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29th March 2009, 08:51 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, 08:59 PM  #2 
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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 flatpack output devices stands on a 0.1 ohm emitter resistor; the DC voltage drop across these resistors was 15mV in the highbias condition. As the output is taken from the common point between each pair of complementary transistors, this voltage is equivalent to a standing outputstage bias current of 1.35A, which in turn implies that, into 8 ohms, the JC 1 operates in classA 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, 10:05 AM  #3 
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Join Date: Jun 2004
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No one... no wizards here to help me out.... ?

30th March 2009, 11:44 AM  #4  
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Join Date: Sep 2006

Re: Calculate class A power
Quote:
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, 11:50 AM  #5 
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Join Date: Sep 2006

"1.35A, which in turn implies that, into 8 ohms, the JC 1 operates in classA up to 29W"
1,35 x 2 = 2,7A (push pull output stage) P rms = 8 * [2,7 / sqrt(2)] ^ 2 = 29W rms 
30th March 2009, 12:12 PM  #6  
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Join Date: Jan 2002
Location: Earth

Re: Calculate class A power
Quote:
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, 12:16 PM  #7 
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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
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30th March 2009, 12:23 PM  #8  
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Re: Re: Calculate class A power
Quote:


30th March 2009, 12:33 PM  #9  
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Join Date: Jun 2004
Location: the Netherlands

Re: Re: Calculate class A power
Quote:
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, 12:37 PM  #10 
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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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