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Diode Da Lima

Discussion in 'General Electronics Discussion' started by Doc_Holiday, Aug 9, 2017.

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  1. Doc_Holiday

    Doc_Holiday

    1
    0
    Aug 9, 2017
    I am not trying to repair an electronic device!!! I have a JLG lift with bad diodes integrated into a 12 volt system. I'm a mechanic with extensive wiring knowledge. But, even tho I have electronic understanding, no understanding of the repricussion of replacing the existing diode with a different valued one. The diode that's factory installed is a 6A60. It is on a relay to a dump valve that can be controlled in two places (on the ground at one controller, or thirty feet in the air at a second controller), and it's bad! (This much about electronic components, I know!!!) My question about this 6 amp, 600 volt, diode, is confusing to me!! Firstly, why is there a 600 volt diode being used on a 12 volt system!! Secondly, is there any thinkable reason that a 6 amp, 1000 volt diode wouldn't work just the same way...(if so, why not) and I guess I'd like to learn, if anyone can tell me, why diodes are integrated into the wiring harness of a 12 volt system anyway?? Why does a relay need a diode on the wire at all??? And what problems should arise from replacing a 1N4004 diode with a 1N4007 diode??? Hence: "Diode Dilemma" Thank you, Doc!!!!
     
  2. kellys_eye

    kellys_eye

    4,282
    1,152
    Jun 25, 2010
    When you connect power to an inductor (coil of wire) you create a magnetic field around it. When the power is removed the magnetic field collapses - as the field collapses it causes a voltage to be induced in the coil (a voltage in the reverse direction of that originally applied) and that voltage can exceed the original supply voltage by many times.
    The diode is meant to 'steer' the #over-voltage' away from the direction of damage. It's connected across coils as 'back e.m.f suppression' device.

    As with most electronic parts, the specification is usually stated as a minimum requirement but fitting devices that exceed that minimum is, in most circumstances, perfectly acceptable. A 1000V diode (1N4007) can replace a 600V diode quite readily but only for the same CURRENT - they are both rate at 1A.
     
  3. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    The need for the diode comes from the fact that you have a circuit that switches an inductive load on /off and vice versa in a "fast manner".
    In this case the inductive load is the relay/solenoid coil.

    The switching element is probably a transistor of the type BJT/MOSFET/Darlington etc.
    They are easily destroyed if voltages exceeding their working specs. are present on them.

    An inductor may produce very high voltages when switched rapidly.
    This may be a good thing, and indeed is/was used in cars with ignition coils.
    If you are familiar with them you may recall the sparks from a car plug ,tens of thousands of volts...
    same thing with now days tasers:eek:

    The formula for calculating how much voltage is produced is basic for inductors

    v= - L *(di/dt) ,don't let it scare you away, it is simple:

    Here:
    (di/dt) is the rate of change in time of the current through the inductor
    (formally the derivative of the current in respect to time).
    v is the voltage produced.
    L is the inductance of the inductor.

    What that means is that for a given fixed value L of the inductor the faster we create a change in current the higher the voltage we get.

    Let's make a numerical example.
    Our inductor coil is 100mH=0.1H,and needs 1A to be energized.

    thus L=0.1H,dt=1A

    If we switch it from on to off and vise versa at a speed of 10msec=0.01sec
    We thus have v=0.1*(1/0.01)=0.1*100=10V

    No let's change the switching time to 1msec=0.001sec
    We thus have v=0.1*(1/0.001)=0.1*1000=100V

    If we would switch ten times faster (0.1mSec) we will get 1000v
    This will destroy a semiconductor switch easily if not protected.

    So,
    Now that we understand the problem we understand the needs for protection diodes.
    They are called "clamping diodes" and sometimes "fly-back diodes"
    What they do is clamp the maximum/minimum voltage on the switching device.
    In mechanical switches they protect against sparks and arcing(that is another issue)but the cause is the same high voltages produced by switching inductive loads.

    The voltages they clamp to is minimum(GND-VD),maximum VCC+VD)
    see circuit below.

    On the issue of diode replacements:
    Assuming the manufacture designed the equipment properly(and they mostly do or they are out of business...)
    stick to the original part if you can!!!

    There are many parameters for a diode which may or may not be important,they are depended on the application.
    Some applications are very forgiving others are not at all...

    this is more so for more complicated devices than diodes.

    untitled.JPG



     
    Last edited: Aug 9, 2017
  4. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    Another way of looking at it is that energy is stored in the magnetic field of the inductor. J = I*I*L/2. This energy has to be dissipated somewhere other than in the switching device so the diode allows the current to continue until it is dissipated by the resistance of the coil

    If the turn off is too slow, a resistor can be added in series with the diode, this will increase the voltage on the switch so it is a compromise.
     
  5. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    All very nice theory, and all wrong. The reverse voltage the diode sees is only the voltage driving the coil. The diode is there to supress the high voltage spike, and the diode will clamp it at the forward voltage of the diode, typically less tha 1V. A 50V diode wiuld be fine if the coil voltage is 12V.

    Bob
     
  6. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    A reversed Zener in series with the diode is sometimes used fior this. This increases the voltage across the coil, allowing the current to fall more quickly.

    Bob
     
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