# calculating output wattage of class AB amp

Joined Feb 21, 2020
239

I took the class AB amplifier circuit from: https://www.electronics-tutorials.ws/amplifier/class-ab-amplifier.html

Anyways, the website states that the power output (to the 16 ohm load) is about 1 watt.

How is this calculated? I tried ohms law by doing 12V divided by 1K (r3) times 12 and I don't even get 1/2 that value.

Also, if I add emitter resistors in series on tip31 and tip32 transistors (like how R4 is connected to the 1st transistor, then how would those values determine the wattage output?

#### Ian0

Joined Aug 7, 2020
9,796
It's simply V^2/R where R is the load and V is the RMS output voltage

OR V^2/(2R) where R is the load and V is the peak output voltage.

The absolute maximum that V could be is 6V, because the output to the speaker must be less than ±6V from a 12V supply.
BUt the output stage is an emitter follower than will drop 0.6V, so no we're at 5.4V
At the most negative point, the voltage across TR3 will be at its minimum, say 0.1V, and the voltage across D1+D2 will be about 1.2V, so the current through R3+R4 will be 10.5mA (12-1.4-0.1)/1.1k. That will result in a voltage drop across R4 of 9.5mA*100Ω=0.95V.
So now we're left with 6V minus 0.6V for the output transistor, minus 0.1V for TR3, minus 0.95V for R4 =4.35V
Output power = 4.25V^2/32 = 0.59W.

#### crutschow

Joined Mar 14, 2008
34,405
Output power = 4.25V^2/32 = 0.59W.
Where does the 32 come from?
The TS stated it was a 16Ω load.

#### Ian0

Joined Aug 7, 2020
9,796
Where does the 32 come from?
The TS stated it was a 16Ω load.
P = Vpeak^2/(2R)

Joined Feb 21, 2020
239
So I'm curious. With a 12V supply (assuming the transistors can handle anything which they won't), and all resistors set to 0 ohms, the absolute maximum wattage output is 2.25W?

Because then that makes the equation: 6V^2/16 ?

#### Ian0

Joined Aug 7, 2020
9,796
So I'm curious. With a 12V supply (assuming the transistors can handle anything which they won't), and all resistors set to 0 ohms, the absolute maximum wattage output is 2.25W?

Because then that makes the equation: 6V^2/16 ?
Almost.
Rephrasing slightly, if you had lossless components, then the output would be 1.125W, because it's Vpeak^2/(2R), not Vpeak^2/R.

#### Papabravo

Joined Feb 24, 2006
21,225
Almost.
Rephrasing slightly, if you had lossless components, then the output would be 1.125W, because it's Vpeak^2/(2R), not Vpeak^2/R.
The estimate is not exact, but it is pretty darn close as the following simulation demonstrates. This is a fun fact of which I was not aware.
Thanks @Ian0

#### Ian0

Joined Aug 7, 2020
9,796
I'm also assuming that it is biassed so that the output before the capacitor is exactly half-supply, and that the impedance of the output coupling capacitor is negligible at the frequency at which the output is being measured, and even harmonic distortion is also negligible.
For a single supply amplifier with a capacitor coupled output maximum theoretical output is Vsupply^2/(8R)

Joined Feb 21, 2020
239
Almost.
Rephrasing slightly, if you had lossless components, then the output would be 1.125W, because it's Vpeak^2/(2R), not Vpeak^2/R.
I guess I forgot to write... I'm dealing with 8 ohms for R because my speaker I'm actually using is an 8 ohm speaker.

#### Ian0

Joined Aug 7, 2020
9,796
I guess I forgot to write... I'm dealing with 8 ohms for R because my speaker I'm actually using is an 8 ohm speaker.
Then, yes, 2.25W - right answer by wrong method!

#### MrAl

Joined Jun 17, 2014
11,453
View attachment 273786

I took the class AB amplifier circuit from: https://www.electronics-tutorials.ws/amplifier/class-ab-amplifier.html

Anyways, the website states that the power output (to the 16 ohm load) is about 1 watt.

How is this calculated? I tried ohms law by doing 12V divided by 1K (r3) times 12 and I don't even get 1/2 that value.

Also, if I add emitter resistors in series on tip31 and tip32 transistors (like how R4 is connected to the 1st transistor, then how would those values determine the wattage output?

Hello there,

It is tempting to use half the supply voltage (12/2=6v) to calculate the average output power, but that would be misleading because the transistors will always drop some voltage.

