulled down the voltage a little, correct?

What is the purpose of diodes D1,D2? Separating the neg/pos bias on TR6 and TR5?

Quote from: Hawkhow does this static EB voltage stay the same when we have full output swing between the rails

Okay, I've been a bit fast and loose with the facts.

A transistor, a "trans-resistor", is a current operated device.

(http://sub.allaboutcircuits.com/images/03104.png)
The current between Base and Emitter determines the current between Collector and Emitter.

The gain or Hfe is the ratio of the Base to Collector currents.

The B-E junction is effectively a forward biased diode, and if we look in detail at a silicon diode characteristic;

(http://www.electronics-tutorials.ws/diode/diode8.gif)

An ideal diode is a dead short for a forward voltage and infinite resistance and voltage withstand for a reverse voltage.

Back in the real world diodes have a cut-in voltage of V_gamma, a certain amount of forward voltage to make them boogie, start to conduct.  For silicon devices this intrinsic junction voltage is variously given as 0.5 to 0.7 volts (0.1 to 0.3 for Germanium devices).

"Variously" because the conduction "knee" isn't an exact point, but rather as the device comes into conduction its resistance rapidly drops over a very small range of voltage, say 0.5 to 0.7V.  At 0.5V the junction is only just starting to conduct, microamps perhaps, but by 0.7V it is turned hard on and could be passing amps (or making smoke).

{Looking at this typical diode characteristic it is important to note that the four quadrants are on very different scales, 100mA forward, a volt forward, microamps reverse and hundreds of volts reverse;
(https://f1c54f8065066eb3e75db11873eeb9e879be507a-www.googledrive.com/host/0B8qXfkNwdWWxLVdnYlViZEVoQms/diode_characteristics.png)
}

The other real world thing a PN junction has is resistance, so apart from acting like a diode, when in forward conduction it also acts like a low value resistance.  Normally we can ignore this as minor, but in some applications such as large motor controllers power diodes may need to go on a big heatsink to get rid of the power being dissipated in this device resistance.  This resistance explains why the current rises rapidly, but not vertically, in forward conduction.

---

"It can be demonstrated" (i.e. I'm not about to cover several pages in hybrid parameter equations - you'll have to take my word for it) that this is the highly simplified circuit of your amplifier;
(http://e2e.ti.com/cfs-file/__key/communityserver-discussions-components-files/247/4336.Transistor_5F00_push_5F00_pull_5F00_follower.gif)

What you basically have is a pair of back-to-back emitter followers (a.k.a. common-collector amplifier (http://www.allaboutcircuits.com/vol_3/chpt_4/6.html)), one for the +ve going signals, and the other for the -ve half, V_in.  In either case the Emitter outputs will follow the Base voltages because they are only one diode drop removed from the voltage on the Bases, one up, one down.  If the Bases go to +10 volts then the Emitter outputs, V_E, will be at 10-V_BE or around 9.4 volts.  Similarly if the input is -10 volts the Emitters will be at -9.4 volts.


Now hopefully it should be obvious that for the upper transistor to conduct at all the input signal needs to be higher than the B-E cut-in voltage (0.5-0.7V).

As the bases and emitters are connected together, if the upper E-B junction is seeing a forward bias of 0.6V, then the lower transistor, being of the opposite gender, PNP, is seeing a reverse bias of 0.6V, in other words it has moved away from conduction and is effectively biased right off by a reverse bias of 0.6V from the upper transistor, plus another 0.6v due to its own B-E drop.

The Bases being connected together this circuit has no bias, so there will be a dead zone for input signals between +0.6V and -0.6V where neither transistor is conducting.  This is known as crossover distortion because it produces a nasty step in the output waveform as it crosses zero.

(http://users.ece.gatech.edu/mleach/lowtim/graphics/Crsdist1.gif)

Why;
(http://www.maximintegrated.com/en/images/appnotes/1760/1760Fig03.gif)

So we put in some bias between the bases so that both transistors are conducting just a little bit, idle current, and the transition from one transistor conducting to the other is much smoother (and much better sounding).

(http://www.learningelectronics.net/images/quiz/00972x01.png)

Great explanation!

