# Class AB Amp - Output Power & Transistor Power Dissipation

#### aac044210

Joined Nov 19, 2019
127
I have simulated a class ab amp.

Vout = 5.14 volts(peak)
IRL = 642.6 mA

I have made the following calculations:
Pout = Vout(peak) * IRL = 5.14 * 642.6 = 3.30 Watts

Pout = Vout(peak)^2 / 8Ω = 5.14^2 / 8Ω = 3.30 Watts

Pout = Vout(pp)^2 / 4 * 8Ω = (5.14 * 2)^2 / (4 * 8Ω) = 3.30 Watts

PDQ = Vout(peak)^2 / (20 * RL) = 5.14^2 / (10 * 8Ω) = 330 mW
PDQ = Vout(pp)^2 / (40 * RL) = (5.14 * 2)^2 / (40 * 8Ω) = 330 mW

Can someone please tell me if these are correct because I have
found conflicting info on the net.

thanks
aac

#### Audioguru again

Joined Oct 21, 2019
872
Peak watts are used only by salesmen who lie about the output power. Your output power is only 1.65W RMS into 8 ohms.
A low power class-AB amplifier like yours needs a quiescent output current of only about 10mA to 20mA.

An LM2896 Stereo amplifier IC has two class-AB amplifiers. With a 12V supply, the quiescent current of the entire IC is only 25mA.

#### Jony130

Joined Feb 17, 2009
5,145
Vout = 5.14 volts(peak)
The output power is P = (Vpeak)^2/(2RL) = (5.14V)^2/16Ω = 1.65W into 8Ω load.

#### aac044210

Joined Nov 19, 2019
127
The output power is P = (Vpeak)^2/(2RL) = (5.14V)^2/16Ω = 1.65W into 8Ω load.
Thanks
Is that rms?

#### Audioguru again

Joined Oct 21, 2019
872
RMS power is real power. Peak power is fake and is simply double the real power.
Many amplifiers have a higher-than-real power number because they are rated when clipping like crazy with 10% distortion.
The 10% distortion creates 20% more power.

#### Jony130

Joined Feb 17, 2009
5,145
Thanks
Is that rms?
RMS ?? No, for sure not.

Term "RMS" refers only to the effective value of an AC voltage or current. So we have an "RMS voltage" and "RMS current".
And we should not use the term "RMS" to describe the real (true) power becouse for example the "RMS voltage" means it is the root of the mean of the square of the voltage. And the "true power" is the average value of instantaneous power. And for this, we don't need to use the root of the mean of the square of the power. So, the "RMS power" is an inaccurate term, and you should never use it.

#### aac044210

Joined Nov 19, 2019
127
RMS ?? No, for sure not.

Term "RMS" refers only to the effective value of an AC voltage or current. So we have an "RMS voltage" and "RMS current".
And we should not use the term "RMS" to describe the real (true) power becouse for example the "RMS voltage" means it is the root of the mean of the square of the voltage. And the "true power" is the average value of instantaneous power. And for this, we don't need to use the root of the mean of the square of the power. So, the "RMS power" is an inaccurate term, and you should never use it.
I guess my last question on this topic is:

Is this the correct formula for transistor power dissipation?

PDQ = Vpk^2 / 10 * RL

Last edited:

#### Audioguru again

Joined Oct 21, 2019
872
When I say RMS power then I mean an undistorted sinewave signal with RMS voltage and RMS current.
Transistor power dissipation is the amount of heat it is producing and is calculated as voltage across it times the current through it. The voltage and current can be AC, DC or both.

You always use the peak of a waveform in your calculations. Power is usually measured using a sinewave and the real power is half the peak power, which is why you divide by 2.

• redrooster01

#### aac044210

Joined Nov 19, 2019
127
When I say RMS power then I mean an undistorted sinewave signal with RMS voltage and RMS current.
Transistor power dissipation is the amount of heat it is producing and is calculated as voltage across it times the current through it. The voltage and current can be AC, DC or both.

You always use the peak of a waveform in your calculations. Power is usually measured using a sinewave and the real power is half the peak power, which is why you divide by 2.
thanks

#### aac044210

Joined Nov 19, 2019
127
I guess my last question on this topic is:

Is this the correct formula for transistor power dissipation?

PDQ = Vpk^2 / 10 * RL
thanks

#### Audioguru again

Joined Oct 21, 2019
872
You wrote, "PDQ = Vpk^2 / 10 * RL" but I do not know what is the meaning of "^". Does it mean "squared"?
Usually "/"means divide and "*" means multiplication.

