Low Power Class AB Amplifier

ericgibbs

Joined Jan 29, 2010
18,855
Hi RRR
To get a meaningful result for the AC analysis set AC=1, the dBs then will be the actual dB response of the amp
E
 

MrAl

Joined Jun 17, 2014
11,487
Hi:

Here is a low power Class AB amplifier.

Two Questions:
1) How can I get more than 180 mW out of this circuit?
2) Can this circuit be used for a small microphone or a cd player?

Thanks,

RS
Hi,

Do you really intend to use an 8 Ohm load or can it be 16 Ohms or 4 Ohms?
Just asking because that will have a lot to do with it.

You can estimate the power by knowing the power supply voltage.
The theoretical would be where the peak of the output sine would be 1/2 of Vcc, and so the RMS of a sine would be sqrt(2)/2*Vcc and so square that and divide by the output resistance and that gives you the output power, theoretically. In practice though you have to subtract maybe 1 to 2 volts for the transistor drops, so just subtract 1 or 2 volts from Vcc before you do that calculation.

To figure out if you have enough gain you would have to test the microphone output and do a simulation with a circuit simulator. Different microphones have different level outputs so you have to think about that, which means a measurement is a good idea. Even different mic's of the same type can put out different voltage levels.

For good bass response you need to have power supply filtering with decent sized capacitors too so that the voltage does not sag too much with low frequency tones. Without good caps there the distortion could be noticeable too.
 

Audioguru again

Joined Oct 21, 2019
6,696
1) The two gain stages produce gain that is too low. The gain of Q1 is roughly RC/RE1= 5 times and the gain of Q2 is roughly RB4/RE2= 20 times for a total gain of about only 100 times.
2) The idle voltage at the output should be half the supply but yours is 12.6V.
3) If you want a peak output voltage to be 8V then the emitter current of Q3 must be about 1A but the resistance of RB4 is too high to provide enough base current (2V/2200 ohms= 0.9mA. 20mA or more is needed for a high output power).
 
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Thread Starter

RRRRSSSS

Joined Jun 16, 2023
128
Hi,

Do you really intend to use an 8 Ohm load or can it be 16 Ohms or 4 Ohms?
Just asking because that will have a lot to do with it.

You can estimate the power by knowing the power supply voltage.
The theoretical would be where the peak of the output sine would be 1/2 of Vcc, and so the RMS of a sine would be sqrt(2)/2*Vcc and so square that and divide by the output resistance and that gives you the output power, theoretically. In practice though you have to subtract maybe 1 to 2 volts for the transistor drops, so just subtract 1 or 2 volts from Vcc before you do that calculation.

To figure out if you have enough gain you would have to test the microphone output and do a simulation with a circuit simulator. Different microphones have different level outputs so you have to think about that, which means a measurement is a good idea. Even different mic's of the same type can put out different voltage levels.

For good bass response you need to have power supply filtering with decent sized capacitors too so that the voltage does not sag too much with low frequency tones. Without good caps there the distortion could be noticeable too.
This is what comes out of this amp with an input of 1 volt. How can it produce a 5 + watt output?
1696509733858.png
 

Thread Starter

RRRRSSSS

Joined Jun 16, 2023
128
1) The two gain stages produce gain that is too low. The gain of Q1 is roughly RC/RE1= 5 times and the gain of Q2 is roughly RB4/RE2= 20 times for a total gain of about only 100 times.
2) The idle voltage at the output should be half the supply but yours is 12.6V.
3) If you want a peak output voltage to be 8V then the emitter current of Q3 must be about 1A but the resistance of RB4 is too high to provide enough base current (2V/2200 ohms= 0.9mA. 20mA or more is needed for a high output power).
There is an input of 50mV and an ouput of 1.2V. Is that not a gain of 24?
So basically, this circuit like all others on the internet and text books is garbage.
 
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Ian0

Joined Aug 7, 2020
9,831
There is an input of 50mV and an ouput of 1.2V. Is that not a gain of 24?
So basically, this circuit like all others on the internet and text books is garbage.
70 years ago it would be the height of technology.
This design (attached) is better than the one you are using, and look at the date on the paper!
Your design is similar, but his is more advanced because he used Darlington and Sziklai pairs to get enough gain to be able to use a low drive current.3CF12A22-2978-4922-BE4C-56A182AF29C6.jpegL
 

Attachments

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crutschow

Joined Mar 14, 2008
34,455
I tried tweaking your circuit some, but could not get good results.