QuoteSo we put in some bias between the bases so that both transistors are conducting just a little bit, idle current, and the transition from one transistor conducting to the other is much smoother (and much better sounding).

So this would explain the presence of DC voltages at the bases of TR6 and TR5, so that both transistors are conducting and therefore we eliminate crossover distortion:
1.3 vdc at base of TR6 (.81 VBE--1.3-.5 = 0.8v)
-1.7 vdc at base of TR5 (-1.1 VBE--1.7-.6 =-1.1v)
Measurements above confirmed measurements with multimeter.
Therefore constant current flow (over B-E cut in voltage of .7/.6/.5v) and therefore no crossover distortion. Correct?
(Not sure why the voltages aren't the same on transistor bases, maybe resistor values have drifted)

Correct?

When I ssuggested watching the collector of TR6 I meant watch it for signal.  That whole up down discussion was trying to explain signal travel through the circuit.  If your collector voltage dropped a little while playing, then also check the V+ supply voltage at the same time, I bet it drops a little under load as well.

Look at your test signal wave form and follow that along the path I described.

Enzo your efforts were not in vain. I understand the signal piece, just wanted to make sure I understood the constant bias to understand that both transistors are conducting and therefore no crossover distortion. I believe that I do.
I used my scope to check the collector of TR6 and, of course, there is  signal, amplitude varied by the volume controls. Scoped all transistors and found the signal there as well.

Thanks again for your efforts!  :tu: :tu:

Quote from: Hawk on March 23, 2015, 09:59:30 AM
Great explanation!

QuoteSo we put in some bias between the bases so that both transistors are conducting just a little bit, idle current, and the transition from one transistor conducting to the other is much smoother (and much better sounding).

So this would explain the presence of DC voltages at the bases of TR6 and TR5, so that both transistors are conducting and therefore we eliminate crossover distortion:
1.3 vdc at base of TR6 (.81 VBE--1.3-.5 = 0.8v)
-1.7 vdc at base of TR5 (-1.1 VBE--1.7-.6 =-1.1v)
Measurements above confirmed measurements with multimeter.
Therefore constant current flow (over B-E cut in voltage of .7/.6/.5v) and therefore no crossover distortion. Correct?
(Not sure why the voltages aren't the same on transistor bases, maybe resistor values have drifted)

Correct?

Well it's only good if you get the right idea, and it seems you have.


There are all sorts of reasons why adding all the V_BE drops and E = I * R drops across resistors don't exactly match (simple) theory.  Even between devices of the same type and under identical conditions we find small differences, but you then connect three different devices together in each output triplet, each operating under different conditions, and the deviations from ideal start to add up.

Just a couple of considerations;

I've mentioned the internal series resistance into the Base, but there is an effective series resistance with every terminal, and the one in the Emitter is a particular trap because it appears not only in series with the controlled main C-E current flow, but also in series with the controlling Base-Emitter flow (and thus results in unwanted local negative feedback within the device). {This is normally called something like r'e, emitter resistance, where the apostrophe implies an internal property.}

For the vast majority of our needs (faultfinding and repair) a very simple model of a transistor is quite sufficient;

(http://www.ibiblio.org/kuphaldt/electricCircuits/Semi/03091.png)

A current through a diode controls a current sink (circle with arrow), but if we want to explore off-Broadway a bit, look in greater detail at what the device actually does, we need to start including minor factors that we have left out in our simplification.

A full model of a transistor actually has quite a few hidden components;

(http://www.maximintegrated.com/en/images/appnotes/697/E37Fig2A.gif)

You may see some different examples of a "full" model because "full" depends on what you are doing.  Particularly when you start dealing in high and very high frequencies factors that are ignorable at audio become very important.  An obvious one would be case capacitance, stray capacity between leads simply due to the encapsulation acting as a capacitor dielectric.  At audio this tiny capacitance is quite insignificant, while at VHF it may need to be carefully accounted for in the design.  {the "full" model above is actually only for the naked chip and still doesn't include realities such as the device packaging, lead inductance, and such.}

A more important stray, particularly in triodes and FET's is called "Miller capacitance" and is the stray between the output and input, typically anode/drain to grid/gate.  This is a capacitance that is internal to the device and results in unwanted negative feedback as the frequency increases.  Again this is not normally a problem at audio frequencies, but where you might be trying to increase the input impedance "seen" by a guitar at the first stage input its effect will be increased.

