# Transistor audio amplifier calculations

#### Uros11

Joined Dec 17, 2018
5
Hello,

I have some problem with my audio amplifier, and I am not sure how to continue with my calculations. I would like if anyone can help write me the right equations that I need to make.
Here you can see the circuit (which I took from "Art of Electronics"), but in my case I have a 8 Ohm load which affects my circuit in a way that the output voltage is cut.

I can experimentally set some values that can give me a solid output, but I would like to know exactly how to find problem in this kind of situation and what should be calculated with what equations?

#### Jony130

Joined Feb 17, 2009
5,212
The problem is that you decided to use as output transistors a small-signal BJT's in the simulation. And the small-signal BJT's do not have enough current capacity and the current gain do drive 8 ohm's load. Ic_max for 2n3904 is 0.2A and power dissipation around 0.6W (for ideal heatsink).
In your circuit the quiescent power dissipation is around 0.64A*15V = 9.6W. Therefore in real-world your BJT's will be well gone and death (magic smoke).

And the clipping occurs (output voltage is cut) due to the lack of enough base current (the Q5 base current is provided by R6 resistor) to drive Q5 hard enough.

We have

IL_neg_max ≈ (Vcc - Vbe)/(R6/β + R9)

As you can see to increase the "negative cycle" current you need to reduce the R6 resistor value.

#### Zeeus

Joined Apr 17, 2019
608
Hello,

I have some problem with my audio amplifier, and I am not sure how to continue with my calculations. I would like if anyone can help write me the right equations that I need to make.
Here you can see the circuit (which I took from "Art of Electronics"), but in my case I have a 8 Ohm load which affects my circuit in a way that the output voltage is cut.

View attachment 178427 View attachment 178428

I can experimentally set some values that can give me a solid output, but I would like to know exactly how to find problem in this kind of situation and what should be calculated with what equations?

#### Zeeus

Joined Apr 17, 2019
608
As you can see to increase the "negative cycle" current you need to reduce the R6 resistor value.
or current source?

how do you get the output power for speaker? How much does the 8 ohm need? Will current source and 2n3904 and 2n3906 work (will post schematic 'soon')?

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#### Jony130

Joined Feb 17, 2009
5,212

#### Zeeus

Joined Apr 17, 2019
608
Yes, good idea.

Every speaker has it rated power so, I don't understand your question.

For low power only.

#### Uros11

Joined Dec 17, 2018
5
The problem is that you decided to use as output transistors a small-signal BJT's in the simulation. And the small-signal BJT's do not have enough current capacity and the current gain do drive 8 ohm's load. Ic_max for 2n3904 is 0.2A and power dissipation around 0.6W (for ideal heatsink).
In your circuit the quiescent power dissipation is around 0.64A*15V = 9.6W. Therefore in real-world your BJT's will be well gone and death (magic smoke).

And the clipping occurs (output voltage is cut) due to the lack of enough base current (the Q5 base current is provided by R6 resistor) to drive Q5 hard enough.

We have

IL_neg_max ≈ (Vcc - Vbe)/(R6/β + R9)

As you can see to increase the "negative cycle" current you need to reduce the R6 resistor value.
Thank you very much, I changed the transistors and edited a circuit a little bit and it is working fine, but my output voltage amplitude is not constant, it varies from 0.02 - 0.5 . Also, I wanted to ask about getting the maximum gain possible from this circuit? Is the main obstacle transistor selection?

Here is my circuit, a little bit changed from last time.

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#### Zeeus

Joined Apr 17, 2019
608
Thank you very much, I changed the transistors and edited a circuit a little bit and it is working fine, but my output voltage amplitude is not constant, it varies from 0.02 - 0.5 . Also, I wanted to ask about getting the maximum gain possible from this circuit? Is the main obstacle transistor selection?

Here is my circuit, a little bit changed from last time.
can increase gain by increasing R4

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#### Uros11

Joined Dec 17, 2018
5
can increase gain by increasing R4
Yes, I know, but how about maximizing it, for example if I put 100kOhms for R4, I would have to adjust other values to get the stable sine wave at the output.

#### Zeeus

Joined Apr 17, 2019
608
Yes, I know, but how about maximizing it, for example if I put 100kOhms for R4, I would have to adjust other values to get the stable sine wave at the output.
sorry, did not understand
"but my output voltage amplitude is not constant, it varies from 0.02 - 0.5" do not understand this

if you check the file, changed R4 to 80k then replaced R6 with current source so it does not clip at the bottom

Is this assignment?

#### Uros11

Joined Dec 17, 2018
5
sorry, did not understand
"but my output voltage amplitude is not constant, it varies from 0.02 - 0.5" do not understand this

if you check the file, changed R4 to 80k then replaced R6 with current source so it does not clip at the bottom

Is this assignment?
I meant that the amplitude is changing in time a little bit, but now everything is good, I changed some values and put bootstrapping. Thank you .

#### Zeeus

Joined Apr 17, 2019
608
I meant that the amplitude is changing in time a little bit, but now everything is good, I changed some values and put bootstrapping. Thank you .
Nice = bootstrapping...You can post file please?

Don't understand bootstrapping well and just asked, this will be for real speaker?

#### Uros11

Joined Dec 17, 2018
5
Yes, sure, here is my circuit. It is just some project that I'm doing for now, I will think about what to do with it. There is probably some wave input and output in my file, I tested it for real audio.

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#### Ylli

Joined Nov 13, 2015
999
Might want to delete D1 and D2 and use a 15 ohm resistor in series with D3. You don't need hundreds of mA of idle current.

