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Old 28th July 2004, 05:46 AM   #1
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Default Thumb Rule for determining Approx output power of class AB amp

Thumb Rule for determining Approx output power of class AB amp
All we need to know is the transformer secondary voltage and the loudspeaker

impedance.
We all know that power = square of voltage divided by resistance.But the voltage

can vary due to various factors, the figure of 8 gives results which are very near to

actuals.
Normally I determine the output power of a class AB amp by substracting 8 from

one side voltage of the power transformer and multypling the square of the result

with the speaker impedance,
for example if the transformer is 40-0-40 then the power output of the amp will be

40-8=32. square of 32= 1024 ,Dividing 1024 by 8 , 4 , & 2 will give the power

output of the amp which is 128W at 8 ohms , 256W at 4 ohms , & 512W at 2

ohms similarly in bridge mode again the approximate output will be the Watts and

ohms X 2 , ie 256W at 8ohms , 512W at 4ohms , & 1024W at 2ohms.
Please note all this is approximate, the output voltage is dependant on the

transformer regulation,value of filter caps,and losses at various stages.
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Old 28th July 2004, 05:55 AM   #2
sam9 is offline sam9  United States
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Since speaker impedance can vary from way over 8 ohms to below 4 ohms within the space of a couple of octaves, the power is also dependant on the content of the input signal. Thus actual power is a very ambiguous concept.
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Old 28th July 2004, 06:07 AM   #3
Jennice is offline Jennice  Denmark
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r.l.,

Basically your way of calculating is ok. However, your result is peak power. You may want to divide the result with sqrt(2) =1,41 to get an RMS value.


Sam9 and r.l.:
I agree, power is frequency dependent, but one has to have some reference point, with a theoretical load.
Thus, I think it's quite allright to use either 4 or 8 ohms for reference in the calculations.

When going as low as 2 ohms, a LOT of current is required
The actual available peak current may be limited by the driver stage (or output impedance of your voltage amplification stage).
Also, at these low-impedance loads, one will need a really serious power supply to avoid voltage drops big enough to take notice of when doing the math on available power.

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Old 28th July 2004, 06:18 AM   #4
sajti is offline sajti  Hungary
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Default Re: Thumb Rule for determining Approx output power of class AB amp

Quote:
Originally posted by rajeev luthra
Thumb Rule for determining Approx output power of class AB amp
All we need to know is the transformer secondary voltage and the loudspeaker

impedance.
We all know that power = square of voltage divided by resistance.But the voltage

can vary due to various factors, the figure of 8 gives results which are very near to

actuals.
Normally I determine the output power of a class AB amp by substracting 8 from

one side voltage of the power transformer and multypling the square of the result

with the speaker impedance,
for example if the transformer is 40-0-40 then the power output of the amp will be

40-8=32. square of 32= 1024 ,Dividing 1024 by 8 , 4 , & 2 will give the power

output of the amp which is 128W at 8 ohms , 256W at 4 ohms , & 512W at 2

ohms similarly in bridge mode again the approximate output will be the Watts and

ohms X 2 , ie 256W at 8ohms , 512W at 4ohms , & 1024W at 2ohms.
Please note all this is approximate, the output voltage is dependant on the

transformer regulation,value of filter caps,and losses at various stages.

OK, let's explain:

If You have a transformer with 40-0-40 AC voltage, You will get +/-55V rectified for the amplifier.
The amplifier has some losses, due the remaining voltage on the driver, and output devices. This is about 5-10V depending the load, and construction.
If Your amplifier is well designed (for high output swing), the peak output voltage sholud be 50V (If the power supply is strong enough!). The rms output is 35V. 35V means 4.375A for 8ohms, and 153W output. It will be almost double for 4ohms (300W), and about 600W for 2 ohms, but in this case the power supply must
be at least 900W....

sajti
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Old 28th July 2004, 08:49 AM   #5
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Transformers are specified in RMS voltage.
peak voltage Vpeak = Vrms x sqrt(2).

Power into a resistive load R is V^2/R. This is an instantaneous value. For a sinusoidal signal, the AVERAGE power into the load will be (Vrms)^2/R or (Vpeak)^2/2R.

