Not sure what kind of diode to use in amplifier output circuit

[Nigel Goodwin]

Your idea of tweaking the output current up to just enough to eliminate visible crossover is interesting, but I was taught that you don't see visible distortion until THD is about 3% and I doubt anybody wants to listen to that.

It's not 'my idea' it's the correct way of doing it, current settings are only given (after first doing it with a scope) to avoid the rerquirement for a scope and a sinewave generator.

I would have thought you would want to dial it up to make the sine wave look good then give it some more to make sure you don't go back down into calss B at low signal levels.

You do just that, adjust it until visible distortion disappears, then tweak it a little higher. For more accurate results, use a distortion meter as well.

If it was mine, I would use diode connected transistors in the bias chain and not a VBE multiplier which does not behave as close to a diode as a DCT does.

Almost all modern (last 25+ years?) commercial designs, including those of the highest possible quality (and prices), would disagree with you.

Why would you even want it to behave as a diode?.

I'm not sure exacvtly what output current is best, I would assume somewhere between about 30mA and 100mA maximum. I have not seen the heatsink so I am not sure how much heating 100mA would create.

Don't guess it, scope it! - as I said before, 100mA on such a small amp is much too high.

 

[ericgibbs]

hi,
The 0.51R resistors in the emitters of the two output transistors make this particular circuit tolerant to a moderate change in base biassing.

Using a 1N4148 diode gives a standing current in the lower transistor of approx 1mA and in the upper transistor approx 3.5mA.

 

[bountyhunter]

It's not 'my idea' it's the correct way of doing it, current settings are only given (after first doing it with a scope) to avoid the rerquirement for a scope and a sinewave generator.///You do just that, adjust it until visible distortion disappears, then tweak it a little higher. For more accurate results, use a distortion meter as well.

"Correct" way is an opinion. I pointed out using visual sine wave on a scope to adjust to minimize distortion was not good because you have to get a lot of distortion to see anything. If you say it should be checked with a distortion meter, we agree on that.

Almost all modern (last 25+ years?) commercial designs, including those of the highest possible quality (and prices), would disagree with you.

If you say so, but matching the output devices with the same device "diode connected" is a common design method and tracks better than a VBE multiplier. Do the circuit analysis and you will see why. A VBE multiplier is OK, but using a matching DCT is better. That's why I prefer it.

Why would you even want it to behave as a diode?.

At the risk of stating the obvious: YOU WANT IT TO BEHAVE LIKE THE P-N JUNCTIONS OF THE OUTPUT DEVICES. That is to say, match them thermally. Which do you think will do that better: a transistor set up as a VBE multiplier where the base current is an error term or the same devices connected as diodes? If the latter are set up for the same idling current (use same value emitter resistors) they will track EXACTLY.

Don't guess it, scope it! - as I said before, 100mA on such a small amp is much too high.

Scope what? For THD or visible notch distortion? You think visual is the way to design for THD? Just set up a matching set of devices and you will know what the output current is because it will be set by the emitter resistors and it will track over temp. You don't have to screw with anything or adjust anything and the current in the bias chain will be exactly the same as the output devices. Set it at 30 mA and measure the THD.

I've covered this topic pretty thoroughly, so I will terminate it.

 

[RCinFLA]

The diode type is not that critical, only that it is normal silicon, not germanium or hot carrier.

Temp comp'g two output transistor Vbe's with one diode is not very effective. If there was two diodes in series then, yes, they are more critical.

There looks to be 60-75 mA of driver current so the output devices are in cutoff at idle. Probably has pretty poor crossover distortion.

 

[bountyhunter]

Well, I suppose I should explain the OP's circuit operation in detail to show why I was saying what I did just in case somebody here wants to learn something. I drew an attached simplified schematic (see below) to explain how the original circuit works and why it is such a piece of junk.

(All the voltage readings shown on the schematic are from the original schematic.)

