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# How would you work out peak voltage at point A? I know the answer

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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
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KCL faggot
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multimeter
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No? Anyone?

FPGA/VHDL jokes?
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>>684194203
Pretty sure no one here can math. You fucked son. And not in the "good" way.
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>>684194543
Please spell it out for me because I am a moron
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>>684195286
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.
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should be your source voltage minus the drops of two diodes
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>>684195797
That's what I thought, but no, because the sim software I'm using says otherwise
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>>684194203
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
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>>684194543
Its KVL you faggot.
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>>684196538
Also your teacher fucked up

AC should never be labeled positive and negative

Should be labeled L and N
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>>684196538
Op asked for peak voltage and not average voltage though.
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>>684197177
Nevermind, i'm dumb, capacitors are involved. Don't know if they're big enough to make the current linear though.
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>>684194203
2 * the drop across the diode. Done
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>>684194203
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.
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>>684194203
ideal diodes?
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>>684197457
They never are

At least for the purpose of school work
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>>684197457
They do a bit, but not completely. The voltage across the load varies by only a milliamp or two
>>684197852
>>684197664
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
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Just make one and hook it up to a scope
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>>684198448
Isn't the impedance of a capacitor 1/RC?
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>>684198716
no the Z of a cap is approx equal to its reactance
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>>684198448
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.
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>>684198716
no it's 1/(2*pi*f*C)
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>>684199283
Yeah i mixed R and omega, my bad
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>>684199218
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
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>>684199752
Where are you getting your answer from?
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>>684199752
>>684199218
The voltage drop is definitely 0.7v
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>>684194203
this is a hard question. Do some differential equations.
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>>684200512
I am running the simulation on NI Multisim
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>>684200882
What model are you using for your diode?
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>>684201459
>>684200643
>>684200565
>>684200512
>>684199752
>>684199483
>>684199283
>>684199218
>>684199038
>>684198716
>>684198703
>>684198448
>>684198383
>>684197922
>>684197852
>>684197664
>>684197457
>>684197177
>>684197168
>>684196903
>>684196538
>>684196273
>>684195797
>>684195596
>>684195286
>>684194852
>>684194773
>>684194651
>>684194543
>>684194471
>>684194203
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
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>>684201459
here is what I got picking a random diode
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>>684202021
Oh, lol glad you figured it out.