Below is a better circuit I believe, with a few more components, which generates over 4W:
It uses a differential input, with DC feedback so the output is always biased at the midpoint voltage without significantly being affected by component variations, and AC feedback to stabilize the gain and reduce distortion (≈0.35% for the signal shown).
Apparently it was designed by someone named Marley. :cool:

Edit: Changed R1 and R8 to increase the input resistance and reduced size of C1.
1696559076541.png
 
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BobTPH

Joined Jun 5, 2013
8,989
Nice. Differential input. Vbe multiplier.

Is C6 a bootstrap? Never seen that configuration before, but it seems better than the usual split collector resistor.
 

Ian0

Joined Aug 7, 2020
9,831
Nice. Differential input. Vbe multiplier.

Is C6 a bootstrap? Never seen that configuration before, but it seems better than the usual split collector resistor.
Yes, that’s a bootstrap, but it is much more commonly seen with two resistors than a resistor and a diode.
A bootstrap load to the VAS works almost as well as a constant current source for distortion and rather better if maximum output voltage is your design parameter.
 

Thread Starter

RRRRSSSS

Joined Jun 16, 2023
128
Yes, that’s a bootstrap, but it is much more commonly seen with two resistors than a resistor and a diode.
A bootstrap load to the VAS works almost as well as a constant current source for distortion and rather better if maximum output voltage is your design parameter.
Thanks Crutschow. Sim works well. Can you tell me what each of the three circled areas are
and their purpose. I need to do some research.

Thanks,
RS
 

Attachments

MrAl

Joined Jun 17, 2014
11,487
This is what comes out of this amp with an input of 1 volt. How can it produce a 5 + watt output?
View attachment 304243
[ADDED NOTE LATER]
This will show you the thinking process that goes behind some of the decisions that you might go through in order to understand this kind of circuit. However, a simple solution is to simply increase the supply voltage from 20v to 25v or even 30v, and change RB4 to 1k. That will probably get you there, or at least close. The real problem lies with the output transistor drive which can not be done very well with just a 20v supply and still get 5 watts output. With 30 volts, this should be easily accomplished.
[END ADDED NOTE]


Hello again,

The approximations i gave you assume that the amplifier it working correctly. If it is not, then the approximations do not apply.

For this amplifier, it looks like the two output transistors are not being driven hard enough to supply the full output peak voltage to the output and thus to the 8 Ohm load.
That will have to be rectified before you can get any use out of this.

If you divide 20-2 by 2 you get 18/2 which is 9v peak.
9v peak into an 9 Ohms load would be 1.125 amps.
If the two output transistors have a current gain of 20 each, then you need to input about 56ma into the base.
With 9v at the output referenced to 10v, that means the output will be 19v. With 19v on the output, the base of the upper output transistor would have to be, if the base emitter drop was really 0.7v at that point, 19.7 volts. Since there is a 20v supply, we have to get 56ma to flow with a drop of only 0.3 volts. That means the resistor RB4 would have to be
about 5 Ohms which is far to small to work correctly because the lower transistor Q2 would have to draw an immense current to be able to get the output to reach the negative excursion peak.
Now we might assign a gain of 40 to the output transistor, and that would lead to an upper resistor RB4 of 10 Ohms, which is still far too small.
How about a gain of 80. That would lead to 20 Ohms. Still way too small.

So, let's say that the output only has to reach to 8v peak, which would mean only 4 watts output. That gives us an extra volt of overhead to work with. Now instead of 0.3 volts we have 1.3 volts. That leads to 23 Ohms with that 56ma requirement.
Double the gain of the transistor again to 40, and we would need 46 Ohms.
Double again to a gain of 80, we need 92 Ohms.
That still probably isnt going to work very well unless we do have transistors that actually have a gain of 80 at the required output voltage and current and power supply levels.

This shows that getting an output power of 4 watts may even be impossible.
What about 2 watts.

If we assign a 4 volt drop for each output transistor that would leave 6 volts peak for the output, and that leads to an output power of 2.25 watts into 8 Ohms.

With that 6 volts positive peak, the transistor base would have to be about 6.7 volts added to 10v dc set point (which we would also have to set) gives us 16.7 volts, and subtracted from 20v we get 3.3 volts. 3.3 volts with the required 56ma, we come out with RB4 value of 59 Ohms, which would still be too small.
Double the given transistor gain to 40 again, we would end up with 118 Ohms, which may be possible.
This means we would have to have transistors with a gain of 40 to get the 2.25 watts output, approximately. If that didn't work either, we'd have to lower the output power requirement again.

So you can see how the output stage drive is so important. We have to maintain the output peak to get the power, which means we have to maintain the transistor base peak, which means we need a certain maximum RB4 value, which means we need a certain minimum Q2 collector current implying the gain and bias of Q2 has to be right, and that implies the gain of Q1 and associated currents have to be right also.

