Welcome to Solid State Guitar Amp Forum | DIY Guitar Amplifiers. Please login or sign up.

July 20, 2024, 10:11:02 PM

Login with username, password and session length

Recent Posts

 

Marshall Valvestate 8100

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

Previous topic - Next topic

Hawk

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.


Roly

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;

(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;


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 1VRMS - 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;

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

J M Fahey

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



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: "

Hawk

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:

Hawk

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!!

Roly

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.).



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

Enzo

Not to brag or anything, but the store shelves here all have condoms with my name on them:

Hawk

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!)

Hawk

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

Enzo

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.

Hawk

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?

Roly

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 Vgamma, 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.




"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, Vin.  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, VE, will be at 10-VBE 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).


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

Hawk

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?

Enzo

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.

Hawk

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: