Audio Amplifier distortion issue

Seems a strange circuit,where did it come from?
Measure the voltages with no input. Check that the speaker does not have significant DC across it.

 

From here I think Duke http://www.circuitstoday.com/15w-class-b-audio-amplifier. The biasing seems a bit strange to me.
Adam

 

That design has two major problems. First, the biasing in the output stage is wrong, and this will cause crossover distortion at low volume. Second, there is not enough current gain in the output stage, and this will cause distortion at higher volume. In other words, it will always distort!


Biasing in the output stage is set by the voltage between the base of Q1 and the base of Q2. For a normal Class AB amplifier, this voltage is set at around 1.4V (for silicon transistors) so that both Q1 and Q2 are conducting slightly.

A transistor is controlled by the voltage between its base and emitter, called V_BE. When this voltage reaches about 0.6V (for a silicon transistor), the transistor starts to conduct, and current flows through the collector-emitter path. This current is called I_C. As V_BE increases above about 0.6V, I_C increases as well. A fairly small change in V_BE can cause quite a large change in I_C.

For proper clean operation of the output stage, Q1 and Q2 must both conduct some current all the time. This is called the output stage's quiescent current, and is typically around 20~50 mA (milliamps).

In that circuit, a single diode, D1, is used to set the bias voltage. A single silicon diode drops about 0.7V (the same as a silicon transistor), not the required 1.4V (approx). Therefore the output stage transistors will not conduct together, and no quiescent current will flow in the output stage.

In this situation, each output transistor will only start to conduct when the signal from pin 7 of IC1 reaches a certain positive or negative threshold where it becomes enough to reach the 0.7V threshold needed to make it conduct. Every time the signal crosses over between one transistor and the other, both transistors "drop out", and this is called crossover distortion. It is most noticeable at low volume, where the signal is small and the output spends a lot of time in the crossover region where it is not working properly.

The situation is actually more complicated than that, because op-amp IC1b is in the feedback loop and will try to correct for the crossover distortion, but that's the general problem. There is not enough voltage between Q1 base and Q2 base for quiescent current to flow, so the output stage exhibits crossover distortion.

Removing C1 will probably help - it definitely shouldn't be there - but the proper solution is to replace D1 with a more complicated biasing arrangement.

Biasing the output stage properly requires more than just increasing the base-to-base voltage, because the output transistor characteristics vary with temperature. As they get warmer, they conduct more current for a given V_BE. The bias circuit needs to adjust for this, otherwise you get a situation called thermal runaway, where the output transistors get hotter and hotter until something somewhere goes pop.

This is not a problem when the output transistors are severely under-biased, which is why that design has no temperature compensation.

The proper biasing arrangement replaces D1 with a circuit consisting of a transistor which is in thermal contact with the output transistor heatsink, and a trimpot (preset potentiometer) to set the initial quiescent current, and a couple of resistors. Have a look at the first schematic on http://www.eetimes.com/document.asp?doc_id=1278305 to see how it's done. This design also shows the two extra transistors between the bias circuit and the output transistors which I will explain now.


The signal from pin 7 of IC1b is "weak", i.e. it can't supply much current. If you connected it to a speaker, you would barely hear anything. The output stage is a current amplifier that boosts the current capability of the little op-amp IC so it can drive the speaker.

This works because a transistor has a characteristic called current gain, denoted h_FE. When voltage is applied across the base-emitter junction of a transistor to cause it to conduct, current flows into the base. The collector-emitter current I_C is roughly proportional to the base current I_B and the ratio between them is h_FE.

Those output transistors, TIP41 and TIP42, have a relatively low h_FE - perhaps 30, typically. Let's say that a certain positive peak in the signal corresponds to +8V at the speaker, and the speaker will draw 1A from the output stage (by Ohm's Law: I = V / R). Q1 will be feeding 1A into the speaker, and because its h_FE is only about 30, it will need about 33 mA of base current. IC1 will try to supply that current, but it's not designed for this, and distortion will occur.

This problem is avoided by inserting another transistor before each output transistor. Actually a simpler solution is to use a different kind of output transistor, called a Darlington transistor, that actually has two transistors inside it. Darlington transistors have a much higher h_FE and this avoids the loading problem on IC1b's output.

Just replacing the output transistors with Darlingtons will fix that problem. Common and suitable Darlingtons are TIP120 (NPN) and TIP125 (PNP).

Darlington transistors need twice the V_BE of single transistors, and this has to be taken into account in the bias circuit.


So you can rescue that design pretty easily by (a) replacing the output transistors with Darlingtons, and (b) replacing D1 with an active bias circuit using a transistor that's in physical contact with the output transistor heatsink, two resistors, and a trimpot.

 

Ok just off the top of my head try this. Add another series diode with D1 and connect another 10K from the base of Q1 to the supply, give that a try.
Adam

 

Ah you beat me again!
Adam

 

LOL but your suggestion will make a quick improvement, along with removing C1.

 

No problem

Well, the other part of that circuit is OK.

You wouldn't normally use TIP41 and TIP42 for the other transistors, because they're bigger and more expensive than you need for just driving the output transistors, and they have a lower gain than smaller transistors, but they will work fine. But you need to make the change to the bias circuit.

As a quick hack, you could replace D1 with four diodes in series, and add a TIP41 and TIP42 to convert the output transistors into Darlingtons. That should give a lot of improvement. (C1 needs to be removed too.)

 

Yes, that's right, connect two TIP41s together like that, and connect two TIP42s together like that too; the only difference is the direction that the arrow on the emitter is pointing. Only the right hand transistor of each pair needs to be on the heatsink.

Yes, keep R1 at 10k, but there's no need for one to mirror it. It's there to make current flow through the diodes, so they will drop about 0.7V each. The current flows from the op-amp output, through the diodes, through R1 and to the negative supply.

Keep C2 across the whole group of diodes. And yes, remove C1. I don't know why the designer put it there but it's not a good idea!

 

One more comment. Just replace the 22K pot with a 10K pot, no series resistors needed. The important thing is the ratio of the two resistors formed by the pot, not the absolute value. 10K should work fine.

Bob

 

Which pot?

 

There is a 22K pot in the circuit as a volume control. He says he used a 10K in series with a 12K resistor in the original post.

Bob

 

Ah, sorry. I missed that. Yes good advice.