How would you work out peak voltage at point A?
I know the answer is around 21.9V, I'm want to know the method
The diodes make up a full bridge rectifier, combined with the capacitors the peaks become evened out a bit
But because it's under load average voltage drops below the RMS of the AC
Pick a case where the source is at peak with +24V at the positive terminal. There are two diode drops between it and the negative terminal. The tricky question is that output voltage (Va) is in reference to ground. Which is placed in between the two diodes. This means that you have to include both of them when finding Va.
They do a bit, but not completely. The voltage across the load varies by only a milliamp or two
The voltage drop of a standard diode is 0.7v
I'm looking for an answer around 21.9v
I know the capacitor is the reason for that extra drop, but I have no idea how to calculate that extra drop
The reactance of those capacitors is really small. You are going to have a lot of current through them.
Xc=1/(2pi*f*C) = 1.06 ohm
A 3mF cap is fucking huge. At 50hz it might as well be just a wire.
Using V=IR we see that we have a lot of current though those poor diodes. Now I cant say, because I dont know if those are ideal diodes or not, but at almost 24 A it would not be unreasonable to see a voltage drop of around 1.05 V across each one.
That said, 1.05 V is in the range of what you can say an "ideal diode is". It depends on the architecture.
LOOKETH UPON ME, WHO BE A MORON
you guys were right and i was doing the maths wrong. 22.6v is the peak voltage. I was doing the confirmation maths wrong. I AM MORON WHO WASTED AN HOUR ON /b/ TO FIGURE THIS OUT
I apologise to all for my idiocy. /thread