ThanUThanUThanU ... your a great audience here tonight.

The only effect of the 56k in parallel with the output control is to reduce the gain of the driving stage by reducing its AC load resistance.  It just doesn't seem very ... erm ... well thought out(?).

Quote from: Ben79Was my analysis correct?

Yes.  +1 Internet.

Quote from: Ben79I should probably audition all options before deciding on the order they run in.

Absolutely.

Quote from: Ben79Any other implementation ideas?

As I keep saying, a major point for building your own over buying commercial is that if you don't like it the way it is you can leap in and give it a rewiring it will never forget.  You are in charge instead of being the hapless victim of the compromises forced by some company bean-counter.

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. 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  )

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 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.

AAAAaarrrggghhhh...   

Where did this come from?   

Deadly problem: If the pot marked "?" is turned right down the Base end, the battery is then connected directly across the second transistor Base-Emitter junction, and you can kiss goodbye to the second BC107 (poof!).

(and I'm not too sure about the 56k in parallel with the 500k Output pot either ... in fact the whole darn thing looks a tad dubious.  8|   If I had my way there would be a license test before anybody was allowed to post circuits on the net.  :trouble )

Quote from: Ben79I want to lose the 'effect' pot from this (and run it at max gain)

1. Make the "?" pot a 150k fixed resistor from +9V to the 2nd Base,

2. change the 100 ohm pot-ground to 10k Base-ground,

3. connect the -ve end of the 25uF from the first Collector directly to the second Base.

That should get you close.  If you want to tweek the voicing then trying resistors for "?" in the range 100k to 220k should cover the available tones.

HTH

Hi resorb, welcome.


Yes - stop randomly replacing parts.  As you can see you have already spent time and money replacing parts and it hasn't got you anywhere.  Here we use diagnosis to first find out what is wrong, then fix that.  It's always quicker and cheaper in the long run.


Fed via a limiting lamp, what voltages do you measure on the +ve and -ve supplies, and on the output (disconnect the speaker)?  These should give us clues.


Schematic here (wait 30 secs for download prompt);
http://elektrotanya.com/fender_stage-100_160_sch.pdf/download.html

Quote from: gbonoI "rewired" the PCB to accept a RC4558 and I appear to have a working preamp with tone controls.

I assume that the degraded U1 could not supply enough gain and my "active" filters were "there" just -20dB down in level???

Yo!   :dbtu:

That would make more sense than not there at all, and if it were way down it might be hard to hear any effect.

So yeah, it looks like you found the old "sick op-amp" trick.

Quote from: gbonodidn't look at phase

The phase is what is critical; need a CRO and using external triggering off the signal source.  The signal on the output of the Baxandall tonestack should be phase-inverted on the input, it is basically an inverting stage.  If the op-amp in the tonestack isn't working it is possible you are seeing direct leak-through from the previous stage.

I must say that this has got me seriously puzzled because I'm trying to see a single failure that could disable the entire tonestack and still pass (flat?) signal.


I got curious enough to introduce some faults into my LTSpice model of a Baxandall control, and shorting the op-amp inputs together causes the control ranges to be seriously reduced, the treble almost entirely and the bass mostly.

So my hit pick is the op-amp in that tonestack.  Maybe.

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 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;


(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

Quote from: Ben79There are LiPo batteries that come with chargers

That sounds like a good option, much more capacity too.   :dbtu:  Most of the laptop and power tool batteries I've investigated seem to have built-in battery management, so this type of battery would be a good choice since it is likely to have all the required protection built-in.  (that's what that other connector with several contacts is about, it talks to its charger).

Quote from: Ben79What effect might a lithium battery have on the behaviour of a fuzz circuit?  I know internal impedance can be a factor but I don't understand it.

Well that's a question I don't think has been asked before.   8|

Happily for simplification the internal resistance (DC) of a battery and its internal impedance (AC resistance) are pretty much the same thing for our needs.  To a first (and even second) approximation a battery behaves pretty much like a pure resistance, not much in the way of reactive (AC) elements such as shunt capacitance or series inductance to bother us even at audio frequencies.



The internal resistance, R_int or r above, isn't a fixed value but varies over quite a wide range, mainly depending on the State of Charge of the battery.  These two are so closely linked that the State of Charge can be implied from the internal resistance - high internal resistance means the "perfect" internal battery can't deliver much current to the outside world, your torch bulb only glows dimly, and thus you reason that the battery is flat.