At maximum output power before clipping, most class-AB amplifiers have an efficiency of 45% t0 60%.
Then for an output power of 100W the total heating of both output transistors is 80W or 40W each.

A class-AB amplifier producing 100W into 8 ohms before clipping has an output of 28.3V RMS with a peak of 40V. Then the amplifier power supply is probably 88V. If the quiescent current is 20W as per your calculation then its idle current is 20W/88V= 227mA which is way too high.
If the idle current is 30mA then its quiescent power is 88V x 30mA= 2.64W for the total.

#### aac044210

Joined Nov 19, 2019
127
You wrote, "PDQ = Vpk^2 / 10 * RL" but I do not know what is the meaning of "^". Does it mean "squared"?
Usually "/"means divide and "*" means multiplication.

At maximum output power before clipping, most class-AB amplifiers have an efficiency of 45% t0 60%.
Then for an output power of 100W the total heating of both output transistors is 80W or 40W each.

A class-AB amplifier producing 100W into 8 ohms before clipping has an output of 28.3V RMS with a peak of 40V. Then the amplifier power supply is probably 88V. If the quiescent current is 20W as per your calculation then its idle current is 20W/88V= 227mA which is way too high.
If the idle current is 30mA then its quiescent power is 88V x 30mA= 2.64W for the total.
hi

^ means "to the power of". eg. 1 x 10^3 means 1 x 10 to the power of 3 or 1,000. It is used in computer
coding and excel.

aac

#### Audioguru again

Joined Oct 21, 2019
872
Of course "something^2" is "something to the power of 2" which is "something squared".

#### Hymie

Joined Mar 30, 2018
760
Many, many years ago as part of my college qualification course we were to determine the maximum theoretical efficiency of class A and class A/B amplifier output stages.

Although I cannot recall the actual maths involved, I do recall the answers were 25% for class A and 75% for class A/B.
So a class A amplifier with a 100W rms output will have a power dissipation in the output stage of at least 300W.

#### MrBluster

Joined Dec 10, 2008
1
Although I cannot recall the actual maths involved, I do recall the answers were 25% for class A and 75% for class A/B.
So a class A amplifier with a 100W rms output will have a power dissipation in the output stage of at least 300W.
It's a pity you can't remember the maths, because if you did you would realise your numbers are wrong.

Remember, all these calculations apply only when certain conditions are met:

1. The signal is not heavily clipped.
2. You are dealing with sine waves.
3. The load is purely resistive.
4. The circuit topology is push-pull.

The actual numbers? Class A efficiency varies linearly between 50% at full power (the precise point of clipping) to 0% at idle. Therefore, a class A amplifier handling music waveforms, will dissipate just under twice its maximum power. This is just the amplifier. If you have a regulated power supply, too, this will have significant dissipation, because the current drain of a class A amplifier is constant, by definition. This latter is the only real benefit class A has.

The behaviour of class B is much more complicated. The maximum (theoretical) efficiency of a class B output stage is actually 78.54% (π/4) at the point of clipping. Therefore, under these conditions the dissipation is about 27% of the maximum output power, or 21% of the total power consumption. Interestingly, the maximum dissipation occurs at 42% of maximum output power.

Many designers also fail to appreciate how much worse the dissipation situation gets when the load is not completely resistive. Clearly, with a completely reactive load, no power is actually delivered to the load even though the power consumption is the same as the resistive case, so it is all dissipated in the output transistors! Real-world loudspeakers are somewhere in between.

#### Audioguru again

Joined Oct 21, 2019
872
Here is what one expert says about heating your home with an audio amplifier:

#### shjacks

Joined Apr 25, 2018
9
Context? Most AB amps are for lower distortion audio amplifiers, typically music or voice. For music general rule is peak output is 20x average output since uncompressed music has a low duty cycle. (Requirement met by high end audiophile equip., note higher voltage power needed for peaks) Voice is frequency limited and even more "dead time" (used to pack more calls per frequency band.) Whereas for a class B amplifier one transistor is non-conducting (no dissipation) for half of each cycle, class AB each transistor continues to conduct as the other transistor begins to conduct. For class A, transistors are always conducting, so power dissipation is not affected greatly by signal output. For sine wave AC 110 Volt RMS power mains the peak voltage is ~150 V.