Piv derivation formula.

[paulmdrdo]

can you help me how to get the PIV formula from the diagram.

And can you please explain why the potential of the cathode in D2 is the same as the potential of the cathode in D1? thanks!

 

[chuckey]

Its the same electrical point. They are shorted.
Frank

 

[paulmdrdo]

where is the electrical point you're talking about?

 

[chuckey]

On your circuit there is a vertical line that shorts the two cathodes together and is connected to Rl.
Frank

 

[paulmdrdo]

can you tell me if my derivation is correct from the diagram. thanks!

 

[chuckey]

Its correct.
Frank

 

[paulmdrdo]

but in my book it says something like this. why is that?
and how will I calculate the PIV across either diode? thanks!

 

[albbg]

According to your schematic the peak inverse voltage is calculated from cathode to anode, then:

PIV = V(cathode) - V(anode) = [Vsec/2) - 0.7] - (-Vsec/2)

you, instead, have considered the PIV as a voltage in series with the diode, in a closed mesh: it's not correct.

 

[chuckey]

Numerically you are correct and your value is equal to the book value, you have chosen to express PIV as a negative quantity, whilst the book and tradition makes it a positive quantity.
Frank

 

[albbg]

Numerically you are correct and your value is equal to the book value, you have chosen to express PIV as a negative quantity, whilst the book and tradition makes it a positive quantity.
Frank

Yes the absolute value is the same, but the sign (with respect to the polarity drawn in the sketch) is wrong. This because the procedure to calculate the PIV is not correct. No matter of tradition, simply the polarity on the drawing doesn't match the polarity of the calculation

 

[paulmdrdo]

According to your schematic the peak inverse voltage is calculated from cathode to anode, then:

PIV = V(cathode) - V(anode) = [Vsec/2) - 0.7] - (-Vsec/2)

you, instead, have considered the PIV as a voltage in series with the diode, in a closed mesh: it's not correct.

should the calculation of PIV always from cathode to anode? can it be the other way around. I thought anode to calthod calculation is more conventional. please explain about this.

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And also I found my mistake in my calculation of my PIV it should be PIV-Vpsec/2-Vout=0.

"you, instead, have considered the PIV as a voltage in series with the diode, in a closed mesh: it's not correct." -what do you mean by this?

I considered the PIV in series with the Vout and Vpsec/2. The only thing I found incorrect in my solution is the sign of PIV.

 

[albbg]

Usually the PIV is a positive number, so it's calculated from cathode to anode. However in your calculation wasn't matter of convention but the problem was that the sign didn't match with the diagram: you put the + sign of PIV on the cathode.

You are rigth in your calculation the mistake was due to the sign of PIV; do not consider my sentence "you, instead, have considered the PIV as a voltage in series with the diode, in a closed mesh: it's not correct.": I made confusion reading your formula so I assumed you did an incorrect reasoning. Sorry.

 

[chuckey]

I stand by my statement, if it was PIVc-a or PIVa-c, then the sign would be important.
Frank

 

[paulmdrdo]

Usually the PIV is a positive number, so it's calculated from cathode to anode. However in your calculation wasn't matter of convention but the problem was that the sign didn't match with the diagram: you put the + sign of PIV on the cathode.

This is what I imagined, as you can see from my diagram I added a "pretend voltmeter" which will give me the PIV across D2. The red lead is connected to cathode(positive) because the potential at that point is higher or positive with respect to the potential at anode. The black lead is connected to anode(negative ) because the potential at that point is lower or negative with respect to the potential at cathode. What I considered here is the potentials at those points not the Diode itself. Am I missing something? please tell me. thanks!