I proposed an idle current of 5 to 10 mA since I haven't seen the heat sink. If it was a 5W amplifier, even in the pre-FTC days, the heat sink was doubtlessly designed to safely dissipate 3W or more. In retrospect, I agree an idle of 10mA (150mW) is much too conservative.

Your suggestion of 100mA (1.5W) idle is much better. My value would probably result in distortion so annoying that the radio would never be used again. Thanks for correcting me.

Originally Posted by bountyhunter

That's a surprising value. I thought the output idling current would be higher? Based on schematic, I think the current flowing down the diode chain is supposed to be about 100mA. I would have guessed the output current would be in the same ballpark. Is there a reason it's so low?

100mA is FAR too high for such a small amplifier - easiest technique is to fit a Vbe multiplier (like I've said all along). Set the bias to zero with the preset, then input a VERY low level sinewave and monitor the output with a scope. Adjust the preset until the crossover distortion just disappears, then check the actual current (it will be a LOT less than 100mA). Leave the amp to warm up, and keep checking it.

easiest technique is to fit a Vbe multiplier

OP hasn't stated whether he wants to improve the amplifier or preserve it.

In either case, he seems to have disappeared.

Originally Posted by bountyhunter

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.

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.

Originally Posted by Nigel Goodwin

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.

Originally Posted by Nigel Goodwin

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.

Originally Posted by Nigel Goodwin

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.

Originally Posted by Nigel Goodwin

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.

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.

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.

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.

Originally Posted by Jony130

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.

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?.