Signal as a pwr diode

Discussion in 'General Electronics Chat' started by Litch, Mar 28, 2013.

  1. Litch

    Thread Starter Member

    Jan 25, 2013
    86
    7
    (Posted in here because the server HTTP 500'd when I try and post in the General Section)

    So I (thought) I ordered a bunch of small power diodes to clamp the EM back charge from a bunch of 5v relays when they disengage - turns out they're 1N4454 Signal diodes Datasheet Here

    Now looking at the datasheet, they can take 4A at 1uS, but will breakdown at 75V reverse volts.

    Now when a relay disengages, the incurred EM voltage could be several hundred volts at barely any current (but with the polarity of the diode, hence it's purpose).

    Anyone see an issue will still using them for this purpose?

    Thanks in advance,

    -KB.
     
  2. richard.cs

    Member

    Mar 3, 2012
    162
    31
    The 1 second pulse rating of 1 amp is more relevant than the 1 μs rating. And I would go by the maximum working voltage of 50 Volts.

    If your relay coil draws less than 1 amp (as this will be the diode forward current at the instant of switchoff), and have less than 50 Volts across it in normal operation (when the diode is reverse biased) then these diodes will be fine.

    As a side note using a simple diode clamp like this results in the relay switching off pretty slowly. If you need to you can speed that up by adding resistance in series with the diode at the expense of applying a greater peak voltage to whatever is driving the relay.
     
  3. Litch

    Thread Starter Member

    Jan 25, 2013
    86
    7
    Yeah, 5v, ~40mA relays...

    My understanding was that upon disengagement, the relay's EM field that held the solenoid actuator in place creates a substantial voltage differential in the opposite direction to the coil's engagement.

    In any case the coils are driver by a ULN2803A Darlington array which has inbuilt diodes, and though these are not specifically designed for the EMF collapse of a relay, they should provide an extra layer of protection.

    Thanks,
    -KB
     
  4. richard.cs

    Member

    Mar 3, 2012
    162
    31
    So long as the COM pin (pin 10) of the ULN2803A is connected to your positive supply rail then I would be happy to rely on the built-in diodes - that is why they are there.

    When the switch opens the current through the relay coil cannot change instantaniously, and the other end of the coil is held at a fixed voltage. Current continues to flow through the coil either 1) through the protection diode2) into stray capacitance or 3) Through the switching element by over-volting it until it breaks down. The voltage at the switched end rises to whatever is neccessary to keep that current flowing, around a volt higher than the supply rail with the diode, possibly hundreds of volts for 2) or 3). Remember that the induced voltage is in the direction which makes your protection diode conduct.

    With a high voltage produced across the coil the current decays rapidly, with the much lower voltage from a forward biased diode it takes substantially longer and can hold the relay in for tens of milliseconds.
     
Loading...