Hello,

I have a question that I cannot find a definitive explanation for. The schematic is a discrete BJT op-amp. The final output is a push-pull amplifier that normally would have crossover distortion. I experimented with eliminating the crossover distortion using resistors, diodes, etc and was unable to in simulation using LTSpice.

After building the circuit and experimenting I discovered that a capacitor between the base-emitter junction eliminates the crossover distortion. I expect that the capacitor holds a charge due to a DC offset that is present from the output of the Sziklai pair. This allows the transistors of the final stage (Push-Pull) to be biased always on which eliminates crossover distortion.

I am also hoping to get more information about the Sziklai pair. Things that I understand about the Sziklai pair:

1. I understand that the base emitter voltage needs to be 0.7V. Compared to a Darlington Pair which needs a 1.4V to put the pair into active region.
2. The compensation capacitor (Ccomp) is connected in parallel with the base-emitter junction of the pair. The capacitance is used to improve the high-frequency response of the amplifier by bypassing the AC signal around the PNP transistor at high frequencies while limiting instability due to gain.
3. The emitter resistor (RE2) in a Sziklai pair provides an additional voltage drop which can be used to set the DC operating point of the circuit. By adjusting the value of the emitter resistor, the DC offset can be set to the desired level.

Can someone provide more clarification and details? Very appreciated!

 

I suspect you have created a low pass filter. Check the frequency response with and without the capacitors.

 

Hey Bob thanks for the response. Yes, it does seem like a low pass filter! But why do the capacitors eliminate crossover distortion? I can provide screenshots of simulation if needed, but it works in simulation and in experimental results.

 

In most amplifiers that do not have crossover distortion there is a DC bias arrangement so that there is no interval where both output transistors are completely off. In the circuit shown there is a range of two Vbe where neither transistor is conducting. In addition, the region of just entering conduction is not so very linear. So of course the circuit will have lots of distortion.

 

Hello Bill,
Thank you for your response.
So, do these capacitors from base to emitter act as a passive DC bias? I included simulations to the circuit which is very similar to the experimental results.

Circuit without capacitors => Output has crossover distortion.

Circuit with capacitors => No Crossover distortion

 

The crossover distortion happens when one transistor turns off before the other turns on. The capacitor keeps it on for a while longer, but that means it cannot follow a high frequency signal.

 

Your amplifier has a 100k load that is almost nothing.
Try using a load of 100 ohms or 200 ohms.
Then it will have lots of crossover distortion.

 

Audioguru again,
Thank you for your reply.

I see what you mean. I should have seen that. So as Bob and Bill noted that it will get rid of crossover but not for a wide range of input frequencies. And as you noted that the crossover is greatly increased when driving a heavy load.

This screenshot shows distortion starts to occur at a load of 1.2k ohm.


This screenshot shows massive distortion at 100-ohm load.

 

Below is the implementation of removing crossover distortion using diodes. This allows for a much heavier load. Thank you everyone for the clarification. I thought I was onto something with the capacitors, but it seems like it does not allow for heavy loads.

Can anyone point me in the right direction to allow for a higher upper frequency cutoff to allow for a larger bandwidth?

Very grateful!

 

Most ordinary opamps have a max output current of 20mA without distortion.
Yours produces only 0.7V at clipping into 1200 ohms which is an output current of only 0.58mA.

 

Now with the diodes you have the 10k resistors turning on the output transistors with a limited voltage swing instead of using the driver transistors to turn on the output transistors with a very wide voltage swing and more output current.

 

Hello Bill,
Thank you for your response.
So, do these capacitors from base to emitter act as a passive DC bias? I included simulations to the circuit which is very similar to the experimental results.

Circuit without capacitors => Output has crossover distortion.
View attachment 288807
Circuit with capacitors => No Crossover distortion
View attachment 288808

YES, the capacitors would be supplying the base current drive and greatly reducing the crossover distortion. BUT that is at an open circuit output condition. With a speaker connected there will be the distortion.

 

All those capacitors are doing is directly coupling the input signal (to the final amplifier stage) to the load.

This is the exact same circuit:



So for small signals, you are just driving the load with the prior stage. Larger signals start will deliver power from the output stage.

