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Medium Voltage Recitifer

  1. Dec 20, 2009 #1
    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. jcsd
  3. Dec 21, 2009 #2
    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
  4. Dec 21, 2009 #3
    Yah, that's why i have to put in the shunt resistor right?
    To prevent the diode from failure.

    Right??? :confused:
     
    Last edited by a moderator: May 4, 2017
  5. Dec 21, 2009 #4
    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
     
  6. Dec 21, 2009 #5

    uart

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    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).
     
  7. Dec 21, 2009 #6
    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.

    High Voltage resistors are less than $2 per pc here for ranges 1M -10M, cost is manageable.
    Space saving and ensuring the rectifier can take the heat generated is a more important issue. Can it still work without the caps? Safety issue?

    Last of all, I AM STILL interested... where did the value of 500V come from??? I doubt it is a constant in an equation like PI=3.142 (very simplified PI). Any idea at all?:shy:
     
    Last edited by a moderator: Apr 24, 2017
  8. Dec 21, 2009 #7

    uart

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    Jestine's rule of thumb is not 500V, it's 500 ohms per volt. So in other words he's just designing around 2mA shunt current for each diode leg. Since 2mA is much larger than any diode reverse leakage that I've ever encountered then his rule of thumb is certainly compatible with the guidelines I mentioned, but perhaps a bit of an overkill at very high voltages.

    I was attempting to use the maximum resistance (and therefore least shunt current and power) that you could get away in your particular application. I was taking into account that you were only using about 55% of the diodes rated voltage and that therefore the leakage current would be considerably less than the 50uA data sheet maximum value. That allowed me to design around a 50uA shut current instead of the 2mA rule of thumb and therefore use much less power.

    BTW. You haven't told us what type of voltage waveform you're rectifying. Is it a sine wave or a square wave or something else - and at what frequency. Don't overlook Mikes advice about also using a small amount of shunt capacitance for dynamic voltage sharing during commutation - this is particularly important if it's a switched waveform instead of something smooth like a sine wave.
     
  9. Dec 21, 2009 #8
    So it means SIMPLY... any resistor values between 650kohm and 13Mohm is good for my application then? Of course the voltage rating of the resistor must be correctly chosen.

    I am using it on a 3 phase sine, without neutral, 50hz
    The input to the bridges will come from a stepup tx, star configuration, limited by a reactor so the inrush current will not trip the MCB.
    Output of tx is 2000V, 2A per phase.
    So two of these bridges will do to make a proper polyphase rectifier

    The output of the bridges will be connected directly to a bank of capacitors rated 2000Vdc
    Mainly for storage of DC energy, to be released as a pulsed energy upon triggering.

    Wana guess what that might be? :devil:
     
  10. Dec 22, 2009 #9

    vk6kro

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    You're making a RAIL GUN? We get about one of those a month. We never hear from them again. Funny about that.

    When you own a power supply like that, your life expectancy drops from about 80 years to about 2 weeks.
    You should make up a sign telling the paramedics how to disable the high voltage so that they don't get zapped.

    2000 volts is instantly and permanently lethal. Believe it or not, you WILL die if you touch it.
     
    Last edited: Dec 22, 2009
  11. Dec 22, 2009 #10
    I intend to live my life beyond 80 years.... :cool:
    Rail guns are super illegal in my country!!! Just carrying it and getting caught, means jail of at least 20 years + 24 strokes of the cane. Seriously not worth it....

    My bridge is for a magnetiser.
    The OEM bridge needs a replacement almost every 6 months.
    The manufacturer have rubbed off the diode part numbers so i can't replace without knowing what's the diode specs.
    So with a bit of trial with my new bridge, maybe we can save some $$$ and downtime.

    Nash78
     
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