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# bridge rectifier -Concept Help

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#### Prisinor

##### Newbie level 4
Hello to every one, i need some one to help me in understanding basic concept about "Bridge Rectification" more specifically about its Peak Inverse Voltage Calculation.

Let’s assume that diode D1 and diode D2 are forward biased and examine the reverse voltage across D3 and D4.

Visualizing D1 and D2 as shorts (ideal model) , we can see that D3 and D4 have a peak inverse voltage equal to the peak secondry voltage. Since the output voltage is ideally equal to the secondry voltage.

PIV = Vp(out)

i'm Clear upto this Point but when diode drops is taken into account i.e.

If the diode drops of the forward biased diodes are included as the peak inverse voltage across each reverse biased diode in terms of Vp(out) is

PIV = Vp(out) + 0.7V

i'm not understanding How is 0.7 is Come ??

This is a case of diode behavior at the threshold of operation. It brings in unusual factors to consider.

A diode starts conducting at a lower volt level than 0.6 or 0.7 V.
At 1 uA it might be .35 V. (That is, the diodes I've tested.)

When you reverse-bias a diode, it does not even conduct as much as 1 uA. Until you exceed its threshold, that is. Then you ruin the diode's operating characteristics (except if you're lucky).

You would need to add many forward biased diodes in line with one reverse-biased diode, to make a difference in the threshold of the reverse-biased diode.

This is a case of diode behavior at the threshold of operation. It brings in unusual factors to consider.

A diode starts conducting at a lower volt level than 0.6 or 0.7 V.
At 1 uA it might be .35 V. (That is, the diodes I've tested.)

When you reverse-bias a diode, it does not even conduct as much as 1 uA. Until you exceed its threshold, that is. Then you ruin the diode's operating characteristics (except if you're lucky).

You would need to add many forward biased diodes in line with one reverse-biased diode, to make a difference in the threshold of the reverse-biased diode.

thanks for your reply, ya i knew about breakdown characteristics my problem is that I'm not understanding mathematical expression of
Peak Inverse Voltage = Pout + 0.7 or PIV= Vpsec - 0.7 (Both are same for bridge rectifier)
How they came?

Peak Inverse Voltage = Pout + 0.7 or PIV= Vpsec - 0.7 (Both are same for bridge rectifier)
How they came?

I would not use the 0.7V amount as a threshold. I can only suppose it comes from a simplistic diode model.

If I were to count forward biased silicon diodes, I might use 0.2 V. Just maybe the current will be 1 pA.
But why count 0.2 V when typical reverse breakdown ratings are 50 or 100 or 200 V?

If I want to calculate a safe breakdown rating for a diode (or diode string), I would add up all the reverse V ratings only.
I would not count forward biased diode junctions, no matter how many there are in series.

PIV = Vp(out) + 0.7V

0.7V is the forward voltage drop of one of the bridge diodes which "terminates" the current flow from the pos. AC terminal through the first conducting bridge diode to +Vp(out) -> load -> 0(out) -> second conducting bridge diode -> neg. AC terminal.

If the neg. output terminal is defined at potential 0 , the neg. AC terminal has potential -0.7V , so the necessary blocking voltage is the difference of the output voltage potential +Vp(out) and this (neg.) potential of the neg. AC terminal, hence PIV = Vp(out) - (-0.7V) = Vp(out) + 0.7V .

Hope this is not too confusing!

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