If we estimate the voltage drop in the two transistors to be 2v each during normal operation, then the peak output voltage (assuming the cap is sized ok for the frequency) will be only 4 volts peak. Since P=V^2/R the estimated peak output power is 4^2/16 which of course is 1 watt peak.
That is a peak output power however not average.
The average would be lower, and the RMS power is of interest with these amplifiers.
Now the output with 4v peak is 4*sin(w*t) and since 4 is the peak the RMS of the sine is 4/sqrt(2) which is sqrt(8) and since the power is Vrms^2/R the power out is 8/16 which is 1/2 watt RMS.

So the estimate is 1 watt peak and 1/2 watt RMS.
Depending on the transistors used it could be a little less than that or a little more than that because they determine the actual peak output combined with the power supply voltage.

What else is interesting is if you calculate the losses in the transistors. It's quite high with these tarnsistors stages.

Last edited:

#### Audioguru again

Joined Oct 21, 2019
6,688
I saw a 16 ohms speaker 50 years ago. Today car and portable speakers are 4 ohms and the amplifier is bridged (a separate amplifier driving each speaker wire) which doubles the voltage swing. 14W with low distortion.

The output power of your amplifier is so low (0.44W) that I simulated it with little transistors since power transistors are not needed.

All audio amplifiers have global (from output back to the input) negative feedback but yours has none to reduce distortion.

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#### Audioguru again

Joined Oct 21, 2019
6,688
I tweaked the amplifier so that its output swing is much more than before. 0.86W output at low distortion.

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Joined Feb 21, 2020
239
I'm still trying to find the full official calculation for the output wattage.

#### Ian0

Joined Aug 7, 2020
9,796
I'm still trying to find the full official calculation for the output wattage.
And my answer in Post #2 is not comprehensive enough?

Joined Feb 21, 2020
239
And my answer in Post #2 is not comprehensive enough?
Papabravo doesn't think its 100% accurate.

But I will try to follow post #2 equation to make sense out of it based on RL being 8 ohms (for 8 ohm speaker), and assuming I use a 9V battery to power the whole thing...

It's simply V^2/R where R is the load and V is the RMS output voltage
OR V^2/(2R) where R is the load and V is the peak output voltage.
ok so peak voltage = (9 squared)/(2 * 8) = 5.06W? That doesn't make sense...

Anyways, carrying on...

The absolute maximum that V could be is 6V, because the output to the speaker must be less than ±6V from a 12V supply.
Ok, so in my case, I work with 4.5 since that's half of 9V battery supply.

BUt the output stage is an emitter follower than will drop 0.6V, so now we're at 5.4V
3.9V for me (4.5V - 0.6V). Are you referring to the top transistor as a common-collector amplifier when you say emitter-follower? and is the 0.6V the assumed value of transistor Vbe value?

At the most negative point, the voltage across TR3 will be at its minimum, say 0.1V
Why not 0V? How is this 0.1V calculated?
I don't know if the left-most NPN base voltage has something to do with your math but when I calculate that base voltage with audio input disconnected, I would get 1.17V (15K/100K times 9V battery). Then subtract Vbe (I'll assume 0.6V) so I get 0.57V

and the voltage across D1+D2 will be about 1.2V,
Are we basing that on the assumption that each diode requires 0.6V to operate (anode >= 0.6V higher compared to cathode)?

so the current through R3+R4 will be 10.5mA (12-1.4-0.1)/1.1k.
I'm confused as to when 1/2 the voltage comes into play. and I'm not sure where you got 1.4 from. I take it that 1.1K is 1K + 100 ohms.

In my scenario, my voltages then are 9 (for battery) minus the diodes (=1.2) minus some low voltage after the transistor (do I use my calculation from the base resistors here?) That's 0.57. So 9-1.2-0.57/1.1K = 6.5mA

That will result in a voltage drop across R4 of 9.5mA*100Ω=0.95V.
Ok so my voltage drop then is 6.5mA * 100=0.65V

So now we're left with 6V minus 0.6V for the output transistor,...
So when do you use 6V vs the 12V here?
...minus 0.1V for TR3, minus 0.95V for R4 =4.35V
Ok so for me its 4.5V minus 0.57V for TR3 minus 0.65V for R4 = 3.28V

Output power = 4.25V^2/32 = 0.59W.
3.28V squared / (8 * 2) = 0.67W.

I must have messed something up.....