A battery in a low State of Charge, flat or dying, has a high internal resistance, while a freshly made or charged battery has a very low internal resistance.  Shorting a fresh D-Cell for example can result in a current of several amps, and applying Ohms Law to 1.5V tells us the R_int must be well under one ohm.

But we don't go around shorting batteries if we can help it.  Internal resistance is determined by accurately measuring the battery terminal voltage for various safe currents between zero and rated maximum, then extrapolating the curve.  (The internal resistance is the change of voltage delta-V (= change in V, divided by current).  It simply isn't safe to directly measure if a car battery will indeed deliver 350 "cold cranking amps" at 12 volts.


(delta-V is 0.9 volts)


What is a bit scary is that battery manufacturers have been beavering away for decades to make "better" batteries, and an important way a battery can be better is to have low internal resistance.  The result is the modern generation of lithium batteries that have internal resistances so low that they can deliver so much current they will catch fire or explode if shorted (or overcharged).  As JMF implied above, if the battery in my drummer friends pocket had been a lithium rather than an common dry battery he would have needed a trip to the hospital burns unit (and new jeans).


{aside: as a theater tech I have been exposed to the fear of on-stage fire that is theater traditional but not shared by rock and roll bands at venues, and the fear is real because there have been some really terrible disasters, cinema, theater, and club fires started by on-stage pyrotechnics.  Even as a sometime accident investigator I could only watch the harrowing start-to-finish YouTube video of the 2003 The Station nightclub fire, Rhode Island, once.  It is just too distressing, but it is also a very salutary lesson in just how fast an on-stage fire can get out of hand.

Onebaldbob has a story about how the gas jets for the Count Down show were fitted backwards (by a very professional crew) when the show moved to Sydney, and instead of framing the drummer (during rehearsals) they ended up flaming the drummer, roasting his arse and causing him to jump over the kit to escape.  Add some of my own near-miss or close-call experiences and I have pretty negative feelings about any potential source of ignition anywhere in a venue.  I've stage managed fire-eating and pyrotechnic acts and I.don't.like.it.  I'm the guy lurking over there with an industrial fire extinguisher.  }


Now in AC Theory all DC voltage supplies are a short circuit to AC, so with a fresh battery the voltage supply rail in your fuzzbox might be at 9VDC but it is also ground for AC.  And if the fresh battery has an internal resistance of say 0.1 ohm, then it appears to AC signals as if ground and the supply are connected by a 0.1 ohm resistor - "ground" close enough.

But when the battery goes flat the internal resistance may now be as high as 100 ohms and this may upset the AC operation of the circuit. (note that portable radios often show low frequency instability with a flattening battery, pumping out pulses of programme with silent gaps.  Generically called "motorboating".)

The normal way that equipment manufacturers deal with this is placing something like a 100uF electrolytic cap across the supply.  When the battery is fresh this cap is effectively shorted by a 0.1 ohm resistor and so doesn't have anything to do, but when the battery goes flat and its internal resistance climbs this cap now provides an alternative low impedance path around the battery to ground for AC signals, hence why it is sometimes called a "bypass" cap.

I'll give you 10:1 odds that if you trace back from the battery snap in your fuzzbox (via the external power connector and on/off switching) you will find the first thing it comes to on the board is a 100uF cap - right across the supply.  And if it's not the first it will be there somewhere.  It's almost universal (except perhaps with home-brew pedals).


So the short answers are;

- lithium batteries typically have much lower internal resistances than the normal dry batteries used in stomps;

- your stomps almost certainly already have taken rising battery internal impedance with falling State of Charge into account;

- and if they haven't it's a 50-cent fix.   :dbtu:


HTH

Hi Ben, welcome.


Google says R$ = 1 Brazilian Real = 0.32US$.


The Ruby is a one watt amplifier, the Tiny Giant uses a TDA7240 which appears to be a 20 watt amplifier.

The important rating for a battery is its amp-hour rating, how much current you can draw out of it for how long before it's flat and needs recharging.

At 12 volts, and cranked flat out at one watt, the Ruby will draw;

P = E*I
I = P/E
1W / 12V = 0.08333333A or 83mA.

With a 12 volt battery rated at 4AHr (4000mA/Hr) the operating time is;

4000mAHr / 83mA = 48.19277108 hours, or a couple of days continuous.