(to a 1st approximation) What is the cutoff frequency of an MPF102 with a 2.7Meg Gate resistor?

f = 1 / 2 Pi CR

1/(2*Pi*2e-12*2.7e6) = 29,473.14Hz

Well 30kHz looks fine, and as it happens the 12AX7 has almost the same Miller capacitance per triode, so the result will be about the same.

Now let us decide (for reasons that we won't examine) that we want to take the input to both triode sections of a 12AX7, so now the Miller capacitance is doubled;

1/(2*Pi*4e-12*2.7e6) = 14,736.57Hz

In any amp with "Hi-Fi" pretensions being 3dB down at 14kHz would not be a particularly good look.  Forget that only bats might notice, in the spec sheet numbers game you are starting off behind.  Meanwhile in the guitar amp workshop we don't give a rats because we know that guitar amps hardly ever make better than 5kHz.

If we now swing in the grandpa of the 12AX7, the 6SN7 with 4pF anode-to-grid it will be;

1/(2*Pi*8e-12*2.7e6) = 7,368.28Hz

7kHz ain't so brilliant, even for guitar, so depending on how we are trying to use a component can make a minor characteristic either ignorable or significant.


The other factor here is that you are looking inside a Negative FeedBack loop.

Ideally there is 100% DC feedback which holds the output to exactly the same voltage as the input.  Ideally.

We assume that under ideal conditions the output would stay at zero if we removed the NFB, and if the whole amp was in critical balance it would, but it isn't in critical balance because it has some huge forward gain without feedback, and the sum of all the effects pulling the output up won't be exactly balance by all the effects pulling the output down.  So if we remove the DC NFB the output will slam to +ve or -ve rail, depending on the overall balance.  The huge gain make it like trying to balance a pin on a knife edge.

When we reapply NFB the output will again go close to zero, balanced and neither +ve nor -ve ... apart from a small residual offset.

Now NFB is not absolute; if there are errors (such as the amp tending to drift +ve or -ve), then an error voltage is produced which (almost) restores balance, but it is subject to the Law of Diminishing Returns.  To correct an error an error voltage must be produced, but there must be some residual error to produce that.  The only time this is not true is when the amp is in critical balance in the forward direction, then there is nothing to correct, and so no error voltage.

The practical result of this is that the amp will tend to "ride" against one of its error limits.  If the accumulation of forward errors makes the output drift up until the NFB corrects it, then inside the NFB loop you will see the voltages tend to rest against their upper limits, and the opposite is also true.


We typically use op-amps with the gain set to somewhere between unity and x100, but the open loop gain of most op-amps is x100,000 or greater.  If you go through a typical power amplifier multiplying all the individual stage gains together you discover that without NFB the power amp has a huge forward or open-loop gain.

The basic trick with NFB is that the improvement in the circuit performance is proportional to the degree of gain reduction by NFB - the bigger the difference between open-loop (without NFB) and closed-loop (with NFB) the greater the improvement in the amplifier characteristics such as distortion, bandwidth, and output impedance.  The no-free-lunch cost of course is gain reduction.

If we compare a typical valve output stage we find that the level of NFB is much lower than in a s.s. amp.  One reason for this is high frequency phase rotation in the output transformer can make the amp unstable, but another good reason is that a similar valve amp has a much lower open-loop gain, so while AC NFB of the order of 30dB is common in s.s. amps it is much more modest in valve amps, say no more than 12dB.  (You can certainly up the forward gain, say by adding stages, but if you are a guitarist you don't actually want the output transformer to go away because it gives you a nice magnetic compression distortion, and in fact many valve amps operate open-loop, particularly small "boo-teek" combos, no NFB at all, and we just love the resulting distortion.)


{a sidebar about the diode "knee".

