Analyse this amplifier for me please :)

[Baluncore]

What is the purpose of the 220k resistors from the op-amp's low-Z output to both supply rails ?

 

[meBigGuy]

I can easily see how it works as class B.
Nobody has actually answered by basic question. I'll ask the question another way. When the output voltage is exactly 0, what significant, controllable, current flows through the PNP and NPN final transistor's collectors to make it class AB (and not just a sloppy class B). And what, EXACTLY, sets that current?

 

[Gingerbunny82]

From your measurements, something is wrong around TIP112/TIP126. The emitter voltage should be -1.18V (the mirror image at the emitter of TIP117).

Thanks for the reply, I thought that also, I thought they were supposed to be identical just opposite polarity. But I have a stereo version of the above circuit and both channels show the same difference in voltage. -0.85 for the NPN and +1.18v for the PNP. I can't see how the circuit could influence these voltages either?? :)

 

[Gingerbunny82]

What is the purpose of the 220k resistors from the op-amp's low-Z output to both supply rails ?

I'll be honest, none I think. When I first started building this amp I only had basic knowledge of electronic design. I think I originally put them into set the mid-point or DC bias for the op-amp output signal. But I'm not sure they make a difference as the bias is controlled by the op-amp and its feedback.

 

[Gingerbunny82]

I can easily see how it works as class B.
Nobody has actually answered by basic question. I'll ask the question another way. When the output voltage is exactly 0, what significant, controllable, current flows through the PNP and NPN final transistor's collectors to make it class AB (and not just a sloppy class B). And what, EXACTLY, sets that current?

Thanks for your reply, I have taken some current measurements, I believe the bias current is set by the 3K3 resistors connected to the emitters of the drivers.
Current measurements are in orange.
The measurements have been taken with no input signal again. :)

 

[Jeff Rosenbury]

You are correct. I didn't check the data sheets on the biasing transistors.

That makes them properly biased. It also makes for a strange current mirror using virtual grounds. A push pull current mirror with (almost) no quiescent current. Clever.

 

[Averagesupernova]

You are correct. I didn't check the data sheets on the biasing transistors.

That makes them properly biased.

Nor did I.

 

[jim hardy]

When the output voltage is exactly 0, what significant, controllable, current flows through the PNP and NPN final transistor's collectors to make it class AB (and not just a sloppy class B). And what, EXACTLY, sets that current?

with zero offset in the opamp,
I = (Vbe of biasing transistor - Vbe of final transistor )/ 0.33Ω ?

 

[Svein]

Now I see one weakness with the construction. The driver transistors do not drive the output transistors, they work by stealing bias current from them.

Example: Assume that the OpAmp goes as high as possible (assume +15V). This brings the bases of TIP112/117 to +15V. This again takes the emitter of TIP112 to about +14V, causing a current through the 3.3k resistor of about 4.2mA. But as long as the base of TIP126 is more positive than the emitter, no current will pass through TIP126. On the mirror image, the emitter of TIP117 tries to go to +16.2V, drawing 0.54mA through the 3.3k resistor. This means that the output transistor (TIP121) will essentially have its base connected to +18V through a 3.3k resistor. The current through this resistor determines the output current!

 

[Averagesupernova]

Now I see one weakness with the construction. The driver transistors do not drive the output transistors, they work by stealing bias current from them.

That is a stretch. The drivers are directly coupled emitter followers. That's it. If you want to consider that emitter followers operate in this manner then I guess that's that. You can say the same thing with a pair of diodes used to bias the outputs. Whatever is driving a pair of diodes being used for biasing is 'stealing current' from the outputs' bases as well.

 

[Svein]

Here is an example of a similar audio amplifier. Observe that the current is amplified through the driver stage, not shared.

Personally, I would have added a few more components, but the principle is clear.

 

[Averagesupernova]

Actually, no it is not amplified in what you call the driver stage since in order for the emitter of Q1 and Q2 to source this current it has to come through the base of Q3. This current won't be a whole heckuva lot. It will be enough to turn on Q3 and Q4. This circuit uses a current boost stage. I have seen and used this scheme in voltage regulators. Use a 2N3055 with a 7805 and you can get 10 to 15 amps out. A person can think of it this way: The emitter followers provide the voltage and the current boost transistors provide the current. Admittedly that is a very simplified and not technically correct description.

Edit: Actually the 2N3055 is an NPN and the voltage regulator scheme requires PNP. So the PNP version of the 3055 would be required.

 

[Baluncore]

Here is an example of a similar audio amplifier. Observe that the current is amplified through the driver stage, not shared.

In that example, Q1 with Q3 and Q2 with Q4 form three terminal Complementary Darlington or Sziklai pairs.
Complementary Darlington transistors have lower base emitter voltages than the standard Darlington configuration that uses transistors of the same polarity. That is because saturation can allow the V_BE of the two elements to overlap.
https://en.wikipedia.org/wiki/Sziklai_pair

 

[meBigGuy]

with zero offset in the opamp,
I = (Vbe of biasing transistor - Vbe of final transistor )/ 0.33Ω ?