For the Tiny Giant on 12 volts flat out at 20 watts;

20W / 12V = 1.66666667Amp

4AHr / 1.66A = 2.41hours

The saving grace here is that the power consumption of the amp is directly related to how hard you drive it, so you might get two or three hours or continuous thrash metal playing, but it could easily go twice as long in a more moderate show.


Caution: you can be pretty casual with most battery chemistries and get away with it, but NOT lithium technologies - they are unforgiving of mistakes and can bite back hard.

- these batteries do not like to be over-discharged, so you really need a low voltage cutoff;

- they are quite picky about their charging regime so a specific charger for the cells/pack used would be a good idea (one aeromodeller burned his kitchen down while recharging and now recharges his Lithium batteries inside some concrete breeze blocks).

- they really do not like physical abuse, being spiked or crushed, or being short-circuited.

https://youtu.be/HCGtRgBUHX8

https://youtu.be/LExMC5buoFg

All battery technologies store very considerable amounts of energy, and the better they are, the more forcefully they can unload that energy back to you.  Particular dangers of lithium batteries are suddenly venting hot material, long jets of flame, taking off like a rocket, and literally exploding - and it doesn't take a lot of mistreatment to trigger this off.


I'm thinking that there are battery powered tool sets with spare batteries and proper charger in the 9-18V range with lots of amp/hours, so I'm thinking something like take the torch/flashlight tool and turn it into a power socket for your amp rig.  As for your 9V stomps I'd just get some 9V rechargables and a charger for those.

HTH

Well none of that fits, Lakes Entrance is way out of my normal orbit, but I certainly have a memory of splinting that very broken circuit board (and something about that screw next to the break, flimsy for some reason (?) ) ; would have been in the Manning River area and around ten years ago, maybe more.

{now I guess I won't be satisfied until I dig out and search back through my records to put my mind at rest - I fixed a lot of brummy gear for bands and musicians just passin' through.}

Quote from: phattI'm now living up the road in Bairnsdale

Be thankful it's not Moe.   We had a fall of snow here in Creswick last winter (but then I once had a fall of snow when I was inland of Taree).  I love the warm and the wildlife, but the ticks are a real bringdown.

 

With the EQ set flat, is the output of IC1A pin 1 the unity gain inversion of the signal at IC1B pin 13?

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;



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 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.

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.


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;



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;

}

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.

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"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;


What you basically have is a pair of back-to-back emitter followers (a.k.a. common-collector amplifier), 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.



Why;


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).

Quote from: JLTfound an old Radio Shack car radio. Don't know the wattage, but it was advertised as "28 Watt" measured god knows how

On the one hand you a quite right to be sceptical about "X watts" (I've got some computer speakers somewhere that say "400 watts" on the front - and take four AA batteries on the back.  :duh   "400 Watt" brand I guess.  )

On the other hand all car audio has to compete with high noise levels (and crap speakers in crap enclosures, i.e. car doors) so it has to actually produce some sort of power, a few real watts at least.  The most reliable way to know is flip the lid and see if you can read the type numbers on the chip amp(s), then we know for sure.  From a battery point of view starting with an amp at the lower end of the power scale is a good thing.

Taking on this path means that you are going to have to use a fair bit of cunning and ingenuity to get your result - there aren't too many trails blazed in the forest in this direction.

Generally speaking small speakers, car speakers, and Hi-Fi speakers are not noted for their efficiency (dB/Watt), and in battery operated gear this is a particular interest because doubling the conversion efficiency (+3dB/W) will double the operating time at a given playing level.

For bass I'd be inclined to go with a 12-inch, but given the low power level you might be able to scrounge something suitable (e.g. I have a few light duty 12's I've recovered from console organs).

A possible alternative if it's a four channel "quad" radio is to actually use four 8-inch speakers (but I'd be a bit dubious about bass).

Remember, an awful lot of different stuff fits in a standard dashboard radio cutout, same for a 12-inch speaker, so there is a lot of latitude to rethink based on experience; in other words it may need a couple of cycles of refinement to get it right.

While the Aux input is about the right signal level (~half a volt) it is typically fairly low impedance (10-47k) so might at least require a FET buffer or the like so the guitar is happy; not difficult, just a small detail.

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 }

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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.).



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?

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.