#### Zeeus

Joined Apr 17, 2019
608
Might want to delete D1 and D2 and use a 15 ohm resistor in series with D3. You don't need hundreds of mA of idle current.
You mean so there's about 2 diode drop between Q3 and Q5? therefore less current through R8 and R7

if so, might as well delete D3 and use 40 ohm resistor?

#### Jony130

Joined Feb 17, 2009
5,212
I changed the transistors and edited a circuit a little bit and it is working fine
What you must understand is that in this class AB power amplifier the power dissipation in the output transistors is equal to arond P = 0.1*(Vcc^2/RL) = 0.1*(15V^2)/8Ω = 2.8W and the peak of instantaneous power is around P_peak = (Vcc^2)/(4*RL) ≈ 7W for a pure resistive load.
Hence you must choose the output stage transistor is such a way that they can handle such a power dissipation (a heatsink is needed) plus the quiescent current power dissipation.

Also, I wanted to ask about getting the maximum gain possible from this circuit? Is the main obstacle transistor selection?
You need to change topology. How large the voltage gain you need?

#### Ylli

Joined Nov 13, 2015
999
With the passive current path in the VAS (Voltage amplifier stage), there is quite a bit of current through diodes D1, D2, and D3. Even with only two diodes, the voltage drop that you get is such that there is more idle current in the output than desired. One diode and a resistor gets you the desired idle current while still maintaining *some* temperature compensation. The next step would be to replace that diode/resistor string with a Vbe multiplier. And while we are making more tweaks..... If you increase the gain by changing the feedback resistor to 80k, you also need to change the input resistor to 80k to keep the LTP balanced. And as now designed, the amp has a large gain peak up near 6 MHz, and needs some compensation. Add a 100 pF cap between the collector and the base of the VAS (center stage). [This is commonly called the miller cap or Cdom) With that 100 pF cap, it looks like the phase margin is OK.

#### Jony130

Joined Feb 17, 2009
5,212
Don't understand bootstrapping well and just asked
Take a look at this circuit:

As you can see without any input signal (DC condition) the bootstrap capacitor (C1) is charged to 7.85V. Also notice that the circuit time constant is very long (t = Rx||Ry * C1 = 0.544s) compared to the audio signal period (1/20Hz = 0.05s). Hence from this, we can tell that the capacitor will act just like a 7.85V DC voltage source (the input signal is changing way too fast to be able to charge/discharge the C1 capacitor).

Now let us assume that voltage at the output is at a positive peak and equal to +10V.

The situation is shown here:

As you can see the voltage at Vx node can be higher than Vcc. And this is why we have a larger voltage swing at the output. And now +Vout_max is now equal (Vcc - Q1Vce(sat) ) ≈ 14.8V.

But we have another benefit from the bootstrap capacitor.
Notice here that now the voltage across Ry is almost constant and equal to Vc1 - Vbe6 ≈ 7.15V. And this means that Ry acts just like a constant current source. And this is why C1 also increase Q4 stage (common emitter stage) voltage gain.

You mean so there's about 2 diode drop between Q3 and Q5? therefore less current through R8 and R7

if so, might as well delete D3 and use 40 ohm resistor?
The VBE multiplier (Rubber diode) is better in this case.

https://en.wikipedia.org/wiki/Rubber_diode
https://leachlegacy.ece.gatech.edu/lowtim/2ndstage.html

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#### Zeeus

Joined Apr 17, 2019
608
And as now designed, the amp has a large gain peak up near 6 MHz.

#### Zeeus

Joined Apr 17, 2019
608
What you must understand is that in this class AB power amplifier the power dissipation in the output transistors is equal to arond P = 0.1*(Vcc^2/RL) = 0.1*(15V^2)/8Ω = 2.8W and the peak of instantaneous power is around P_peak = (Vcc^2)/(4*RL) ≈ 7W for a pure resistive load.
Hence you must choose the output stage transistor is such a way that they can handle such a power dissipation (a heatsink is needed) plus the quiescent current power dissipation.
Yes this was my question earlier : Although not sure why you uses Vcc in your calculation...Don't answer, will ask again when i post thread(hope to understand before then)

Take a look at this circuit:

As you can see without any input signal (DC condition) the bootstrap capacitor (C1) is charged to 7.85V. Also notice that the circuit time constant is very long (t = Rx||Ry * C1 = 0.544s) compared to the audio signal period (1/20Hz = 0.05s). Hence from this, we can tell that the capacitor will act just like a 7.85V DC voltage source (the input signal is changing way too fast to be able to charge/discharge the C1 capacitor).

Now let us assume that voltage at the output is at a positive peak and equal to +10V.

The situation is shown here:

As you can see the voltage at Vx node can be higher than Vcc. And this is why we have a larger voltage swing at the output. And now +Vout_max is now equal (Vcc - Q1Vce(sat) ) ≈ 14.8V.

But we have another benefit from the bootstrap capacitor.
Notice here that now the voltage across Ry is almost constant and equal to Vc1 - Vbe6 ≈ 7.15V. And this means that Ry acts just like a constant current source. And this is why C1 also increase Q4 stage (common emitter stage) voltage gain.

The VBE multiplier (Rubber diode) is better in this case.

https://en.wikipedia.org/wiki/Rubber_diode
https://leachlegacy.ece.gatech.edu/lowtim/2ndstage.html
Thanks professor Jony....There's something else which is not Vbe multiplier, will try that to see..