The main point here though is the output transistor drive because if we do not have the ability to supply the right output transistor drive then we cannot use this construction as is. We'd have to increase the drive or go to better output transistors such as a pair of Darlington transistors or something similar, then recalculate.
 
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Thread Starter

RRRRSSSS

Joined Jun 16, 2023
128
Hello again,

The approximations i gave you assume that the amplifier it working correctly. If it is not, then the approximations do not apply.

For this amplifier, it looks like the two output transistors are not being driven hard enough to supply the full output peak voltage to the output and thus to the 8 Ohm load.
That will have to be rectified before you can get any use out of this.

If you divide 20-2 by 2 you get 18/2 which is 9v peak.
9v peak into an 9 Ohms load would be 1.125 amps.
If the two output transistors have a current gain of 20 each, then you need to input about 56ma into the base.
With 9v at the output referenced to 10v, that means the output will be 19v. With 19v on the output, the base of the upper output transistor would have to be, if the base emitter drop was really 0.7v at that point, 19.7 volts. Since there is a 20v supply, we have to get 56ma to flow with a drop of only 0.3 volts. That means the resistor RB4 would have to be
about 5 Ohms which is far to small to work correctly because the lower transistor Q2 would have to draw an immense current to be able to get the output to reach the negative excursion peak.
Now we might assign a gain of 40 to the output transistor, and that would lead to an upper resistor RB4 of 10 Ohms, which is still far too small.
How about a gain of 80. That would lead to 20 Ohms. Still way too small.

So, let's say that the output only has to reach to 8v peak, which would mean only 4 watts output. That gives us an extra volt of overhead to work with. Now instead of 0.3 volts we have 1.3 volts. That leads to 23 Ohms with that 56ma requirement.
Double the gain of the transistor again to 40, and we would need 46 Ohms.
Double again to a gain of 80, we need 92 Ohms.
That still probably isnt going to work very well unless we do have transistors that actually have a gain of 80 at the required output voltage and current and power supply levels.

This shows that getting an output power of 4 watts may even be impossible.
What about 2 watts.

If we assign a 4 volt drop for each output transistor that would leave 6 volts peak for the output, and that leads to an output power of 2.25 watts into 8 Ohms.

With that 6 volts positive peak, the transistor base would have to be about 6.7 volts added to 10v dc set point (which we would also have to set) gives us 16.7 volts, and subtracted from 20v we get 3.3 volts. 3.3 volts with the required 56ma, we come out with RB4 value of 59 Ohms, which would still be too small.
Double the given transistor gain to 40 again, we would end up with 118 Ohms, which may be possible.
This means we would have to have transistors with a gain of 40 to get the 2.25 watts output, approximately. If that didn't work either, we'd have to lower the output power requirement again.

So you can see how the output stage drive is so important. We have to maintain the output peak to get the power, which means we have to maintain the transistor base peak, which means we need a certain maximum RB4 value, which means we need a certain minimum Q2 collector current implying the gain and bias of Q2 has to be right, and that implies the gain of Q1 and associated currents have to be right also.

The main point here though is the output transistor drive because if we do not have the ability to supply the right output transistor drive then we cannot use this construction as is. We'd have to increase the drive or go to better output transistors such as a pair of Darlington transistors or something similar, then recalculate.
ok thanks
 

crutschow

Joined Mar 14, 2008
34,455
Can you tell me what each of the three circled areas are
and their purpose.
#1 is called a bootstrap circuit.
It provides a signal from the output through C6 to boost the R12 bias voltage above the supply voltage on the positive signal peaks to allow a higher peak output before Q5 saturates.
You can look at that node in the simulation to see how it works.

#2 shapes the negative feedback from the output:
C2 blocks DC so the DC bias negative feedback is not attenuated, making the DC output bias equal to the voltage at Q1's base (here 1/2 the supply voltage).

C2 is essentially a short for audio AC signals, so the AC negative feedback to control the gain and reduce distortion is determined by the divider consisting of R4 and R5.
If not for the divider, there would be 100% AC negative feedback and the gain would be essentially one.

#3 provides negative feedback gain rolloff at high frequencies to reduce noise and the possibility of high frequency oscillations.

All make sense?
 

MrAl

Joined Jun 17, 2014
11,487
ok thanks
Hello again,

You probably missed the note i added later at the beginning.
It basically states that if you increase the power supply voltage to 25 or even 30 volts, you can get better results. You may have to change RB4 to 1k and go from there.
 

Ian0

Joined Aug 7, 2020
9,831
Thanks Crutschow. Sim works well. Can you tell me what each of the three circled areas are
and their purpose. I need to do some research.

Thanks,
RS
Think of the bases of Q1 and Q2 as the non-inverting and inverting inputs of an op-amp.
You can then see that R5,R4,C2 form the feedback for a non-inverting amplifier.
 
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