As it happens the curvature of the knee of a germanium diode is a very good approximation of a square-law y = kx² i.e. the current rises as the square of the voltage over the initial conduction range.  Since the power in a resistor is proportional to the square of the voltage;

P = E² / R

... I have made use of this characteristic to fit a linear scale wattmeter to my dummy load.  It's a bridge rectifier, but the voltage is scaled down by resistances so that it operates over the early part of its conduction curve, and the result is a current of 0-1mA for 0-100W in 8 ohms (calibrated to be exact at 50 watts, leaving an error of about 5% at 10 and 90 watts.}

---

Revisit: and just to add; the effect of doing this;

(http://www.learningelectronics.net/images/quiz/00972x01.png)

...adding bias, is to move these traces (right side);

(http://www.maximintegrated.com/en/images/appnotes/1760/1760Fig03.gif)

...so the left hand V_EE moves 0.7V to the right, and the V_CC trace moves 0.7V to the left, resulting (hopefully and ideally) in a perfectly straight line between V_EE and V_CC, in other words this transfer characteristic has been linearised, made into a straight line.

But since we are digging into real world detail, the thing that really matters is the cut off curvature for the pair that are opposite gender, the NPN/PNP drivers or pre-drivers.  The only time that crossover distortion will be totally eliminated is if the curvature of both these transistors matches exactly, as one turns off the other turns on by exactly the same amount, and in the real world that doesn't happen even with "matched" pairs.  As with other (less subjectively objectionable) distortions we can greatly reduce them to insignificance, but until we get perfect devices we can't totally eliminate them.

Hey Roly, thanks for going the distance with this stuff, I'm learning a ton!! So with the knowledge gained so far I'm going to dive into my transistor book and study and re-read more and then I will go back to your full model of a transistor and make more sense of it--I don't think it's fair to go further and ask more of yourself  or anyone until I've done more work on my own. (except for question at bottom of this post). :-[ :)

Quote... I have made use of this characteristic to fit a linear scale wattmeter to my dummy load.  It's a bridge rectifier, but the voltage is scaled down by resistances so that it operates over the early part of its conduction curve, and the result is a current of 0-1mA for 0-100W in 8 ohms (calibrated to be exact at 50 watts, leaving an error of about 5% at 10 and 90 watts.}

Sounds great, I'd like to see a picture if you have one.

QuoteI've mentioned the internal series resistance into the Base, but there is an effective series resistance with every terminal, and the one in the Emitter is a particular trap because it appears not only in series with the controlled main C-E current flow, but also in series with the controlling Base-Emitter flow (and thus results in unwanted local negative feedback within the device). {This is normally called something like r'e, emitter resistance, where the apostrophe implies an internal property.}

When you say "and thus results in unwanted local negative feedback within the device". Could you elaborate on that a little more. I find it interesting but I can't quite picture it.

Thanks again for this great information!! :dbtu:

Quote from: HawkI don't think it's fair to go further and ask more of yourself  or anyone until I've done more work on my own.

"Library 101".  It's the old fashioned way, read, study, try to wrap your head around the concepts, struggle with it, finally ask for help with what you still can't quite "get".  {My local Uni library was heated in the winter, cooled in the summer, and was open late most nights, still a great resource.  :dbtu: }

Exploring simple circuits on the bench can also teach you a lot, and helps to develop a real world "feel" for what is going to fly, and what isn't.

Quote from: HawkSounds great, I'd like to see a picture if you have one.

(http://www.ozvalveamps.org/optrans/fenderbm40w.jpg)

... well, part of it, anyway.

Quote from: HawkWhen you say "and thus results in unwanted local negative feedback within the device". Could you elaborate on that a little more. I find it interesting but I can't quite picture it.

First step: a perfect device in a voltage follower:

This applies generally to transistor emitter-followers, FET source-followers, and valve cathode followers ('tho there are some differences).

(http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/ietron/npncc2.gif)
A transistor emitter-follower (elemental, bias assumed for clarity).

We have two current loops;

- an input controlling current I_B from Base to Emitter, down through R to ground;

- an output controlled current from I_C Collector to Emitter, and also down through R to ground.

- note that I_E = I_B + I_C

Hello, these two current loops, one controlling and the other controlled by it, both share a path through R.

Since we know that the B-E drop is about 0.6V it should be obvious that the output voltage can never be other than the input voltage, less the V_BE drop - the "amplifier" must therefore have unity voltage gain.

