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Two pairs of 1N4007 in series really necessary in a 12w amp's rectifier?

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  • #16
    Originally posted by Alan0354 View Post
    From my understanding, I don't think it is particular important to have resistors across the diodes. I don't think diodes are like capacitors. It's the power that burn the diode. ...

    In the example one diode leak a little and put most of the voltage across the second diode. As long as the second diode does not conduct, there should not be stress. If the second diode start leaking, then the voltage drop across the diode will decrease and both diodes share the voltage.

    That's my understanding in semiconductor.
    It's correct from one stand point. However, if you ever have a situation where the applied voltage is bigger than either diode by itself can withstand, and a voltage transient causes the "strong" diode to break over, the power dissipated in that diode burns it shorted. Then the entire voltage is across the "weak" diode and it kills that one too.

    Admittedly, this is not common, and the physics are slightly different from capacitors. However, it can happen. Ask me how I know that.... Resistors are so very cheap that conservative engineering would spread the loading out over all the devices so there is no weak spot.

    VERY conservative engineering would also put swamping capacitors across each diode so that the dynamic voltage distribution is forced to be equal with voltages moving around quickly. That way a diode with high capacitance would not let it's better brothers be damaged.

    Note that this may not happen instantaneously. Zenering only happens in the voltages below about 8V as I remember dimly. Above that, it's avalance breakdown. Avalanche breakdown can cause hot spotting, where small areas of junctions get hotter, conduct better, and carry even more current and get hotter, especially in silicon bipolar junctions. Over time, you can get "aging" from the accumulation of crystal defects caused by the breakovers. So it can be a slower-than-instantaneous death.

    And resistors are very cheap.
    Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

    Oh, wait! That sounds familiar, somehow.

    Comment


    • #17
      Originally posted by potatofarmer View Post
      So if I suggested ditching FWCT rectifiers in favor of bridge rectifiers that would be, what, the overalls view?
      The FWCT connection forces twice the voltage across each diode as the FWB, because each diode must withstand the sum of the filter cap voltage plus having its anode pulled to the negative peak of the input AC. The current in each diode is less, as each diode only conducts alternate charging pulses.

      However, the transformer feeding a FWB versus a FWCT sees different heating currents as well.
      Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

      Oh, wait! That sounds familiar, somehow.

      Comment


      • #18
        Originally posted by R.G. View Post
        It's correct from one stand point. However, if you ever have a situation where the applied voltage is bigger than either diode by itself can withstand, and a voltage transient causes the "strong" diode to break over, the power dissipated in that diode burns it shorted. Then the entire voltage is across the "weak" diode and it kills that one too.

        Admittedly, this is not common, and the physics are slightly different from capacitors. However, it can happen. Ask me how I know that.... Resistors are so very cheap that conservative engineering would spread the loading out over all the devices so there is no weak spot.

        VERY conservative engineering would also put swamping capacitors across each diode so that the dynamic voltage distribution is forced to be equal with voltages moving around quickly. That way a diode with high capacitance would not let it's better brothers be damaged.

        Note that this may not happen instantaneously. Zenering only happens in the voltages below about 8V as I remember dimly. Above that, it's avalance breakdown. Avalanche breakdown can cause hot spotting, where small areas of junctions get hotter, conduct better, and carry even more current and get hotter, especially in silicon bipolar junctions. Over time, you can get "aging" from the accumulation of crystal defects caused by the breakovers. So it can be a slower-than-instantaneous death.

        And resistors are very cheap.
        Of cause in case of the +B bigger than one diode. You have to be more careful. This is not in this case. But I can tell you we do this all the time all these years. It is one thing people have the luxury to put caps or resistors across. In our designs, we deal with really high voltages that we need multiple transistors/diodes/TVS to stand off the voltage. And we were doing 5KV pulsing with rise time in less than 100nS. Just driving the capacitance of the 5' coax used a lot of power. We cannot afford to have either resistor or caps across. We used multiple transistors in series count on the TVS to avalanche to clamp each transistor on every pulse. We were running in over 10KHz for years, hours a day without problem.

        Of cause it is a better practice to put all the protections. In this case, the max is less than 1000V, , but I think it is save without it as Fender did that and it's ultra reliable.

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        • #19
          Every situation is different. For multi-KV, sub-microsecond rise time applications, yes, I fully understand why driving any excess parasitic parts could be prohibitive. Fortunately, such applications are esoteric enough you can do a more complete job of characterizing the components and the signals.
          Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

          Oh, wait! That sounds familiar, somehow.

          Comment

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