The two output transistors Q1 and Q2: they have D1 and a 4.7 Ohm resistor connected across their bases. 19.2V is across the 300 ohms which puts 65mA in the D1 side bias chain. The output current is "set" by D1 and the 4.7 ohm resistor which forces a voltage across the base-emitter junctions of Q1 and Q2 which forces a current through them. This perverted "pseudo current mirror" is what is supposed to control the output stage idling bias current. It can only do it poorly at best because a resistor does not track a diode junction: as the output transistors heat and cool, the output stage bias current will change. Note that the DC output voltage is at 13V which is only about 1/3 of the positive rail voltage of 33V. That's odd, usually you set the output's DC point center range to allow symmetric signal swing going positive and negative from the middle. But this design has another bad feature: that 300 ohm resistor is not a current source, and as the output voltage moves more positive toward the 33V rail, the current through the 300 ohm resistor diminishes, and that current is the base drive for Q1. In other words: the top output transistor Q1 gets "starved" off as the output stage signal swings positive. Setting the output bias point at 1/3 rail voltage may be because of this effect, cenetering the operating point in the useful range of output voltage excursion.

What sets the DC output voltage under no signal? Basically the 3.3k resistor in series with the 200 ohm resistor at the base of Q3. The no signal output voltage will roughly be the VBE of Q3 "gained up" by the ratio of 3300/200. As Q3 pulls current in it's collector, it moves D1 and the 4.7 ohm resistor up and down. Q1 and Q2 move along with D1 and the 4.7 resistor because they are tied at the bases.

The output DC bias voltage point is also the equilibrium where Q3's base current can just drive the Q3 collector current coming down the bias chain which is about 65 mA (Note: since Q3 has a 200 ohm base resistor to ground which eats up about 3.2 mA of the 3.8 mA coming back through the 3.3k resistor, the base drive for Q3 is what's left which is about 0.6 mA).

There is negative feedback on the output voltage DC set point since as the output moves down, the current through the 3.3k resistor goes down (and vice versa). The worst aspect of this design which makes it blow up is that the output stage current is very poorly related to the bias chain current and will not track over temperature.


TEMPERATURE COMPENSATED FIX FOR THE BIAS PROBLEM

The schematic below the first one shows how to fix it using diode connected transistors which are the same devices as the output transistors, so their characteristics will be very close. In this case the bias chain and the output stage form a true current mirror, and the adjustable resistor in the bias side allows the user to adjust the output current. Since the transistors are the same type/maker and operating at the same temperature, the output bias current should track very well over temperature. Once set, it should not wander around.

Basically, you still force the same current in the bias chain but in this design, the output stage will mirror it but at a lower magnitude as ratioed by the emitter resistors of both sides. If the total resistance in the output transistor emitters is equal to the resistance in the emitters of the DCT's, the currents will be the same. If there is more resistance in the output emitters, they will have less current. The variable resistor is included to allow adjusting the output current. If my calculations are correct, the output current should be adjustable from about 15mA to maybe 80mA.

 

[bountyhunter]

ADDING A TRUE CURRENT SOURCE TO DRIVE THE POSITIVE OUTPUT TRANSISTOR:

It's easy to replace the 300 ohm resistor with a 65 mA true current source which will provide a constant 65 mA of drive capability to Q1 regardless of output voltage swing (see below).

DCT3 and Q4 are same type of transistors, DCT3 acts as a diode with a 3.3K resistor to ground which creates 10mA of current through it. That current mirrors over to Q4 but is gained up to 65 mA because the emitter resistors are not equal and that forces a different current through Q4. The higher power dissipation in Q4 (about 1.9W) would require a small heatsink on it.

 

[Jony130]

But this design has another bad feature: that 300 ohm resistor is not a current source, and as the output voltage moves more positive toward the 33V rail, the current through the 300 ohm resistor diminishes, and that current is the base drive for Q1. In other words: the top output transistor Q1 gets as the output stage signal swings positive. Setting the output bias point at 1/3 rail voltage may be because of this effect, cenetering the operating point in the useful range of output voltage excursion.

Do you forgot about bootstrap capacitor C111?
So, Q1 will not "starved" off.
And current that is flow through R127 will not diminish, the voltage across R127 is kept constant by bootstrapping "effect".
Nevertheless very good post.

 

[bountyhunter]

Do you forgot about bootstrap capacitor C111?
So, Q1 will not "starved" off.
And current that is flow through R127 will not diminish, the voltage across R127 is kept constant by bootstrapping "effect".
Nevertheless very good post.

I see the cap. It will provide some current, but that cap is driving two parallel resistors of 150 ohms each when it swings positive: it feeds R127 and also the other 150 (R126) resistor bact into the 33V rail and whatever capacitors are on that line. That makes the time constant of the cap and resistors about 3.7 milli seconds. It certainly will not hold the voltage across R127 constant, it will support Q1 with some transient drive as long as the frequency is high enough.

