How would you work out peak voltage at point A?

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

DigiKey thread? DigiKey thread!

KCL faggot

multimeter

No? Anyone?

FPGA/VHDL jokes?

Pretty sure no one here can math. You fucked son. And not in the "good" way.

Please spell it out for me because I am a moron

The other guy is wrong (unless he just wasn't specific enough) you would need to know the characteristics of your diodes first i think.

should be your source voltage minus the drops of two diodes

That's what I thought, but no, because the sim software I'm using says otherwise

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

Its KVL you faggot.

Also your teacher fucked up

AC should never be labeled positive and negative

Should be labeled L and N

Op asked for peak voltage and not average voltage though.

Nevermind, i'm dumb, capacitors are involved. Don't know if they're big enough to make the current linear though.

2 * the drop across the diode. Done

Vpk -2*Vd

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.

ideal diodes?

They never are

At least for the purpose of school work

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

Just make one and hook it up to a scope

Isn't the impedance of a capacitor 1/RC?

no the Z of a cap is approx equal to its reactance

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.

no it's 1/(2*pi*f*C)

Yeah i mixed R and omega, my bad

The voltage drop of the diodes is 0.7v per diode. I think. That might be the case. Let me go fuck with my sim a little to find out

Where are you getting your answer from?

The voltage drop is definitely 0.7v

this is a hard question. Do some differential equations.

I am running the simulation on NI Multisim