still making dc offset filters gbart? i i need some!
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
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