(not so) Simple opto-coupler/MOSFET question :)

Discussion in 'The Projects Forum' started by Marcus2012, Mar 3, 2015.

  1. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    Hey again everyone

    Just a quick Q about opto-couplers. I need some for my modified sine wave generator but they will need to be able to handle high frequency switching. I've found a couple of Vishay ones that look fine @ 250kHz cut-off frequencies but..... the two I have found seem identical but they have different phototransistor configurations. Is this just a non-issue really and I just run opposite tracks? Apart from that they even have the same descriptions but different titles.

    Here they are, can some please tell me which of these is the best? If one even is.

    http://www.farnell.com/datasheets/1696905.pdf (This one does have a notice at the top that says "not recommended for new designs" anyone come across this before?)
    http://www.farnell.com/datasheets/7983.pdf

    This is my first time using these components so some advice would be greatly appreciated.

    Thanks guys
     
  2. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    yeah the pins are different but it doesn't matter as long as your circuit attaches as needed..

    "not recommended blah blah" means exactly what it says..

    Or basically..
    "Dont use this for a new design where you expect the part to be available for a while as we are obsoleting this soon enough and you won't be able to buy it and your purchasing department will be annoying you trying to find a replacement as your boss and sales department are yelling at you because that end of life product you chose can't be found and now your customers are pissed and production is mad because the line is now shut down"
     
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  3. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    Hi
    Thanks for the quick reply :D
    LOL yeah I can see how that situation would be very frustrating, probably happened before then I'm guessing haha.

    I'll have a think about those then depending on prices, I can alter my configuration. I'm using them to isolate the MOSFETS from the signal generator so am I right in assuming the CTR shouldn't be too important as MOSFETS are voltage driven? The CTR would be 100 - 200%.
     
  4. ronv

    AAC Fanatic!

    Nov 12, 2008
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    You need to be a little careful. While FETs are voltage controlled they do have gate capacitance that has to be overcome by the driver to make them switch.
    If they are very large the low current of the opto won't turn them on and off at the high frequency.
     
  5. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    Hey ronv and thanks :)

    I have been looking into these ones

    IRFB4710PBF
    and possibly these (but I think they're a bit hardcore for 12v 40A)
    IRFP4668PbF

    but I am unsure as to which capacitance value applies to my query though. The MOSFETs (both N and P channel) gate will be driven of the same supply as the source/drain terminals respectively in each channel device, the other potential worry here is the gate voltage will exceed the GS threshold value. Is this ok if it is by a small amount, say 10%? because then I could get +/- 10V threshold voltage versions rather than 5 but this could impede full saturation due to the voltage drop across the N-Channels.

    I attached the circuit diagram so you can see the application (the bulb is just arbitrary). Due to differences in power dissipation between P and N channels the Ps (Q10,12,13 &14) are in parallel but I may do this with the Ns anyway just to distribute the current (and maybe get cheaper MOSFETs :) ).

    Untitled.png
     
  6. #12

    Expert

    Nov 30, 2010
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    The way I see this, you're going about 54 KHz and 10k to 20k ohms is a really bad choice for a gate driver.
    The, "threshold" voltage is the beginning of conduction, not the end. Every MOSFET I know will be very happy with 12 volts on its gate.

    Warning, I am not the best person here on MOSFETs.
    What you need to be looking at is total gate charge at 170 nano-coulombs.
    That's 170 nanoamp-seconds, really quick, like 1.8 microseconds or quicker.
    From V2 = dV e^(-t/RC)
    7.5V = 12V x e^(-t/10k x 100e-12)
    7.5V/12V = e^(-t/1e-6)
    log .625 = -t/1e-6
    1e-6 log .625 = -time
    - 20.4 microseconds = - time
    20.4 microseconds = time

    See? Really bad compared to 1.8 microseconds.
    And that is the minimum current required for your mosfets to spend 10% of the time going up and 10% of the time going down. If you can, I would try for 1% rise time and 1% fall time, and that requires some hefty current for 180 nanoseconds. That's why gate drivers were invented. Go looking for a dedicated chip called a Gate Driver.

    Edit: I've already forgotten where I got 100 picofarads for Cgate,but it seems too small. That means my numbers are way conservative for the requirements you have. You need a gate driver to slap some serious current into that gate so the mosfet doesn't smoke.
     
    Last edited: Mar 4, 2015
  7. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    Thanks for the reply but "The main advantage of a MOSFET transistor over a regular transistor is that it requires very little current to turn on (less than 1mA)"
     
  8. MaxHeadRoom

    Expert

    Jul 18, 2013
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    Think of a Mosfet as voltage operated rather than current controlled as in a Bi-polar transistor.
    Max.
     