One way of looking at this is that the input loop suffers 100% NFB from the output loop via the shared emitter load resistor R.  As the Base current injection tries to pull the Emitter voltage up, so the output loop passes just enough current to exactly balance that, and the Emitter remains stubbornly at Base-0.6 volts, the Emitter follows the Base voltage.  (in real world single-device followers the actual gain is slightly less than unity, perhaps x0.98.  The higher the device gain, H_FE, the closer it gets.  Op-amp followers with their vast open-loop gain get so close we generally consider them to be perfect x1.00...)


Second step: now we introduce the fact that the Emitter has its own resistance internal to the device;

(http://i.cmpnet.com/planetanalog/features/Fairchild_Tutorial_Part1/Fig1.gif)
(for beta above read H_FE or current gain.)


Now hopefully we can see that this internal resistance, r_E, produces the same situation of a perfect transistor with an Emitter resistor as in the follower above, but it is in-built, we can't get to the actual perfect Emitter because this resistance is an inherent imperfection, like it or not it comes with the turf.

And it has much the same effect in generating a NFB voltage in the controlling B-E loop due to the main controlled current flowing from Collector to Emitter.

With the typical emitter follower this normally has little impact because the external emitter load resistor will be very much greater than the internal r_E, but when we want to operate the transistor in grounded Emitter mode so we can get some voltage gain on the Collector, it becomes a large proportion of the total Emitter circuit resistance and produces significant unwanted NFB in series with the input loop. 

Semiconductor manufacturers have long since reduced the effects of these unwanted "parasitics" to the point where we can happily build a transistor amp that actually works, but they are still there in the more detailed view, and every once in a while a situation arises where they become significant.  The transistor model you have to use to design a 20 watt power amp for 144MHz is vastly different to the model you use in designing a 20 watt guitar amp because these parasitic fleas at audio become elephants at VHF.


HTH

Quotebut when we want to operate the transistor in grounded Emitter mode so we can get some voltage gain on the Collector,

Sorry, couldn't resist another question. Firstly, excellent description and I understand most of it! Great stuff and thanks Roly! So, as for the quote  above, my mind was going along beautifully with your description until I read the  bit about the voltage gain on the Collector if we grounded the emitter. Could you elaborate on how we achieve voltage gain on the collector? Is it because more current is flowing through the collector due to the removal of the emitter resistor and therefore the increased current multiplied by the Collector's internal resistance creates a greater voltage? That's what messes me up about Transistors-- I understand the base current loop creating current flow from Collector to Emitter, like a valve letting more water through, and then I think I've got this stuff, but when we start talking from Emitter to Collector I start thinking about holes filling valence shells missing electrons and conventional current. With transistors should we be thinking neg to pos/conventional current at the same time? Or should we always think neg to positive--N type material to P type material-- just to keep our heads on straight? Should we not overly concern ourselves with current flow from emitter to collector? Thanks!   :-\

Some of the books I study ask me to build circuits to prove their point. I have electronics workbench from 14 years ago but it doesn't work on my operating system.  Circuit Logix Pro looks great but not cheap. Is there any good freeware/cheap software to use to help build simulated circuits to help with my understanding of electronics? Thanks.

Instead of looking to simulate circuits, consider buying a few cheap transistors and some cheap resistors and a few caps, and actually making little circuits.  Then a basic voltmeter can take readings in the circuit.

Quote from: HawkCould you elaborate on how we achieve voltage gain on the collector?

{I had a funny feeling when I wrote that ... you ever consider doing an electronics course?  8| }


The reason that so much is described in terms of "conventional" current flow, positive-to-negative, is that it is normally easier to visualise what is going on than if you use electron flow, negative-to-positive; so the "current" goes in the direction of the little arrows on diodes and transistor Emitters.