HOORAY --- A coherent answer. And, given that answer, what determines the difference between the Vbe of those two transistors? Maybe the bias resistors have some effect. Certainly does not seem like a robust design with respect to class AB quiescent final stage current. I consider it a sloppy class B design. Some will work well, some not so well. Depends on Vbe matching of discrete devices.

As for the 220K resistors, I don't can't assign a purpose to those either.

Another quality of design issue is that the opamp is not sourcing current at 0V. I generally add a load resistor from the opamp output to the negative supply to keep the output transistor conducting. Helps with crossover distortion.

The lack of bypass on the feedback resistor has been mentioned before.

I don't have an answer for why lowering the bias resistors to 1.5K made the sound poor. I'd have to see the waveform. These transistors have internal resistors that are not on your schematic. Maybe that is part of it. ( see fig 1 in http://www.onsemi.com/pub_link/Collateral/TIP120-D.PDF )

 

[Svein]

Actually, no it is not amplified in what you call the driver stage since in order for the emitter of Q1 and Q2 to source this current it has to come through the base of Q3. This current won't be a whole heckuva lot.

It has already been answered:

In that example, Q1 with Q3 and Q2 with Q4 form three terminal Complementary Darlington or Sziklai pairs.
Complementary Darlington transistors have lower base emitter voltages than the standard Darlington configuration that uses transistors of the same polarity. That is because saturation can allow the V_BE of the two elements to overlap.
https://en.wikipedia.org/wiki/Sziklai_pair

 

[jim hardy]

what determines the difference between the Vbe of those two transistors?

at quiescent, perhaps their temperatures ?
Those Vbe curves are Onsemi
From TI:

 

[Gingerbunny82]

HOORAY --- A coherent answer. And, given that answer, what determines the difference between the Vbe of those two transistors? Maybe the bias resistors have some effect. Certainly does not seem like a robust design with respect to class AB quiescent final stage current. I consider it a sloppy class B design. Some will work well, some not so well. Depends on Vbe matching of discrete devices.

As for the 220K resistors, I don't can't assign a purpose to those either.

Another quality of design issue is that the opamp is not sourcing current at 0V. I generally add a load resistor from the opamp output to the negative supply to keep the output transistor conducting. Helps with crossover distortion.

The lack of bypass on the feedback resistor has been mentioned before.

I don't have an answer for why lowering the bias resistors to 1.5K made the sound poor. I'd have to see the waveform. These transistors have internal resistors that are not on your schematic. Maybe that is part of it. ( see fig 1 in http://www.onsemi.com/pub_link/Collateral/TIP120-D.PDF )

- Thanks for the reply, I've removed the 220k resistors from the circuit. And added a bypass for the feedback. The Vbe of the output transistor increases as the collector current does, so the difference in Vbe between the driver and output transistors increases as the load does?
I think I understand, how does this Vbe difference affect the amplifier though? Does it cause increased excess current to flow through the output transistors?

 

[meBigGuy]

No, there is no issue with wasted current through the output transistors.

I want to focus on your first post where you asked whether this was class A or Class AB. My opinion is that it is class B, but sometimes (depending on transistor matching) it might turn out class AB. Being class B doesn't mean it will sound "horrible". It means there can be some non-linearity in the crossover region which your feedback will mostly fix. "Fix" in this case could mean reduce it to "not audible to most listeners" levels.

 

[Gingerbunny82]

Thanks for all your replies folks, I'll take what you all said on board - and thanks for taking the time to help me out, cheers Gingerbunny82

 

[rude man]

• Since there is almost no local feedback, only an overall feedback, I see a possible problem with TIM (Transient Intermodulation Distortion).

That is what struck me at once.
If you can get away with the one feedback from output back to the input as you do, that is nice, but there is a lot of circuitry inbetween. I would test trasient response with a square-wave drive to look at the leading & lagging edges. I suspect you're close to oscillation.

 

[jim hardy]

I would test trasient response with a square-wave drive to look at the leading & lagging edges.

i would make that test using for load a speaker in its enclosure, not just a 6 or 8 ohm load resistor.http://www.edn.com/design/consumer/...periority-of-current-drive-over-voltage-drive

 

[meBigGuy]

Am I completely out of whack here? Is it a class B or a class AB?

 

[jim hardy]

Am I completely out of whack here? Is it a class B or a class AB?

Quiescent readings across the 0.33 ohm resistors would say which it is at the moment.
I suspect that'll change with temperature, especially if driver and output transistors aren't real close together on the heatsink.

 

[Gingerbunny82]

Quiescent readings across the 0.33 ohm resistors would say which it is at the moment.
I suspect that'll change with temperature, especially if driver and output transistors aren't real close together on the heatsink.

- I'm thinking it primarily operates in Class-B to be honest, the driver transistors aren't mounted on the same heatsink as the output transistors, so the output transistors get hot whilst the drivers barely get above ambient. :P

- I took a look at the output waveform on an oscilloscope, I can see a slight line in the waveform near the crossover point, very small though.

- The negative feedback must be the main reason it sounds acceptable, because if I remove any form of feedback the sound is horrific.

- It's good enough for my application though and the low quiescent power consumption of 2.5W for a stereo board is a bonus :P

I think If I was to design it again the first thing I'd do would be to replace the darlingtons with regular transistors, giving me more control over the design :)