Calculate class A power

[Sebastiaan]

Dear All,

How to calculate the exact power delivered in class A in a power amplifier?

Let's say I have one pair of transistors and each of them have a Re of 0,1 Ohm.

I have a voltage drop of 70mV across each Re so I can calculate the current is 0,070/0,1=0,7A

So I have a current draw of 700mA from each transistor. Makes total 1,4A for the pair.

Now I want to know the class A working array into an 8 Ohm load.

The voltage will be

1,4x8= 11,2V

11,2X1,4= 15,60 watt in class A.

Is this correct?

I feel I miss something here... please help me out.

With kind regards,
Bas

[Sebastiaan]

Dear,

In a stereophile review John Atkinson comment on the measurement of the Parasound JC1

"Each of the Halo's 18 flat-pack output devices stands on a 0.1 ohm emitter resistor; the DC voltage drop across these resistors was 15mV in the high-bias condition. As the output is taken from the common point between each pair of complementary transistors, this voltage is equivalent to a standing output-stage bias current of 1.35A, which in turn implies that, into 8 ohms, the JC 1 operates in class-A up to 29W"

I just miss the boat there.

How he get the 29 watts of class A power if the claims the current is 1,35A?

P= 1,35x135x8=14,58watt. That is the halve of 29 watt.

Where I get wrong?

With best regards,
Bas

[Sebastiaan]

No one... no wizards here to help me out.... ?

[roender]

Originally posted by Sebastiaan
Dear All,

Let's say I have one pair of transistors and each of them have a Re of 0,1 Ohm.

I have a voltage drop of 70mV across each Re so I can calculate the current is 0,070/0,1=0,7A

So I have a current draw of 700mA from each transistor. Makes total 1,4A for the pair.

Now I want to know the class A working array into an 8 Ohm load.

The voltage will be

1,4x8= 11,2V

11,2X1,4= 15,60 watt in class A.

Is this correct?

I feel I miss something here... please help me out.

With kind regards,
Bas

No.
The RMS power is:
P rms = 11,2/sqrt(2) * 1,4/sqrt(2) = 15,6/2 = 7,8 Wrms
or
P rms = 8 * [1,4/sqrt(2)]^2 = 7,8 Wrms

[roender]

"1.35A, which in turn implies that, into 8 ohms, the JC 1 operates in class-A up to 29W"

1,35 x 2 = 2,7A (push pull output stage)

P rms = 8 * [2,7 / sqrt(2)] ^ 2 = 29W rms

[traderbam]

Originally posted by Sebastiaan
Dear All,

How to calculate the exact power delivered in class A in a power amplifier?

Let's say I have one pair of transistors and each of them have a Re of 0,1 Ohm.

I have a voltage drop of 70mV across each Re so I can calculate the current is 0,070/0,1=0,7A

So I have a current draw of 700mA from each transistor. Makes total 1,4A for the pair.

Now I want to know the class A working array into an 8 Ohm load.

The voltage will be

1,4x8= 11,2V

11,2X1,4= 15,60 watt in class A.

Is this correct?

I feel I miss something here... please help me out.

With kind regards,
Bas

If your bias current is 700mA then the max current swing in class A will be +/- 1.4A. This is peak current. So peak power will be I^2 x R = 1.96 x 8 = 15.68W.

In audio, the peak power is not normally used. Instead, average power is used which is half the peak power for a sine wave and a resistive load. So when a product claims to be 100W into 8 ohms they mean 100W average sinewave power into an 8 ohm resistor.

So your average power = 7.84W.

Be aware that average power is often, incorrectly, referred to as RMS power. This is a term commonly used in audio parlance but is not an engineering term (mathematically, RMS power is not the same thing as average power). The basis is that the average power can be calculated by multiplying the RMS voltage by the RMS current (both being root 0.5 of their peak values - hence 0.5 of peak when multiplied together). So the prefix RMS seems to have slipped in front of power too.

[Williams Audio]

Supposing the Iq current is 1.4 amp dc and is a push pull topology the current will be exactly equal for the N and P out devices if the DC Offset is zero volts.
What can be withdrawn out as call class a is this current as RMS versus an effective load
Po= (1.4*0.7)²*load, in case of 8 ohms is 7.68 watts as mention in previous answer.
Is that simple
Best regards
Williams

[roender]

Originally posted by traderbam


Be aware that average power is often, incorrectly, referred to as RMS power. This is a term commonly used in audio parlance but is not an engineering term (mathematically, RMS power is not the same thing as average power).

I know you're wright ... but this is a well know term in audio, and if is used enough time by many people it will become "de facto", even if is incorrect from mathematical POV

[Sebastiaan]

Originally posted by roender


No.
The RMS power is:
P rms = 11,2/sqrt(2) * 1,4/sqrt(2) = 15,6/2 = 7,8 Wrms
or
P rms = 8 * [1,4/sqrt(2)]^2 = 7,8 Wrms

Thank you Sir,

There I went wrong. I told you already I feel I miss the boat somewhere. Really stupid I never considered the RMS value. Thanks a lot now I am one illusion lessen

With best regards,
Bas

[Sebastiaan]

Thank you all!

It is clear to me.

A little offtopic, but this calculation makes me realize that for example the Musical Fidelity A1 wasn't a real class A amplifier

All the best,
Bas