There will still be distortion. How much depends on how well your next-to-last stage can drive the load. But with a 100 kΩ load, do you really need that output stage at all? Anything much above 10 Hz is going to be driven from the prior stage and as long as it can deliver 100 µA of current, the final stage will never be asked to do anything.

My guess is that if you look closely at your simulations, you will discover that the output transistors are never being turned on. To turn either one of them on would require that there be at least 0.6 V across the capacitor and to get that would involve enough current that the voltage across the load would exceed the supply voltage. That would be especially true at higher frequency signals, so sufficiently low-frequency signals, if big enough, might turn on the output stage transistors a little bit.

See how much distortion you get if you lower the load resistance to something like 100 Ω.

Then look into a Vbe-multiplier circuit and how it can be used to largely eliminate crossover distortion, as well as provide thermal compensation if mounted to the same heatsink as the output stage transistors.

 

Try this simulation:

ADDED:

I think you would find this book useful https://amzn.eu/d/2CElZ8o

This book can be downloaded here: https://www.pdfdrive.com/audio-power-amplifier-design-e167210849.html

 

Another alternative is to bias the output transistors using two forward biased diodes between the ir base connections, with the diodes being forward biased from the collector supplies. That is done in my Marantz 2235 amplifier/receiver. It works well until the diodes open up. The diode bias scheme came first, and it does work, but it also does have "issues".

 

<snip> I experimented with eliminating the crossover distortion <snip> I am also hoping to get more information about the Sziklai pair.

I think you would find this book useful https://amzn.eu/d/2CElZ8o

 

The Push-Pull Sziklai Pair design is used the vintage Marantz Model 18 and is a classic from the late ’60s. I recently picked one up and I am in the process of trying to restore it. I have encountered a problem in the Left channel power output stage.

The problem I am having is exactly how the bias circuit (comprised of Q2) works to provide the correct Q point for the output stage. Exactly how does this transistor perform the function of setting the desired bias for the amplifier? The Marantz service manual states that the voltage across R43 should be set to 80mv without an AF signal. I have had problems in trying to set it to the desired level and experienced failures in both of the power transistors (Q13/Q14) while attempting this adjustment. It appears to me that the way the R27 is connected is a potential failure mode. R27 is a wire wound pot and there exists a possibility that during an adjustment the resistance could drop to zero between the Base and Emitter of Q2. This would put Q2 in the cutoff but is this condition which would lead to an overcurrent condition in Q13/Q14 causing them to fail? It would help considerably in the troubleshooting of this circuit if someone could explain the best way to approach how to determine which part of this circuit might be causing the problem.

I have attached the schematic for the power output section and the associated driver circuit; left channel only. I have tested Q2 and R27; the diode stack (CR18) was only showing a 1.7v drop and I replaced it with 3 x 1N914s; after installation, this turned out to be 1.99 volts. Posts about this problem indicate that it should be around 2.1 to 2.2. I don’t know if I need to select 3 new 1N914s and try to achieve a higher very drop closer to the 2.1 value.

 

The temperature compensating diodes have been a Marantz problem for many years. They fail. To get the voltage drop specified you may need to add another diode or a different type of diode. The voltage drop versus current in diodes is very non-linear at low forward currents. Wit the R27 issue, the workaround is to do the adjustment with the power off. Tedious but safer.

 

CR18 is a small package about 14mm x 14mm x 3mm. My tester shows that it’s Silicon and has a Vf of 2.038V and it was mounted next to R27 and in contact with the main heat sink. Several posts on audiokarma.org indicate that it could be replaced with 3x 1N4148 diodes. The original Q2 had an Hfe of 50 and I have tried others (recommended subs) and it did not help.

 

CR18 is a small package about 14mm x 14mm x 3mm. My tester shows that it’s Silicon and has a Vf of 2.038V and it was mounted next to R27 and in contact with the main heat sink. Several posts on audiokarma.org indicate that it could be replaced with 3x 1N4148 diodes. The original Q2 had an Hfe of 50 and I have tried others (recommended subs) and it did not help.

Just because it was posted is no reason to believe it is correct. Consider that the diodes are supposed to be compensating for th temperature gain changes in those power transistors. Are you sure that the original device had failed?? What leads you to think that it had failed???