When you are looking down at atomic level at what is going on inside the semiconductor (or vacuum tube) then it is normally easier to visualise it in terms of electrons (and when explaining why different gasses in discharge lighting have different characteristic colours and spectra, neon red, mercury blue, sodium yellow, &c) - but most techs most of the time don't operate at that level, in fact some techs never operate at that level, that's for the designers of transistors, we just deal with them as three-wire, two-port, black box.  Only occasionally do we delve into them as deeply as we are here, mostly we use a very basic model;

(http://www.ibiblio.org/kuphaldt/electricCircuits/Semi/03091.png)
... it serves well enough for faultfinding.  {designing a circuit is a different matter; you need to at least be aware that these other factors like r'e exist}


Okay, a less mathematical, more intuitive model of a transistor or "trans-resistor". (can you see where I'm going with this?  ;) )

Think of the Collector-Emitter as a variable resistor with its value controlled by the Base current - more Base-Emitter current = lower Collector-Emitter resistance (more current).


(http://i.stack.imgur.com/ESIy6.gif)

First consider the two limiting cases;

1. V_in is zero and the transistor is therefore off, so no current is flowing C-E, there is no drop across the 1k Collector load, therefore the Collector voltage is the same as the supply, 10V.

2. V_in is high enough to saturate the transistor hard on, so the C-E is now effectively a short circuit, 10mA is flowing through 1k and the C-E of the transistor, and the Collector voltage is now zero.  (actually the saturation voltage, V_sat, of the transistor, typically around 0.2-0.5V).

Now let us assume that we want a clean audio amplification stage, and that we have a transistor with a known H_FE or current gain of 100.  (note that this device parameter accounts for the effect of r'e.)

For maximum signal headroom without clipping the collector has to be at half the 10V supply, or 5 volts, and given the Collector load is 1k that gives us a current of 5mA through 1k and C-E.

If the C-E current is 5mA and the gain is 100 then the Base current must be 5mA/100 or 50uA.  Since we have 10k in series with the Base, V_in has to be;

E = I * R

50e-6 * 10e3 = 0.5 volt higher than the Base voltage (meaning with a real transistor V_in will have to be biased to somewhere around 0.5 + 0.6 = 1.1 volts).

Now we can work out the voltage gain of the amplifier using a bit of The Calculus.  (relax, this wont hurt a bit  8) )

The trick is to assume a small change in V_in and work out how the Collector voltage will respond.

Let us assume that we increase V_in by 0.1 volt (with the V_BE = bias either assumed, or a perfect transistor with V_BE = 0 volts).

This will cause the Base current to rise to 60uA, and because of H_FE the Collector current will also rise to 6mA.  6mA through 1k is 6 volts, so the Collector voltage will drop by 1 volt.

A 1V change in output voltage for 0.1 volt change in input is a voltage gain of;

Vgain = Vout/Vin

1V/0.1V = 10 times

So while the transistor itself has a current gain of 100, in this configuration is has a voltage gain of only x10.

I'd encourage you to run through this again setting the Collector load 1k to 10k.


Circuit simulation
Forget EWB, it was a brave effort, but truly awful.  You can now get LTSpice free (http://www.linear.com/designtools/software/#LTspice) and there is a great library of component models at the Yahoo LTSpice group.  Sure, learn how a sim works, and this is a good one and Spice generally is the main game in town, BUT what Enzo said is very good advice, tinker with actual components.

You can't blow anything up in LTSpice and you learn a lot about real-world electronics by blowing stuff up (and thankfully bits are not only cheap, if you do a bit of "dumpster diving" there are a wealth of components in unloved equipment to be had for the cost of ripping them off old boards, whole power supply modules, all sorts of neat stuff).

I use LTSpice a lot, and it can normally get you close enough, but the ultimate sim is to actually build the thing with real-world components, strays, external noise, all the stuff that LTSpice leaves out.

QuoteI'd encourage you to run through this again setting the Collector load 1k to 10k

Thanks Roly, and I did just that. If interested, please see attachment. Hopefully I have it right! I had to post it as a .pdf so it spreads over three pages but it still quite readable.

Quote6mA through 1k is 6 volts, so the Collector voltage will drop by 1 volt.

I'm assuming we're using variables to understand the workings of a transistor and wouldn't really be applicable as I remember that we wanted a max. 5 volts at the collector, half the 10 v supply so we could have a clean audio amplification stage.

Quote{I had a funny feeling when I wrote that ... you ever consider doing an electronics course?

