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How to calculate the efficiency of an amplifier?

by Pagedown
Tags: amplifier, efficiency
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Pagedown
#1
Jan21-11, 01:27 AM
P: 92
How to calculate the efficiency of an amplifier e.g. op-amp? Been thinking about that, and have no answer. It should not be Vout / Vin ?
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vk6kro
#2
Jan21-11, 03:56 AM
Sci Advisor
P: 4,044
Efficiency always refers to the ratio of (power out / power in).

For an amplifier, this means the DC power in, not the AC drive signal. It may be given as a number from 0 to 1, or a percentage from 0 to 100.

Normally, this is not considered with opamps, but it is important with audio amplifiers intended to drive a speaker or radio frequency amplifiers intended to provide a signal to an antenna.
Pagedown
#3
Jan21-11, 07:17 AM
P: 92
So shouldn't be the audio amplifity efficiency more than 1 since its driving power?(generating greater signals)

I know its not, but just dunno how to prove in mathematically. So usually current output is very low although voltage is high? Something like transformer?

vk6kro
#4
Jan21-11, 08:05 AM
Sci Advisor
P: 4,044
How to calculate the efficiency of an amplifier?

No.

The power out is power delivered to a load. For an amplifier, this would normally be as an AC waveform.
If the load was a resistor, you would calculate this power from Power = (RMS voltage squared) / R (of load)

The power in is the power taken from a battery or a power supply. This is DC power supplied to the amplifier.

In some amplifiers it is possible for the input drive power to appear as part of the output. This is normally the case with vacuum tube grounded grid amplifiers. These are not at all obsolete.

This may give the impression that the amplifier is more efficient than it actually is. This is not free power, though, and it is obtained originally from a DC supply in a previous amplifier.

Do not confuse efficiency with voltage gain. They are very different .
skeptic2
#5
Jan21-11, 08:51 AM
P: 1,822
Quote Quote by Pagedown View Post
So shouldn't be the audio amplifity efficiency more than 1 since its driving power?(generating greater signals)

I know its not, but just dunno how to prove in mathematically. So usually current output is very low although voltage is high? Something like transformer?
What vk6kro says is correct, what I think you want is called gain. Voltage gain is, as you suggest, calculated from Vo/Vi.
Pagedown
#6
Jan21-11, 10:52 AM
P: 92
okay, thus power in does not depend on the input voltage/characteristics.

For an example, a non-inverting op-amp. It has +5v and -5v input power supplies. Thus is 10Vx Iin(input current) in power. Say 100mA max current.Power out should be 1W.

If 1V input is given to the inputs of the opamp, with voltage gain configuration of 100, thus output voltage is 100. Say load is 100 ohm. Thus power out should be 100W.

I know this is wrong, but i duno how to calculate in the right way.
Thanks for the guidance and time you all make for me. =)

P.S - must we use root mean square always when calculating power? It means average power right? Why is it always root mean square, and sometimes square root of 3 for 3 phase.. that is another question popping on my head..
yungman
#7
Jan21-11, 11:20 AM
P: 3,904
You have to specify what kind of signal, is it peak to peak?

People specify using sine wave. If you are talking about 1V peak to peak, then the peak input voltage is only 0.5V and for gain of 100 the average output power is only W = 100 X 0.707 X0.5V X I. Power usually means average power or RMS ( root means square ). Peak power should be 100 X 0.5V X I. Or something like that!!!

Double check my work, I think you really should do some reading on amplifier power definitions and voltage gain.
Pagedown
#8
Jan21-11, 11:22 AM
P: 92
Okay, I have reading on those definitions, but no clue of calculating the efficiency of the opamp yet.

Simply, all I need is just an example calculation of the efficiency of an amplifier circuit.
berkeman
#9
Jan21-11, 12:58 PM
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P: 41,303
Quote Quote by Pagedown View Post
Okay, I have reading on those definitions, but no clue of calculating the efficiency of the opamp yet.

Simply, all I need is just an example calculation of the efficiency of an amplifier circuit.
Why? What is the point of your question? What amplifier are you working with? What is its power source, its signal source, and what is the load it is driving?

The efficiency will be the output power divided by the input power. For most linear audio amplifiers, that will be about 50% or less.
Averagesupernova
#10
Jan21-11, 01:34 PM
P: 2,536
Put quite simply, efficiency of an amplifier is power out/power in * 100%. The power in is whatever power (in watts) the power supply, battery, whatever your source is, is delivering to the amplifier. The power out is whatever power in watts the amplifier is delivering to the load. Power is usually going to be average. It is often referred to as RMS, however that is incorrect. Voltage and current are typically spec'd in RMS, but power sholuld not be.
vk6kro
#11
Jan21-11, 07:14 PM
Sci Advisor
P: 4,044
Simply, all I need is just an example calculation of the efficiency of an amplifier circuit.

Suppose you have an 8 ohm speaker and there is 43.8 volts peak to peak across it.

Divide by 2.82 (2 * sqrt{2}) to give 15.53 volts RMS.

Power out = E^2 / R = 15.53 volts *15.53 volts / 8 ohms = 30.1 watts

The amplifier has a DC supply of 60 volts and it is drawing an average current of 1.5 amps DC.

Power in = 60 volts * 1.5 amps or 90 watts.

Efficiency = (30.1 watts / 90 watts) *100 = 33.44 %
vk6kro
#12
Jan21-11, 08:57 PM
Sci Advisor
P: 4,044
Quote Quote by Pagedown View Post
For an example, a non-inverting op-amp. It has +5v and -5v input power supplies. Thus is 10Vx Iin(input current) in power. Say 100mA max current.Power out should be 1W.

If 1V input is given to the inputs of the opamp, with voltage gain configuration of 100, thus output voltage is 100. Say load is 100 ohm. Thus power out should be 100W.

I know this is wrong, but i duno how to calculate in the right way.
Thanks for the guidance and time you all make for me. =)

P.S - must we use root mean square always when calculating power? It means average power right? Why is it always root mean square, and sometimes square root of 3 for 3 phase.. that is another question popping on my head..
Incidentally, an opamp with a plus 5 and minus 5 volt supply cannot deliver more than plus 4 volts and minus 4 volts out. It cannot deliver more voltage out than it is supplied with and there are drops in the chip that stop it delivering even those voltages.

Some chips are better than this and may deliver 4.5 volts or even 5 volts but none can deliver more than 5 volts with a +/- 5 volt supply despite what you might calculate from the gain equation.

Root mean square voltage is not the same as average voltage.
Root mean square voltage is the DC voltage which would have an equivalent heating effect to this voltage.
It allows for the fact that power is proportional the the SQUARE of the voltage, so the top part of a sinewave voltage has a much greater heating effect than the lower parts.

Average is just what it sounds like. Take a lot of slices and average their absolute values.


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