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  9. #12

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    Yeah. Let me know how that turns out.
     
  10. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    Well seems fine in simulation at 1 microsecond steps so yeah I guess I will let ya know
     
  11. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    I'm gonna be totally honest here, that didn't make a lot of sense but I couldn't ignore a potential flaw so I think I get it now I have read this.

    "The advantage of using gate charge is that
    the designer can easily calculate the
    amount of current required from the drive
    circuit to switch the device on in a desired
    length of time because Q = CV and I = C
    dv/dt, the Q = Time x current. For
    example, a device with a gate charge of
    20nC can be turned on in 20msec if 1ma is
    supplied to the gate or it can turn on in
    20nsec if the gate current is increased to
    1A. These simple calculations would not
    have been possible with input capacitance
    values."

    So thanks, I guess it would probably have fecked me up otherwise


     
  12. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Danger Will Robinson!

    While it will eventually turn on with 1 ma. of current it will take a long time (in electronic terms) to do so. The easy way to get the turn on time close is to use Nc as #12 suggests. The 170 Nc number indicates it will turn on in 170 ns if your driver can supply 1 amp. To get the time in your case divide 170 by .001 amps you get something like 178 usec. Since your frequency of 250 KHz is much higher than this it won't work, You need a FET driver with the opto going to it's input.
    Maybe your simulation used an ideal FET?
     
  13. #12

    Expert

    Nov 30, 2010
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    It appears that you redeemed yourself. I don't know whether another simulator shot itself in the foot or you were using it wrong, but I can say that simulators give a false sense of confidence to a lot of people.
     
  14. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    Ok thanks this starting to make sense slowly :) . Yeah I think it must be an ideal value then so I can't rely on that. I have found this MOSFET driver

    ICL7667

    Which I think is up to 1-1.5A output, so all good there if that's ok with the I/Os but just to check, I take the supply for this from my unregulated rail, is that right? Also am I right in assuming this is for N-channels only? I'm a bit confused with this concept for P-channels at the moment. Is it inverted so I need to overcome the capacitance to close the channel?

    EDIT: Found this for my P-channels
    LTC1693-5 so if I have it in inverted output mode the output should go low when the input goes high I think. Do I need pull up/down resistors with drivers?
     
    Last edited: Mar 4, 2015
  15. ronv

    AAC Fanatic!

    Nov 12, 2008
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    You can power the chip from your +12, but it wants to see a 5 volt input so you need to hook the optos to 5 volts. If you arrange your logic carefully you could drive the optos with the 74LS, but it cannot source much current so you need to arrange the logic so a ground signal turns on the diode.
    A negative voltage with respect to the source will turn on the PFET, while a positive will turn on the NFET.
    You also need to worry about the time when they are switching. The way it is now they will both (top and bottom) be on for a short period of time so the current will be very high for this short period. Google up shoot thru.
    What is the purpose of the diode in the ground line of the 7805?
     
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  16. #12

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    and don't forget to add a 10 uf aluminum capacitor and a 0.1 uf ceramic capacitor to the power terminals of each gate driver chip. They provide the current for that fast pulse.
     
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  17. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    Good point thanks, so I'll take the anode and collector of the optos from the same 5V rail and the driver Vcc from the 12V. I should be ok with the ICL7667 for my P-channels too if I have the input connected directly from my 5V, if I then pull the input to ground with the opto I'm hoping the current logic will suffice. I wasn't clear on this shoot through though because I've worked the logic to give a clearly defined OFF period. Let me read about that somemore see I can understand it properly.

    The diode, lol, that was a suggestion from a member, I didn't see the need but it seems harmless :)
     
    Last edited: Mar 4, 2015
  18. #12

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    It jacks up the 5 volt output to about 5.6 volts, and I don't know why this circuit needs 5.6 volts instead of 5.0 volts.
     
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  19. Marcus2012

    Thread Starter Member

    Feb 22, 2015
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    Ok thanks #12 and ronv I've been trying to apply what I've learned to my circuit diagram and I think I'm getting somewhere and I've attached it for critique. I wasn't sure on the need for pull up/down resistors with the drivers so the downs are still in it for now for the n-channels. I do hope I am getting this lol. Untitled.png Untitled.png
     
  20. #12

    Expert

    Nov 30, 2010
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    No pull up or pull down resistor can compare to the speed of a Gate Driver, so they don't need the resistors. If I had any suspicions, I would add a 10k resistor from gate to source of all mosfets just to stop them from accumulating a static charge when the board is unplugged.
     
    Last edited: Mar 4, 2015
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