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reverse polarity protection

Discussion in 'Electronic Design' started by Jenalee K., Apr 19, 2006.

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  1. Jenalee K.

    Jenalee K. Guest


    I tried to make a MOSFET reverse polarity protection for a 24V power
    supply. The MOSFET I had available has a Vgs max of +/-20V so I came up
    with the design below (I only had a 7V5 zener). The OK LED serves as
    power indicator, the other LEDs are test load.

    When the power supply is connected correctly it works as expected
    (measured Vgs of -7V4). But when I reverse the polarity the red LED
    goes on and I measure a Vgs of -3V6. How is this possible?

    Jenalee K

    | | ||| |
    | z === |
    | 7V5 A | |
    | | | |
    | o--------' .--o--.
    | | | |
    | OK V -> green V -> - red
    /+\ - led - ^ -> led
    24V ( ) | | |
    \-/ | '--o--'
    | | |
    | .-. .-.
    | | |1k5 | |2k2
    | | | | |
    | '-' '-'
    | | |

    (created by AACircuit v1.28.6 beta 04/19/05
  2. Take a look at that MOSFET symbol. Notice the reverse diode poking out from
    the substrate connection.

    A power MOSFET has a parasitic diode (that is, a diode that results from the
    semiconductor arrangement necessary to make the MOSFET, not one that they
    put in on purpose). So, it will conduct in reverse even if the gate is
    biased off.

    Does the power supply itself have crowbar overcurrent protection? If so,
    you might just be able to put a beefy reverse-biased diode across the
    terminals. Reverse the polarity, lots of current flows, power supply
    voltage-limits or shuts off.
  3. Jim Thompson

    Jim Thompson Guest

    See "OverAndReverseVoltageProtection.pdf" and
    "PerfectDiodeForChargerIsolation.pdf" on the S.E.D/Schematics page of
    my website.

    ...Jim Thompson
    | James E.Thompson, P.E. | mens |
    | Analog Innovations, Inc. | et |
    | Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
    | Phoenix, Arizona Voice:(480)460-2350 | |
    | E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
    | | 1962 |

    Old Latin teachers never die...they just decline
  4. Slavek

    Slavek Guest

    Same components but mirror vertically the zener with resistor - works
    well - tested.
  5. Slavek

    Slavek Guest

    Zener and resistor on the other side of FET.
  6. Jenalee K.

    Jenalee K. Guest

    I know about the parasitic diode, that's why I put the MOSFET in with
    its source connected to the load. The behaviour is the same when I swap
    the source and the drain. I've tried several MOSFETs of the same type
    and they all do the same thing. Maybe they are all faulty?

    In fact I've traced the problem to the LED in the gate path. If I
    connect the zener to the source instead of the drain (as suggested by
    Slavek) *and* if I short the OK LED it works fine. With OK LED it

    Anybody care to explain?

    Jenalee K
  7. Jenalee K.

    Jenalee K. Guest

    Zener and resistor on the other side of FET.

    And without the OK LED. Why doesn't it work with the LED?

    Jenalee K
  8. Slavek

    Slavek Guest

    It relys on the load - when load consumes less than leakage then it may
    not work.
    same like a diode.
    Try simulate it.
  9. Mac

    Mac Guest

    Yeah. Keep dreaming. ;-)
    Every time anybody posts the FET reverse polarity protection circuit here,
    at least one person fails to notice that the FET is "backwards," and that
    the body diode will therefore be under reverse bias when the power supply
    polarity is reversed. This time it was Walter Harley's turn, I guess. ;-)

    The problem with your circuit is that the PFET gate must be connected to a
    voltage which is negative (with respect to the source) when polarity is
    correct, and positive or zero (with respect to the source) when polarity
    is backwards. If you look at your original circuit, there is no reason to
    expect the gate to be positive or zero with respect to the source when the
    supply polarity is reversed. In fact, as the OK LED goes into reverse
    breakdown (another problem with your original circuit) the Zener will
    be under forward bias, thus guaranteeing that the gate will be only
    one diode drop higher than the lowest Voltage node in the whole

    So, the gate is at, let's say, 0.6 V. If the FET were off, then the source
    would be at 24V. But a Vgs of -23.4 would turn the FET on strongly, so the
    FET must not be off. If the FET were completely on, then Vds would be 0,
    and Vgs would be positive, so that isn't possible, either. So the FET must
    be somewhere in between on and off. Which means that the source will be
    at around 0.6 + Vth. Which means significant current will flow backwards
    through the load. Oops.

    In the classic version of the circuit, where the power voltage is less
    than Vgs max of the FET, you just connect the gate to the negative rail,
    the drain to the positive rail, and the source to the load.

    In your case, the 24 Volt supply is too much for the gate, so you need the
    Zener. The problem is that you put the Zener in the wrong place. Put the
    Zener from the gate to the negative rail. The Zener current-limiting
    resistor should probably go to the source, not the drain, as you have it.

    Let's see if I can draw it:

    | ||| | |
    | === \ |
    | | /1k5|
    | | \ |
    | | / |
    | +-+ |
    | | |
    | o--------' .--o--.
    | z | |
    | 7V5 A green V -> - red
    /+\ - led - ^ -> led
    24V ( ) | | |
    \-/ | '--o--'
    | | |
    | | .-.
    | | | |2k2
    | | | |
    | | '-'
    | | |

    When polarity is correct, the body diode conducts, causing the 1k5
    resistor to drop around 15 or so volts, which will turn on the FET before
    the body diode can fry.

    When the polarity is reversed, the drain will be at zero volts, the DC
    resistance of the load will pull the source up to 24 Volts, and the Zener,
    under forward bias, will pull the gate up to 24 Volts. With Vgs=0, the
    FET will be off.

    You don't need the OK diode, and if you put it in series with the Zener,
    it may cause the FET to turn on slightly when polarity is reversed. This
    is because it could keep the gate from getting pulled up all the way to 24
    V. Although, even then, I think the circuit will eventually turn off under
    reverse polarity.


  10. Jenalee K.

    Jenalee K. Guest

    Thanks a lot for your help! I get it now.

    Thanks a lot,
    Jenalee K
  11. Whoops, yes it was. I blame it on the ASCII ;-)
  12. Mac

    Mac Guest

    Yeah. Pretty hard to tell. Well, anyway, it will probably be my turn next. ;-)

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