Roly so here's my story: fifteen years ago I completed an electronics certificate at the college level, had ideas of getting into the audio repair business, but kids, mortgage and other pressures put that on the back burner, way back! so now, with adult kids and at age 53 I'm gradually dusting off my knowledge and asking many questions. There are definitely holes in my knowledge and it's amazing what you can forget in fifteen years--usually the small connecting details/equations/theory but it's coming back and I pretty much spend a couple of hours  a day with my head in this stuff with an eye to start up a part-time repair business down the road. I'm having a blast and loving it!

QuoteInstead of looking to simulate circuits, consider buying a few cheap transistors and some cheap resistors and a few caps, and actually making little circuits.  Then a basic voltmeter can take readings in the circuit.

Enzo, that makes total sense. So should I be thinking about getting a DC Power Supply? any recommendations in terms of max current/voltage?

Roly thanks for taking the time, I feel like a real taker on this forum and hope that one day I can give back. :cheesy:

I actually meant just changing the collector load from a 1k to a 10k, not stepping through "1k to 10k", however by going through it in steps of 1k the picture should be all the clearer. 

Quote from: HawkI'm assuming we're using variables to understand the workings of a transistor

Quite so.  Remember that the bias is to set the device idle condition, and in use it will be receiving an input signal that is superimposed on the bias, so while the resting Collector voltage will be half the supply (roughly), when driven with an actual signal the Collector voltage will swing up and down in sympathy.

{half the problem with bias networks is getting the bias in there without stuffing up the desired operation too much.  Early transistor guitar amps often had a real problem maintaining a reasonable input impedance for guitar (1M min) due to the effects of the required bias networks, and I've seen some as low as 47k.}

Quote from: Hawk
here's my story:
...
I'm having a blast and loving it!

Well I wish you the very best of it.   :dbtu:   May your days be filled with interesting faults.   ;)

I survived a 240VAC belt as a breedling by sticking my fingers in a live light socket, and it seems to have set down a path of curiosity that lasted a whole career and into retirement.  Some people do crosswords, some climb mountains, but I get my rush from designing, building, and repairing electronic/electrical stuff.

We have a community-run "Infolink" in town which provides walk-in Internet access, and occasionally one of the punters will do "something" to one of the machines that needs to be sorted out.  Mostly the resident SysOp is on top of things, but a couple of months ago he mentioned that one of the machines had "forgotten" it had a CD drive.  He had tried all the obvious things like replacing the drive but to no avail, and asked me if I'd take a look.  After a lot of stuffing around covering the obvious, then a fair bit of Google research, I found mention of a possible problem with the registry entries, so taking the bit between my teeth (and my heart in my hands) I got stuck into editing the Registry, and presto, the drive was back.  Yo!

Particularly with obscure faults, or "stinkers" that have been hanging around for some time, there is a real sense of having won one over entropy (https://www.google.com/search?q=entropy&ie=utf-8&oe=utf-8) (and a certain amount of dancing around and air punching).  YES!  Humans - 1, contraption - nil.  "Take that!  Who's King of the @$^&ing castle then?"  It's a great feeling.

Quote from: HawkRoly thanks for taking the time, I feel like a real taker on this forum and hope that one day I can give back. :cheesy:

My pleasure.  I also get a charge out of teaching (electronics, physics, some maths), particularly when it falls on fertile ground (and your trainees are coming back from Tech waving "Distinction" results).  My tutors and mentors struggled to teach me, and now it's my turn to give back (forward).

Quote from: Hawkshould I be thinking about getting a DC Power Supply?

Build your own test gear.   :dbtu:

Thanks Roly! 

:tu: :tu:

A power supply must be whatever you need.  If you are making little one or two transistor preamp circuits, hell a 9v battery will do for experimenting.  Or adapt a 9v wall wart.  A little basic power supply is simple enough, a little transformer, a rectifier bridge, a couple filter caps, et voila.  Power supply circuits are easily found.

For transistors, a 9v or 12v supply would be common.  Anything like a power amp of course needs higher power supply.  If you get into op amps, then a split supply (two voltages of opposite polarity) is useful, 12v or 15v, both polarities.