What Schottky diode for protection?

Discussion in 'General Electronics Chat' started by Nicholas, May 1, 2016.

  1. Nicholas

    Thread Starter AAC Fanatic!

    Mar 24, 2005
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    Hi guys

    Thus far I have used a 1n1001 normal diode in series from my controller (5V) to my MOSFET switching something like 30-50V.
    I want to use a Schottky diode to get a smaller voltage drop from the controller, but I would like a recommendation as to which one.

    The Schottky should protect the micro controller in case something goes wrong, and 50V is going out of the gate of the mosfets and
    try to fry the micro controller. I have been adviced that this is very rare, but I want to be on the safe side :)

    So, what Schottky should I use?

    Thanks!
     
  2. Lestraveled

    Well-Known Member

    May 19, 2014
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    @Nicholas

    Oh come on, we are not mind readers. A schematic please.

    While you are posting a schematic look up a SS110 Schottky diode.
     
  3. Nicholas

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    Mar 24, 2005
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    Coming right up!

    The 40V in the drawing could be anything from 5V to 50V

    Thanks!
    mosfet2.JPG
     
  4. Nicholas

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    Mar 24, 2005
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    About the SS110 Schottky diode, I use through-hole components only, as far as I can see
    the SS110 is surface mount, right?
     
  5. Nicholas

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    Mar 24, 2005
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  6. Sensacell

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    Jun 19, 2012
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    While I understand the instinct to "protect" your microcontroller, a series diode in this case is not really a good idea.

    1) You need all the gate voltage you can get to turn your FET on, the diode subtracts ~0.5 v from the gate drive

    2) With the diode in place, the fall time of the gate drive will be extended, the only path for current flow is the 10K resistor. Without the diode, you have symmetric rise and fall performance.
     
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  7. ElectronicMotor

    Member

    May 1, 2016
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    Your flyback suppression diode will short, or the IRL540 will short (never open). The worst that can happen is that your PIC output port will be grounded if it is set as an output. The 1k base resistor will limit the output port current to under 25mA, but they can handle up to 100mA, I've found.
     
  8. Nicholas

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    Mar 24, 2005
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    What is my flyback suppression diode in this case?

    I get the gist from a lot of people that I should not use a diode for protection at all. It just makes me a little nervous :)
     
  9. OBW0549

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    Mar 2, 2015
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    It's the diode connected across the inductor.

    I don't think you really need protection here; but if you're sure you want some, then a simple Zener diode, such as a 1N4735A, connected between the MOSFET gate and GND should suffice.
     
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  10. Nicholas

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    Mar 24, 2005
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    OBW0549, I was recommended using that a zener between gate and ground earlier, this might be the way to go.
    Does it matter if it is connected to micro controller output-pin side of the 1K resistor or the gate side? See drawing below:

    mosfet2_zen.png
     
  11. OBW0549

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    Mar 2, 2015
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    Connect it to the gate side, NOT the microcontroller side. The Zener will clamp any incoming positive spike to ≈6V, allowing the 1kΩ resistor to limit any current that might find its way into the μC port pin. Connecting the Zener directly to the port pin accomplishes nothing.
     
  12. Nicholas

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    Mar 24, 2005
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    Okay, so like this:

    Also, this should take of any voltage drop from the 5V from the controller, like I had with
    a normal diode, right?

    mosfet2.JPG
     
  13. OBW0549

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    Mar 2, 2015
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    I don't understand what you mean by that. Try re-phrasing your question.
     
  14. Nicholas

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    Mar 24, 2005
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    To prevent voltage from going back into the micro controller, I had a normal diode in
    series with the 1K resistor. This cut the 5V to 4.3V, which I thought was too much of
    a loss.

    Using the zener method, the 5V from the micro controller should reach the gate of the
    MOSFET, right? Not being lowered to 4.3V or 4.5V.

    Thanks!
     
  15. OBW0549

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    Mar 2, 2015
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    Correct.
     
  16. Picbuster

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    Dec 2, 2013
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    Do not use zener protect your mpu output. This is slow and uses a lot of power. (>300 micro amp)
    The best way is to clamp to MPU's VDD with an ultrafast schottky and one to GND to eliminate the negative part.
    Make sure that the unwanted signal is consumed by a resistor.
    So, at dangerous site via resistor two diodes one to mpu vdd one to gnd. junction to mpu you want to protect. (see attached)
    Resistor = (vmax-vdd)/ ( acceptable current ) (for positive pulse)
    Resistor =(Vmax)/(acceptable current) ( for negative pulse) ( breaking a wire will create a diffentiator (serial capacitor) as result a negative pulse)
    Energy in resistor (W)= I^2 x R
     
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  17. Nicholas

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    Mar 24, 2005
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    Okay, I am confused about which direction to take, now :)

    If I were to use a Schottky diode in series from pin output to gate, what value should I use?

    Thanks,
    Nicholas
     
  18. wayneh

    Expert

    Sep 9, 2010
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    The scheme provided by Picbuster (middle right box) is very thorough for protecting the MPU from any voltage placed on its output that is not within the power rails supplying it. However I would leave the pull-down resistor in place (gate to ground) to ensure against the MPU failing to "open", or "float", some undefined state. The resistor shuts off the MOSFET if that happens. It's a small risk but a very small cost for the insurance against it.

    As suggested, don't put any diode in series with your MOSFET gate. Leave only a current-limiting resistor in that position. Your 1k should be fine.
     
  19. Nicholas

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    Mar 24, 2005
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    Does it make any difference to the solution that I am driving a coil(probably) at the mentioned ~50V and 10A?

    The reason for not using a diode in series is the voltage drop, right? But with a schottky, this should still leave something
    like 4.6V or a little more, which should be enough to saturate a logic level MOSFET or?

    On picbusters scheme, titled Nicolas :), one schottky is going from the output pin to Vdd, what voltage is this? Is it the voltage
    used to drive the solenoid? (50-ish). Also, what exact value schottky diodes should I be using? By the way, thanks for the picture
    picbuster!!

    So, the zener method is not good then?
     
  20. wayneh

    Expert

    Sep 9, 2010
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    No, those two diodes go to the two power rails of the MPU. This ensures that the output pin voltage cannot get more than one diode drop higher or lower than the range between the power rails. Any voltage outside that range is then conducted by one diode or the other to the power rail itself, and away from the MPU output.

    Placing a zener on the pin to protect (as in #10) is commonly done for input pins and I don't see anything wrong with it here. As long as the output is normally below the zener voltage, I don't see the zener doing any harm and it does offer protection.

    I do not understand the objection to #10 from OBW0549. The zener in #12 would pop like a fuse if any significant voltage came along to challenge it. There's nothing to limit the current. Arrange as in #11, the 1kΩ resistor limits the current that the zener needs to sink to protect the MPU.
     
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