In this country, amplifiers are specified from 20Hz to 20kHz operation. That capacitor might help at higher frequencies, at lower frequencies it will not do much. It is a cheap substitution for a better design.

My point was, it's very easy (and doesn't cost much) just to use a current source that will provide constant base drive current to the top transistor regardless of output swing. The output stage has strong drive at the bottom from Q105 for swinging negative. If the top transistor had strong and constant drive to go positive, the value of R124 could be increased to move the output's DC set point to halfway which would be about 16-17V, not 13V as shown. That would give more signal swing room and allow more output power.

 

[MrAl]

Hi there,


Another idea is to simply get hold of another transistor the same as that output
NPN transistor, and connect the collector to the base and use that as the new
diode (emitter as cathode). This will have similar characteristics to the output
transistor but as Nigel says it still wont be perfect, especially since the new
'diode' will not be biased the same as the output transistor.
I've seen lots of amplifiers built with regular diodes too and although they are not
perfect they do work and last quite a long time.

 

[Nigel Goodwin]

A current source is one way, and certainly the more modern way - however (and again) even some top amplifiers still use bootstrap capacitors.

The ONLY problem with this one is it's size, but even then it's no where near as bad as you're imagining - at 100Hz it's rectance is only 32 ohms. Up it to 470uF or 1000uF, and it would perform much better at low frequencies. Bear in mind also, the speaker coupling capacitor is only 1000uF as well.

Interesting that you advocate the use of the modern current source, but not the use of the modern Vbe multiplier?.

 

[MrAl]

Hi there,


Another idea is to simply get hold of another transistor the same as that output
NPN transistor, and connect the collector to the base and use that as the new
diode (emitter as cathode). This will have similar characteristics to the output
transistor but as Nigel says it still wont be perfect, especially since the new
'diode' will not be biased the same as the output transistor.
I've seen lots of amplifiers built with regular diodes too and although they are not
perfect they do work and last quite a long time.

Hi again,


I forgot to mention something interesting...

In the past i have done a lot of work in control electronics, using
feedback to control some parameter of the circuit or to help eliminate
what are called "disturbances" in the input or injected somewhere
else in the system.

For example, i once designed a temperature compensated zero voltage
current shunt, which as the name suggests is a current shunt with
zero voltage drop and zero temperature drift. The idea was to figure
out what bothered the common current shunt (resistive) and try to
develop a feedback system that would eliminate all the disturbances
(in that case it would be temperature and voltage drop). What came
of this was a pretty cool shunt that was near perfect.

In the case of the audio output stage, the main disturbance is the
base emitter voltage, or at least the change, and the change comes
over temperature as well as when the amplifier is driven by a voltage
source, but in any case it's mostly the voltage change that we are
worried about because as you all know a change in voltage means
a change in dc bias. When diodes are used they can track the Vbe
change to some extent, but really fail short as Nigel pointed out.
What is really needed then is a system that can track this Vbe change
and can do it near perfectly. Enter once again the feedback system.
Using op amps and resistors a bias scheme can be developed that
perfectly tracks ANY change in the base emitter voltage of both transistors
and makes the necessary adjustment in the bias current. The drawback
of course is that it requires the use of some op amps and so this
introduces some complexity in the circuit. What it doesnt fix is the
zero offset problem which still needs to be adjusted for a perfect zero
out for zero in, and also there is still some crossover distortion using
bipolars, but of course the crossover problem can be almost eliminated
by using MOSFETs instead. Since the design gets more complex however
it may be better to simply redesign the entire output stage using op amps
and MOSFETs and get a really clean output that is always biased perfectly.

 

[Jony130]

but of course the crossover problem can be almost eliminated
by using MOSFETs instead. Since the design gets more complex however
it may be better to simply redesign the entire output stage using op amps
and MOSFETs and get a really clean output that is always biased perfectly.

Why it is easier to eliminate crossover distortion in MOSFET then in BJT?
I don't see any difference.

 

[Nigel Goodwin]

Why it is easier to eliminate crossover distortion in MOSFET then in BJT?
I don't see any difference.

Because there isn't, it's just as easy to eliminate with bipolar as FET.

Depending on the type of FET's used it 'may' be easier to prevent thermal runaway - as some have negative temperature coefficients, others though are positive like bipolar, and can just as easily runaway.

