Calculate class A power - diyAudio
 Calculate class A power
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 29th March 2009, 07:51 PM #1 diyAudio Member   Join Date: Jun 2004 Location: the Netherlands Calculate class A power 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
 29th March 2009, 07:59 PM #2 diyAudio Member   Join Date: Jun 2004 Location: the Netherlands 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
 30th March 2009, 09:05 AM #3 diyAudio Member   Join Date: Jun 2004 Location: the Netherlands No one... no wizards here to help me out.... ?
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Join Date: Sep 2006
Re: Calculate class A power

Quote:
 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

 30th March 2009, 10:50 AM #5 diyAudio Member     Join Date: Sep 2006 "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
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Join Date: Jan 2002
Location: Earth
Re: Calculate class A power

Quote:
 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.

 30th March 2009, 11:16 AM #7 diyAudio Member   Join Date: Apr 2004 Location: far away 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 __________________ williams audio
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Join Date: Sep 2006
Re: Re: Calculate class A power

Quote:
 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

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Join Date: Jun 2004
Location: the Netherlands
Re: Re: Calculate class A power

Quote:
 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

 30th March 2009, 11:37 AM #10 diyAudio Member   Join Date: Jun 2004 Location: the Netherlands Re: Re: Re: Calculate class A power 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

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