Note that RMS (root mean square) voltage (or current) is the equivalent dc voltage (or current) that will give the same AVERAGE heating power into a resistive load.

Example:
Say your transformer is 25-0-25 and the amp has BJT darlington outputs with a simple common-emitter VAS stage.
The Vpeak from the transofrmer will be about 35V. At maximum positive output swing, you lose 1.4V for rectifier, maybe 0.5V for the nearly-saturated VAS, and maybe 1.5V across the darlington, perhaps 0.5V across the emitter resistor. This leaves 31V peak across the speaker. If the speaker is 8-ohms resistive the average power will be 60W.
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Old 28th July 2004, 10:40 PM   #6
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Maximum power transfer therom states that max power is transfered when the load impeadence equals the output impeadence of the output stage. If you consider the Thevenin equivalent circuit, a current source in series with the output impeadence, in series with the load impeadence, max power is transfered when the two Z's are equal. This should make the calculations simple.


figuring out the output Z is another story...
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Old 28th July 2004, 10:57 PM   #7
sam9 is offline sam9  United States
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"Sam9 and r.l.:
I agree, power is frequency dependent, but one has to have some reference point, with a theoretical load.
Thus, I think it's quite allright to use either 4 or 8 ohms for reference in the calculations."

I wasn't knocking the method, just wanted to be sure it was realized that the figure is nominal. For published figures there is also the problem at what distortion level the figure is calculated. For instance the old Sereo Review metod first involved a Variac to get a standard 120VAC main power input, then cranked up the input signal until a THD figure of 1% was reached then measured the voltage drop across the dummy load. Although this may relate poorly to actual operating conditions, it may have had the virtue of a "level playing field".

A slight nuance the the original nominal calculation: the value of the RE resistors (and RC resistors in some configurations) factors in a little. Basically the more resistance the lower the power - about 2% per 0.1 -ohm of RE+RC resistance, I think. Fortunately, as a practicle matter, very few people will notice the difference between a 75W amp and a 100Wer.
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Old 29th July 2004, 12:31 AM   #8
sam9 is offline sam9  United States
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This is the best on-line explanation of this calculation I'm aware of.

http://www.signaltransfer.freeuk.com/powerout.htm
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Old 3rd August 2004, 06:12 PM   #9
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Default Maximum power transfer

Forget maximum power transfer. People who harp on that don't really understand audio. Maximum power transfer is not the issue. While the person who says that maximum power transfer occores when the impedances are equal is correct, remember that this is a red herring whan trying to determine power output.

an audio power amp will have an output impedance of under 0.1 ohms but we are not going to laod it to that level.

actually the person who did the original posting has come up with a good way of estimating the power output of an amplifier. By using the RMS AC voltage of one side of the transformer secondary, they have eliminated the effect of the filter capacitors charging to the peak voltage of the supply (theoretically root 2) and then having to divide by 2 root 2 of the end to end supply voltage. What he has done is eliminate the converstion back and forth. Then the subtracting of 8V is a decent rule of thumb for a rectifier and transistor losses. A little much for lower powered amps but effective if the transfprmer is say 30-0-30V or higher.

However, for single ended amps with a capacitor couples output you have to divide the voltage by two if there is no center tap.

also, this does not work for class D ampliers.

As for those who throw in the maximum power transfer stuff, stick with your RF stuff. In audio, we use massively unequal impedances as we want maximum voltage transfer. even with a power amp. We just want that maximum voltage transfer with a lot of current behind it. But equal impeadnce, never.That would destroy damping factor.
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Old 3rd August 2004, 06:31 PM   #10
mjarve is offline mjarve  United States
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I found that using this calculation (as rough as it is) my harman/kardon 930 receiver, nominally rated for 45 watts per channel output in to 8 ohms, would put out 55 watts ( (29-8)21 / 8 ohms), which seems close to the mark but it is more than a 10% difference, ignoring for the moment that harman/kardon is known to "conservatively" rate their amps.

It's a nice rough estimate to be sure.
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