But commercially bipolar amps are far more popular than FET ones, presumably for reasons of price, and also probably reliability - certainly in my experience, FET's are generally LESS reliable than bipolar.

 

[bountyhunter]

A current source is one way, and certainly the more modern way - however (and again) even some top amplifiers still use bootstrap capacitors.

The ONLY problem with this one is it's size, but even then it's no where near as bad as you're imagining - at 100Hz it's rectance is only 32 ohms. Up it to 470uF or 1000uF, and it would perform much better at low frequencies. Bear in mind also, the speaker coupling capacitor is only 1000uF as well.

Let's try some reality: what does it cost to chuck the huge "bootstrap" capacitor in favor of a dual transistor current source? maybe fifty cents at retail? And remember, we are not having to pay for your 1000uF capacitor which proabably costs at least half that. Give me any cogent reason why anybody sans brain damage would choose to use a drive circuit which is frequency dependent when a good one is so cheap. I just don't understand using a wimpy design when a better one is so cheap and easy. Of course the 8 Ohm speaker is coupled with a 1000uF capacitor, and that gives it a cutoff frequency (3 dB down) of 20 Hz. Did you actually do the math?

The point I made about the bootstrap cap is that it's time constant looking into those 150 Ohm resistors in parallel is about 3.7 milliseconds. That means it is going to go useless WAYYYYYY before you ever get down to 20 Hz. You said:

at 100Hz it's rectance is only 32 ohms.

The point is, how much is the cap going to discharge? The point of the cap is to hold a voltage to shoot current through the resistor to drive the top transistors base. If (as will happen at 100 Hz) most of the voltage has discharged off, what difference does the reactance make? If the RC time constant is 3.7ms, how well will it handle a waveform period of TEN milliseconds? Not well, it will lose most of it's charge.

Do the math.

 

[bountyhunter]

Interesting that you advocate the use of the modern current source, but not the use of the modern Vbe multiplier?.

Neither the current source or VBE multiplier is "modern". They were both staples when I started designing back in 1975. I am not saying a VBE multiplier is terrible, it's better than the garbage stock circuit using a resistor and diode. However, I don't have confidence in it because it doesn't track as well as diode connected transistors. You have one NPN transistor in a gain circuit mirroring a PNP and NPN device output stage. Matches sort of, not as well as the design I posted. I have no idea how long it would live with a VBE multiplier, but when I fix things I use best available circuits and that's what I posted. I don't check the time stamps on them.

 

[bountyhunter]

Hi there,


Another idea is to simply get hold of another transistor the same as that output
NPN transistor, and connect the collector to the base and use that as the new
diode (emitter as cathode). This will have similar characteristics to the output
transistor but as Nigel says it still wont be perfect, especially since the new
'diode' will not be biased the same as the output transistor.
I've seen lots of amplifiers built with regular diodes too and although they are not
perfect they do work and last quite a long time.

I already posted a design that uses that.

Hi there,


Another idea is to simply get hold of another transistor the same as that output
NPN transistor, and connect the collector to the base and use that as the new
diode (emitter as cathode). This will have similar characteristics to the output
transistor but as Nigel says it still wont be perfect, especially since the new
'diode' will not be biased the same as the output transistor.
I've seen lots of amplifiers built with regular diodes too and although they are not
perfect they do work and last quite a long time.

Using a DCT will match way better than a VBE multiplier will.

 

[bountyhunter]

Hi again,


When diodes are used they can track the Vbe
change to some extent, but really fail short as Nigel pointed out.
What is really needed then is a system that can track this Vbe change
and can do it near perfectly.

Same devices from same manufacturer track very well. I have measured them for this. If you want them dead on, buy a dozen and measure all the VBE's and pair them up for values. A small amount of resistive degenration handles the mismatch. This is well known stuff.

Hi again,


Enter once again the feedback system.
Using op amps and resistors a bias scheme can be developed that
perfectly tracks ANY change in the base emitter voltage of both transistors
and makes the necessary adjustment in the bias current. The drawback
of course is that it requires the use of some op amps and so this
introduces some complexity in the circuit. What it doesnt fix is the
zero offset problem which still needs to be adjusted for a perfect zero
out for zero in, and also there is still some crossover distortion using
bipolars, but of course the crossover problem can be almost eliminated
by using MOSFETs instead. Since the design gets more complex however
it may be better to simply redesign the entire output stage using op amps
and MOSFETs and get a really clean output that is always biased perfectly.

