Please someone help my math ....
One amplifier has 100+100 V rails
In total 20 output transistors
each transistor 8mv bias over one balast resitor
ballast resistors are 1R
so total bias is 160ma on idle
how many watts produced in Class A ?
One amplifier has 100+100 V rails
In total 20 output transistors
each transistor 8mv bias over one balast resitor
ballast resistors are 1R
so total bias is 160ma on idle
how many watts produced in Class A ?
P(rms) = (1/2)* (I_idle)^2 * R_load
SE:
205mW_peak_8
102,5mW_peak_4
PP:
410mW_rms_8
205mW_rms_4
assuming that the quiescent current is equal to ipeak (sin omegaT) and 2*I_idle for ip in push-pull operation.
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SE:
P(rms) = (1/2 )* (I_idle)^2 * R_load
PP:
P(rms) = 2 * (I_idle)^2 * R_load
I hope this helps you.
P(rms) = (1/2 )* (I_idle)^2 * R_load
PP:
P(rms) = 2 * (I_idle)^2 * R_load
I hope this helps you.
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Are you sure about that?
The bias on an AKSA 'Alpha Nirvana 4R' amp is 3a; this is nominally a 39w amp.
3 x 3 x 4 = 36w ... so I think the "1/2" in your equation is in error?
Yes, i'm shure.
P = U * I, P = U^2/R, P=I^2*R we all know -> valid for DC, or else RMS.
For simple sinewave it is also correct:
(I_Idle/sqrt(2)) °2 ---> (1/2) * I_Idle^2 ---> if we declare the DC current, the quiescent current at the operating point, as the maximum possible peak current (in Class A).
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If we belive that 3Adc automatically corresponds to the RMS signal current, then you would be right. However, this is not the case (despite all the fancy math), which is probably why there are so many definitions of Class A on the market.
@east electronics
could you please tell us what was important to you about the knowledge (the information and the value /number)?
That would be nice.
kind regards,
HBt.
could you please tell us what was important to you about the knowledge (the information and the value /number)?
That would be nice.
kind regards,
HBt.
