# Question for Atmasphere re: class A output calc

Hi Ralph,

Let me start by saying that, like some other members here, I've learned quite a bit from your excellent posts. Now to the point.

On another forum, we were discussing calculation of class A output level of a class AB amp before it starts running in class B. I always understood that total bias current squared times load resistance gave you the class A output (IxIxR=W).

Another member says this is not entirely correct and made the following statement:

"Lets assume a 0.1556A standing current on a push pull output stage. Peak current before either of the positive or negative rail output transistors will be double this, above this one half will be cutoff.

Just say we have 0.3112A peak before one half of the output stage actually turns off we can get the peak voltage, 0.3112 (Amps) x 8 ohm resistance = 2.4896V according to ohms law. That's the peak voltage and current though, not the RMS voltage or current.

2.4896V peak is 1.7606V RMS (peak voltage divided by 1.414), square that & divide by the load resistance. You get 0.387W RMS".

What is the correct method of calculating the number of watts an amp runs in class A, before running in class B ?

Thanks

Glen
6 responses
 Let me jump in. I'm not sure where double current comes from. I would think that when whole 0.1556A of bias current flows thru the speaker then the other transistor has to be off (Il=Ip-In).You cannot calculate RMS because you don't know the shape (1.41x relationship applies only to sinewave).Number of watts before jumping to class B might be not so important. Main purpose of bias is to avoid switchover distortion and not the nonlinearities of transistors. Increasing bias current increases distortion (overbias) since it is also increasing area of double transconductance (gm doubling) where gain is doubled since both transistors conduct. Of course emitter resistors can be set to minimize it but tests showed (Douglas Self) that increasing bias increases distortion. Problem is that holding amplifier on the edge of class B (minimum bias to make both transistors conduct) is difficult to do. I'm not sure what happens when bias is set much higher, but not pure class A.read page 138-140: http://books.google.com/books?id=Qpmi4ia2nhcC&pg=PA140&lpg=PA140&dq=Douglas+Self+class+b+overbias&source=bl&ots=hAE7f6LFUj&sig=MTAeHQeKHRn17SEEUnMv03DQprg&hl=en&ei=ORDJTLSBK8rOnAfWz7XhDw&sa=X&oi=book_result&ct=result&resnum=6&sqi=2&ved=0CCwQ6AEwBQ#v=onepage&q&f=false Gbart,Lets assume a 0.1556A standing current on a push pull output stage.Can you tell me, is this the total current between two output devices, or is just one? There is an interpretive issue there. Are we talking about transistors here? Is the resistance of the devices assumed to be zero?I'm trying to get a feel for the example in this exercise... in the meantime Kijanki's post seems valid enough to me. Atmasphere,We are talking solid state. 0.1556A is the total current between two output devices (bias of 42mV with emitter resistance of 0.27ohm).Glen Hi Glen, given your numbers above I would say that's pretty close, without knowing more about the amp. For example, for the purpose of this example, the base current has been ignored. Gentlemen, thank you for your responses. still making dc offset filters gbart? i i need some!