Main Question or Discussion Point
What is the voltage leaving the rectifier?
What is the voltage at the two 22 ohm resistors?
What is the effect of the 4.7 volt zener diode?
It is hard to tell the exact number due too many unknowns. But the the peak voltage will be around 3.1V * 1.41 - 0.7V = 3.6V But we have a bleeder resistors, so the voltage will be probably around 3V as noted in the diagram (with respect to ground ).What is the voltage leaving the rectifier?
6V ( the voltage seen across 22 ohm's resistor and a Zener diode) minus the Zener diode voltage (6V - 4.7V) = 1.3VWhat is the voltage at the two 22 ohm resistors?
Zener Diode together with 22 ohm's resistors forms a Zener shunt voltage regulator. The output voltage will be Vz ≈ 4.7V with respect to negative rail. But because ground is up by 3V from negative rail. The Vout with respect to ground is around 4.7V - 3V = 1.7VWhat is the effect of the 4.7 volt zener diode?
great link CWatters posted thereWatters...I think I understand everything you said...and thanks
Remember my soapbox - voltage is potential difference ie between two points.What is the voltage leaving the rectifier?
((6.2 * √2) - 2Vd) = 6.77V rail to rail (for Vd=1)+/- ((6.2 * √2) - 2Vd)/2
The cathodes of the two right hand diodes should have around +3V DC on them.Is there any case where a 4 diode bridge rectifier would not have AC on all 4 diodes??
Other people have worked on this amp or that is to say they have messed up so much stuff I not sure I will ever get it fixed. I ask the question about the rectifier because one diode's cathode is not connected to the AC trace.
To finish CWatters' thoughtIs there any case where a 4 diode bridge rectifier would not have AC on all 4 diodes??
and their other ends ought to have 3.1 VAC with respect to circuit common.The cathodes of the two right hand diodes should have around +3V DC on them.
The anodes of the two left hand diodes should have around -3V DC on them.