#### MrChips

Joined Oct 2, 2009
30,792
4V rms into 16Ω load = 1W

#### Ian0

Joined Aug 7, 2020
9,796
Papabravo doesn't think its 100% accurate.

But I will try to follow post #2 equation to make sense out of it based on RL being 8 ohms (for 8 ohm speaker), and assuming I use a 9V battery to power the whole thing...

ok so peak voltage = (9 squared)/(2 * 8) = 5.06W? That doesn't make sense...

Anyways, carrying on...

Ok, so in my case, I work with 4.5 since that's half of 9V battery supply.

3.9V for me (4.5V - 0.6V). Are you referring to the top transistor as a common-collector amplifier when you say emitter-follower? and is the 0.6V the assumed value of transistor Vbe value?

Why not 0V? How is this 0.1V calculated?
I don't know if the left-most NPN base voltage has something to do with your math but when I calculate that base voltage with audio input disconnected, I would get 1.17V (15K/100K times 9V battery). Then subtract Vbe (I'll assume 0.6V) so I get 0.57V

Are we basing that on the assumption that each diode requires 0.6V to operate (anode >= 0.6V higher compared to cathode)?

I'm confused as to when 1/2 the voltage comes into play. and I'm not sure where you got 1.4 from. I take it that 1.1K is 1K + 100 ohms.

In my scenario, my voltages then are 9 (for battery) minus the diodes (=1.2) minus some low voltage after the transistor (do I use my calculation from the base resistors here?) That's 0.57. So 9-1.2-0.57/1.1K = 6.5mA

Ok so my voltage drop then is 6.5mA * 100=0.65V

So when do you use 6V vs the 12V here?

Ok so for me its 4.5V minus 0.57V for TR3 minus 0.65V for R4 = 3.28V

3.28V squared / (8 * 2) = 0.67W.

I must have messed something up.....
You'll not get 100% accurate - the limits of error are too large.
For instance, diode voltage drop = 0.65V±0.1V not only does it vary from type to type, from batch to batch and depend on the current, it also varies with temperature.
Same applies to voltage drop across an emitter follower. For a transistor with low Hfe working at high current it could be as much as 1V, but the generic 0.6V for both diode and emitter follower voltage drops is a good start.
Then there's your battery. It says 9V on it, but it's 10V when it's new and still working at 8V.
Voltage drop across TR3: it says 0.4V for Vce(sat) in the datasheet, but that's at 150mA. No figures for 10mA, but you can be certain that it's NOT zero. Some transistors might manage 10mV, but most don't, unless they are designed for switching, and you need a transistor designed for amplification.

6V is the hypothetical maximum peak output from a 12V supply, because the hypothetical maximum peak-to-peak output is 12V.

@Papabravo 's simulation is not "exact" either, it uses the values for voltage drops contained is SPICE. It's only as accurate as the SPICE models - and they are based on the typical values in the datasheets.
I'm sure he could make SPICE give you minimum and maximum values, calculate the standard deviation and the limits of error if he wanted to; but if the answer is accurate within 1dB (±12%) then audibly no-one will be able to tell. Anyway, it assumes that we all listen to 1kHz sinewaves, not music.

#### MrAl

Joined Jun 17, 2014
11,453
4V rms into 16Ω load = 1W

Hello,

You need to review posts numbers 7 and 11. The PEAK output power is about 1 watt as you can glean from both of those posts. One is a simulation, the other is a straight up calculation.

Both of those posts suggest the output power is much less than 1 watt RMS and that is because the non ideal transistors both drop a significant voltage. So instead of 6v peak we end up with around 4v peak output.
In the ideal case we could see slightly over 1 watt but that's not going to happen with 12v and 16 Ohms output and this kind of output stage.
It's also interesting that if the output is not set to full possible output the efficiency drops and that is because the transistors are forced to drop even more voltage.

#### MrAl

Joined Jun 17, 2014
11,453
I saw a 16 ohms speaker 50 years ago. Today car and portable speakers are 4 ohms and the amplifier is bridged (a separate amplifier driving each speaker wire) which doubles the voltage swing. 14W with low distortion.

The output power of your amplifier is so low (0.44W) that I simulated it with little transistors since power transistors are not needed.

All audio amplifiers have global (from output back to the input) negative feedback but yours has none to reduce distortion.
Hi,

Today 16 Ohms and 32 Ohms are common in wired headphones.
8 and 4 Ohms are for larger speakers usually but i think some headphones have 8 Ohm speakers.