5V down to 4.3V after diode...correct with TTL?

Discussion in 'General Electronics Chat' started by Nicholas, Feb 16, 2016.

  1. Nicholas

    Thread Starter AAC Fanatic!

    Mar 24, 2005
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    Hi guys

    I am making a MOSFET(logic level) circuit, and I was adviced(correctly :)) to put a diode in series with
    the controller and gate of the mosfet, to protect the controller. This results in the 5V being knocked to
    4.3V. I know this is normal, but is there any way to get it back to 5V before reaching the mosfet?

    I was thinking about something like a TTL with its own 5V supply....would that be a good way to
    get the voltage corrected? That should take the the 4.3V in and output 5V.

    Any special considerations? (there probably is)

    Thanks,

    Nicholas
     
  2. Picbuster

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    Dec 2, 2013
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    please provide schematic
     
  3. Papabravo

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    Feb 24, 2006
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    Interesting idea, but TTL is not known for it's ability to output 5V from a +5V supply. Read the datasheet carefully. A CMOS part has a better chance. With the diode inserted, how do you switch the MOSFET off? A schematic would be oh so helpful if you can manage it.
     
  4. Nicholas

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    Mar 24, 2005
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    Here you go! Thanks:)

    ttl.jpg
     
  5. Papabravo

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    Feb 24, 2006
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    A schematic usually has values for the components, or haven't you got that far in the design yet?

    A diode and a resistor, that makes your voltage level problem even worse. TTL still won't do what you want it to do. Please read the datasheet for a TTL part to understand why a TTL totem pole output cannot source current into a load and maintain the Vcc level. BTW that is not a standard symbol for a MOSFET. Did you just make that up on your own?

    I suppose you could make it a Schottky diode and the drop would be 0.2-0.3V

    I would be looking for a CMOS part with TTL thresholds. That's the ticket.
     
  6. Alec_t

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    Sep 17, 2013
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    .... or, you could use a NPN as a level-shifter/buffer between the controller and the FET gate, and invert the controller logic appropriately.
     
  7. Picbuster

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    Dec 2, 2013
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    One diode will clamp to vcc the other will avoid negative signals at Gate.
    The serial resistor R1 is needed to 'consume' when negative or > vcc+v_diode is applied
     
  8. AnalogKid

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    Aug 1, 2013
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    Or eliminate the diode. Unless the gate drive goes below -20 V, the diode serves no function. Other than catastrophic failure, the FET cannot back-drive the gate and cause problems for the uC. Based on the information so far, I disagree with the "correct" advice. Where did it come from?

    ak
     
  9. ErnieM

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    Apr 24, 2011
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    Just what are you trying to protect your controller from?
     
  10. Nicholas

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    Mar 24, 2005
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  11. AnalogKid

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    Aug 1, 2013
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    Other than post #2, which characterized the failure possibility as unlikely and was discussing a bipolar driver transistor, none of the other responses mention a diode. If you want protection without a series voltage drop during normal operation, place a 5.1 V or 3.3 V zener diode directly across the output of the uC, and a series resistor to the gate or base of the driver transistor. No voltage drop with a FET or appropriate current limiting with a BJT during operation, and the resistor and zener form a protection network for excess voltage coming in the out door.

    ak
     
  12. Nicholas

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    Mar 24, 2005
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    AnalogKid, something like this? :

    (Maybe the zener should point the other way :)

    zener.jpg
    Thanks,

    Nicholas
     
  13. AnalogKid

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    Aug 1, 2013
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    Reverse the zener.
    No need for the 10 K pulldown to GND
    Increase the series gate resistor to whatever is appropriate to limit the worst case zener current. For example:
    36 - 5 = 31V
    0.5 W max zener power / 5 V zener voltage = 0.1 A worst case zener current
    R = E / I = 31 / 0.1 = 310 ohms
    Therefore R = 1 K or more.

    ak
     
  14. Nicholas

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    Mar 24, 2005
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    AnalogKid, thanks! I'll get the breadboard working!

    Can you explain to me how the protection in that works?
     
  15. Nicholas

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    Mar 24, 2005
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    Cautious bump :)

    Anyone got an 'easy' explanation as to how the zener-method above works? :)
     
  16. AnalogKid

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    It is a zener diode, acting like one. Nothing special or tricky.
     
  17. ScottWang

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    Aug 23, 2012
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    If you don't know how a zener diode works then take a look this:
    [​IMG]
     
  18. ian field

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    Oct 27, 2012
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    For starters - you've got that Zener upside down.

    The problem you're trying to guard against mainly happens in SMPSUs - the chopper transformer may saturate and the drain current goes exponential. Usually the MOSFET fails completely short circuit all ways round, its the source resistor bursting open that lets the fault current to deck through the chip. That probably isn't going to happen driving a relay.

    A Zener has dynamic resistance, so a large fault current can develop more than Vz. The data sheet will specify a voltage not to exceed on any pin outside Vcc/Vdd, normally you'd use clamp diodes so the pin can't be pulled more than Vf further than either rail (Schottky barrier diodes have lower Vf). You can get fairly robust Vcc/Vdd clamp diodes to make sure the fault current doesn't pull the rail up in extreme cases - or you coud protect it with an SCR crowbar circuit.
     
  19. ian field

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    You can get MOSFET driver chips in an 8-pin package.

    Some can be fed from the higher voltage, that means you don't have to use a LL MOSFET.

    Using an optocoupler would give you complete isolation - the MOSFET could be a melted blob and nothing happen to the MCU.
     
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