I suppose given enough money and size, anything is possible. I was keeping it real as something a person could buy and install into the existing assembly.

 

[Nigel Goodwin]

Let's try some reality: what does it cost to chuck the huge "bootstrap" capacitor in favor of a dual transistor current source?

Let's try some more reality, in the days that bootstrap capacitors were originally used a capacitor was a tiny fraction of the cost of using an extra VERY expensive transistor. Transistors were VERY, VERY expensive devices back then.

The high quality of the sound such amplifiers give (and many of the best sounding amplifiers ever made used bootstrap capacitors) means that they are still used today in some designs, including very highly rated ones.

I've designed amps using both types over the years, and never noticed any differences - if I was designing an amp tomorrow, I don't know which I'd use?, as I see it there's no real difference in performance.

As for your hatred of Vbe multipliers, why this fixation on 'perfect tracking'?, the bias components have TWO functions, to prevent crossover distortion, and to prevent thermal runaway - which Vbe multipliers do perfectly. Which is why they are still usually used in the highest quality amplifers.

 

[bountyhunter]

Let's try some more reality, in the days that bootstrap capacitors were originally used a capacitor was a tiny fraction of the cost of using an extra VERY expensive transistor. Transistors were VERY, VERY expensive devices back then.

So, when are we fixing (or advising the OP on a course of action to fix it)?

Oh, yeah... today. So I should advise an inferior circuit because it was a better choice 60 years ago? OK, I'll stick with mine.

The high quality of the sound such amplifiers give (and many of the best sounding amplifiers ever made used bootstrap capacitors) means that they are still used today in some designs, including very highly rated ones..

Look, using a current driver with a terrible frequency dependency doesn't contribute to hi fidelity. It doesn't do anything good or better than using a current source, it's just cheaper but not cheaper enough to interest me in using it.

If high end designs are actually using it, I think that's pathetic. It's just an inferior design technique and that really isn't debatable. Other than saving a few cents, it has no advantage. I suspect if there are high end designs using it, that's because they used a "legacy circuit" or a design from an old guy who did it because he always did it that way. But it's not superior, it's inferior.

I've designed amps using both types over the years, and never noticed any differences - if I was designing an amp tomorrow, I don't know which I'd use?, as I see it there's no real difference in performance.

You seriously can't see how that output stage would have a drive problem at rated voltage swing at frequencies below 100 Hz? You can't be serious. At 20 Hz, the wave period is 50 milliseconds (12.5 ms to go from zero cross to the peak amplitude) and your bootstrap cap's time constant into it's resistors is less than four ms. There isn't going to be any usable voltage across the cap by the time the output tries to swing to the top of the peak of the sine wave. You can't see that driver is going to go away at the low end?

As for your hatred of Vbe multipliers, .

Non sequitur. Just because they are not the best design, doesn't mean I hate them.

As for your hatred of Vbe multipliers, why this fixation on 'perfect tracking'?,

There is no perfection, there is the one that tracks best. Let's see, these amps tend to blow up as evidenced by:

a) the crappy design

b) the fact the OP is trying to fix one that blew up

Why would you NOT go for best tracking to prevent thermal runaway which is clearly what is killing them?

the bias components have TWO functions, to prevent crossover distortion, and to prevent thermal runaway - which Vbe multipliers do perfectly.

NO, NO, NO NO.... and NO. VBE multipliers do prevent runaway better than a resistor/diode, they do NOT do it perfectly and they do not do it as well as a design with DCTs. Draw the circuit for a VBE multiplier, or model it, and notice a couple of things:

1) The VBE multiplier uses an NPN which should match an NPN, not so well the PNP. At least not as well as a PNP would.

2) The multiplier is forcing a voltage across two VBE's, one PNP and one NPN.
The base current flowing into the VBE multiplier's transistor is an error term in it's action as a diode.


A VBE multiplier might work. To do it, you add a transistor and resistors. Add one more transistor and you get a better design which will track better. Nothing personal, it's just a better design.

As for eliminating notch distortion, the key is to set the output bias at the optimal current and keep it there. Which one keeps it there better? The one that tracks better and drifts less.

 

[bountyhunter]

As for "modern" ways to do this, they now build power transistor with on board diodes to use for biasing. If you want perfect tracking, that's as close as you will get.

https://www.electro-tech-online.com/custompdfs/2009/09/App20Note2010120rev1p2.pdf