Calculating bias current for ACA

I am in the process of building 4, ACA amplifier boards. I plan to use 2 per chassis, and parallel the channels ,for a set of 16 WPC monoblocks. I have all the boards stuffed, except for the output mosfets. I have modded the boards according to the post by Tungsten Audio here. I am somewhat confused on how the bias current is calculated. As explained in Pappas' original article, Q3 needs .65 volts to turn on . .575 ohms is the equivalent resistance of ((R1//R2)+(R3//R4)). The bias current would then be .65V/.575 ohms , or 1.13 Amps. The Tungsten mod lowers the value of R4 to .5 ohms, instead of .68 ohms. Giving a new equivalent resistance of .523 ohms. .65V/.523 ohms yields 1.24 amps. Tungsten schematic lists his new bias current as 1.6 amps. I am sure Tungsten is correct, but would like to better understand the circuit. Can someone show me where I am making my error?
Thanks,
mg16
 
Looking closer at the schematic, Vbe is across R15, and there is R8 that drops some voltage before R15. So Vbe is still 0.65V, but voltage drop at R1,R2,R3,R4 would be higher. If R15 was not there, then Vbe would also be across R1,R2,R3,R4.

So my initial thought of a different Vbe was incorrect.

If you look at the original ACA writeup and schematic, there was no R15.

https://www.firstwatt.com/pdf/art_amp_camp_1.pdf

My original post stated that the original drawing showed 1.45A. That is incorrect. It was a later version that showed 1.45A. And that current is most likely correct and based on the addition of R15.
 
Last edited: