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Marshall Valvestate 8100

Started by Hawk, March 13, 2015, 01:53:11 PM

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Roly

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 VBE 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;



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;



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;



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



...so the left hand VEE moves 0.7V to the right, and the VCC trace moves 0.7V to the left, resulting (hopefully and ideally) in a perfectly straight line between VEE and VCC, 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.
If you say theory and practice don't agree you haven't applied enough theory.

Hawk

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:


Roly

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.



... 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).


A transistor emitter-follower (elemental, bias assumed for clarity).

We have two current loops;

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

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

- note that IE = IB + IC

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 VBE 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, HFE, 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;


(for beta above read HFE or current gain.)


Now hopefully we can see that this internal resistance, rE, 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 rE, 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
If you say theory and practice don't agree you haven't applied enough theory.

Hawk

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!   :-\

Hawk

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.

Enzo

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.

Roly

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;


... 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).




First consider the two limiting cases;

1. Vin 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. Vin 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, Vsat, 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 HFE 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, Vin has to be;

E = I * R

50e-6 * 10e3 = 0.5 volt higher than the Base voltage (meaning with a real transistor Vin 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 Vin and work out how the Collector voltage will respond.

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

This will cause the Base current to rise to 60uA, and because of HFE 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 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.
If you say theory and practice don't agree you haven't applied enough theory.

Hawk

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:

Roly

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 (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:
If you say theory and practice don't agree you haven't applied enough theory.

Hawk


Enzo

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.