# Medium Voltage Recitifer

1. Dec 20, 2009

### Nash78

A full bridge rectifier for 2A, 2000vdc is required.
It is used for a capacitor charging circuit.
A full bridge rectifier consists of 4 diodes in theory.

But due to physical size limitations that i can install this item.
Diodes that are above 3kV does not fit my sizes, so i have another option.

What i know is, diodes can be put in series to increase the voltage.
So i want to put 3 diodes, namely BY255 in series, to form the single diode.
Reason? Cheap, reliable, axial type
It is rated 3A, 1300V. So 3 in series would give me 3900V, a good safety factor.

So my final step is, how to calculate the shunt resistance per BY255???
Online searches tell me to take the Rated Voltage of the diode x 500V
Which in my case would be 650kohms....
I cannot verify that is a good value with my current knowledge and experience.

Anyone can direct me to any source of writeups so i can understand more on how this value of 500V came about?

Thanks,
Nash78

2. Dec 21, 2009

### mugaliens

Er... Diodes in series aren't like resistors in series. Something about cascading failure, where voltage buildup over a series of diodes is uneven. http://hobby_elec.piclist.com/e_diode.htm" [Broken]may help.

Last edited by a moderator: May 4, 2017
3. Dec 21, 2009

### Nash78

Yah, that's why i have to put in the shunt resistor right?
To prevent the diode from failure.

Right???

Last edited by a moderator: May 4, 2017
4. Dec 21, 2009

### Mike_In_Plano

The deal with putting resistors in parallel with the rectifiers is to prevent leakage current imbalance from driving a rectifier into avalanche mode. The difficulty is that the resistors must be high voltage types and there are a goodly number of them. That adds up to a lot of space and cost for resistors.

Many high voltage supplies simply have caps in parallel with the rectifiers. Something on the order of 4.7nF - 10nF. At 60Hz, the leakage current of the rectifiers isn't sufficient to build up excessive voltage during the short time blocking time. During forward conduction of the rectifiers, the caps are all discharged.

Best of Luck,

Mike

5. Dec 21, 2009

### uart

Yes that's right.

Ideally you want to choose the resistors such that their current at full reverse voltage is large compared with the diode reverse leakage current. But you don't want to go too far as you obviously also want to limit the power dissipation in the resistors.

For the part you mention "BY255" for example the typical reverse current (at max voltage @100C) is only a few uA, but the datasheet listed maximum reverse leakage is 50uA. Since you got a bit of voltage headroom on that part then 50uA at your nominal reverse operating voltage (667 V) should be plenty. That gives R = 667V/50uA which is about 13 Meg.

So either 10M or 12 M would be fine but make sure they are rated for the voltage. The wattage is quite manageable. At 10M you've got V^2/R about 45mW (and obviously less when taking into account the typical 50% duty cycle in this type of application).

6. Dec 21, 2009

### Nash78

Yo MIKE and UART,

http://ezinearticles.com/?Easy-Meth...lel-To-Get-The-Desire-Specification&id=502120

This is the article that i am refering to.... this guy makes it sound so simple.
Which was also my idea. The difference is, HE seems to understand this subject more than i do.

What UART is doing, is a different approach. Jestine's calculations gives a shunt resistor of 650k ohm, while UART gives a value of around 13M ohm. Tatz a way big big big difference man.... Anyway, yours make more sense in using the reverse leakage current for calculation.

Nash78