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Confusion calculating Amplifier efficiency

ahas1234

New Member
Hi, I designed a simple B-Amplifier.

Output power across load= 3.91W
Power delivered
by +15V DC = 6.35W
Power delivered
by -15V DC = -6.35W


How should i calculate efficiency ?

Add both input Powers in formula ? or just use one ?

i.e.
Output/ Input power = (3.91Wx100)/6.35W or = 3.91Wx100/(6.35+6.35)W
 

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rjenkinsgb

Well-Known Member
Most Helpful Member
Yes, it's total of both, if you measure each supply separately to 0V.
Your efficiency is around 31%, which is not too bad for a small class AB amp; then can never be high efficiency - often the 40% range for higher power ones.

[Digital amps - Class D etc. - can reach 90%+, but not conventional Class AB].
 

ahas1234

New Member
Yes, it's total of both, if you measure each supply separately to 0V.
Your efficiency is around 31%, which is not too bad for a small class AB amp; then can never be high efficiency - often the 40% range for higher power ones.

[Digital amps - Class D etc. - can reach 90%+, but not conventional Class AB].

Thank, i was worried.

I tried it with transformer coupled B Amplifier. for same output power 3.91W.

Surprisingly, each DC output power fell to 4.58W.

Efficiency
according to 3.91Wx100/(4.58+4.58)W : 42.7 %

It got efficient ? or Why is that so ?

:\
 

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rjenkinsgb

Well-Known Member
Most Helpful Member
As you have it in the diagram, there is a 1.2V "dead" range between one transistor turning off and the other turning on. With no bias at all, each transistor is technically in Class C, as it's conduction is less than 180'

Class B would have a clean changeover between conduction of the two transistors, but without any overlap.

The higher amplifier class you go to, the more efficient, though less linear (more waveform distortion). That does not matter in some applications, but it's pretty important for audio.

See the article here:
 

audioguru

Well-Known Member
Most Helpful Member
Most class-AB amplifier ICs show a graph on their datasheet showing the amount of heating vs output power. Then the efficiency is calculated with simple arithmatic:
 

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ahas1234

New Member
As you have it in the diagram, there is a 1.2V "dead" range between one transistor turning off and the other turning on. With no bias at all, each transistor is technically in Class C, as it's conduction is less than 180'

Class B would have a clean changeover between conduction of the two transistors, but without any overlap.

The higher amplifier class you go to, the more efficient, though less linear (more waveform distortion). That does not matter in some applications, but it's pretty important for audio.

See the article here:
I got it, really thanks.

But i'm still little unsure, adding transformer coupling to the input and output has lowered the power consumption of B-Amplifier from DC source in my simulator. According to efficiency formula, the efficiency has increased >10%.
Why is that change ?
Is this practically possible ?
 

rjenkinsgb

Well-Known Member
Most Helpful Member
adding transformer coupling
... With no bias, has changed it to a class C push-pull amp; think of it as being half way towards switch-mode; it's no longer linear or suitabel for audio as the output distortion would be terrible.

It's nothing to do with the transformers, it's the change in (or lack of) bias in that setup.
 

ahas1234

New Member
... With no bias, has changed it to a class C push-pull amp; think of it as being half way towards switch-mode; it's no longer linear or suitabel for audio as the output distortion would be terrible.

It's nothing to do with the transformers, it's the change in (or lack of) bias in that setup.
Sorry if i'm incorrect, but Biasing the Class-B would make it Class AB ? Isn't it so ?
I've to do it with simple class-B configuration. :confused:
 

audioguru

Well-Known Member
Most Helpful Member
Yes, biasing the class-B transistors will convert the amplifier into low distortion class-AB like most amplifiers use.
Why does your teacher want you to simulate a class-B very distorted amplifier when nobody makes one like that? With a normal low level input then its output is ZERO!
 

rjenkinsgb

Well-Known Member
Most Helpful Member
In Class B, one or the other transistor is on, but with no overlap, while Class AB has an overlap where both are conducting.

Try emulating the adjustable bias circuit in this design - T1 and the preset potentiometer.
That allows the voltage between the two bases to be adjusted; for class B it should be as near the point the transistors conduct as possible, without them taking any standing current.

 

rjenkinsgb

Well-Known Member
Most Helpful Member
Or this style, for the transformer coupled version - again, set it just below the point there is standing current for class B.

Note that a properly biassed class B amp has some crossover distortion, but not "very distorted" - that's what the zero bias one would be.


 

LacieSmith

New Member
Best is to measure it. For example some equipment may have class-D for bass and class-AB for treble.

Also power draw will depend on exactly what you are playing (not only on the volume) - there is something called the "crest factor" of sound (peak power to average power ratio) for most modern music it is between 9dB and 6dB, but for classical music it can be higher (20dB is very easily possible).

My advice - look at the PMPO rating and at least double it (in any case your electrical systems will have to work under peak load, not just the average load) preferably tripple it. If you know which ones are class AB and which ones are class D then again you can look at PMPO of each, and multiply the AB PMPO by 2.5 while the D's PMPO by 1.5 and add them together - this will take care of all classes of amplifiers along with their supply's losses.
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
Output power across load= 3.91W
Power delivered
by +15V DC = 6.35W
Power delivered
by -15V DC = -6.35W
I have some concerns as to how Power was calculated.
Quiesient?
Signal?

The current delivered by the rail will change, so, I think, RMS power should be considered. e.g. DC+AC and the frequency response of the RMS meter needs to be considered. i will be a function of t for each rail.

A scope with math functions should be able to do the power calculations. Power delivered is by definition negative.
 

unclejed613

Well-Known Member
Most Helpful Member
often the 40% range for higher power ones.
for class AB, efficiency can approach 50% at full rated power, but 70% rated power gives maximum heat dissipation, so is often the power level chosen for thermal testing.
 

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