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MOSFET Gate Driving Problems

Discussion in 'General Electronics Discussion' started by BlackMelon, Jun 24, 2014.

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  1. BlackMelon

    BlackMelon

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    Aug 7, 2012
    Hello,

    Right now I'm doin a gate driving hobby.... My target is to replace the gate driver with some appropriate discrete components (Of course, I don't want it to be as perfect as the commercial gate driver but not so lousy)

    As the circuit shown in the picture, I want to know that why the voltage signal at Q(A)'s collector is so ugly?.... (I saw this from an oscilloscope), I've tried to reduce gain to get a higher bandwidth by reducing the its collector resistance.. but it helps just a little bit.....

    PS: Voltage at Q(B) shown some capacitor charging in the rising edge but it doesn't distort the signal from the square wave so much
    PS2: The MOSFET gate signal has some parts similar to the Q(A)'s collector.... some is similar to the Q(B)'s collector to... I guest it's the summation of the two signal.


    (The circuit is redesigned from http://tahmidmc.blogspot.com/2012/12/low-side-mosfet-drive-circuits-and_23.html)

    BlackMelon
     

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    When the output of the optocoupler is low, QB is turned off via R3 and QA is turned on via R2. Thsi works fine.
    When the output of the optocoupler is high, QB is turned on via R1 and R3, that should be o.k, too. QA, however, is not turned off completely. The voltaeg at the output of the optocoupler is Vopto=(Vdrive-VbeQb)*R3/(R1+R3). This is considerably lower than Vdrive and therefore QA is biased into a conducting state.

    You can use either 2 optocouplers, one for each transistor, or decouple QA and QB by another transistor.
    Adding a suitable zener diode in series with R2 may also help (by reducing the voltage available to drive QA via R2), can't check this right now in detail.
     
  3. BlackMelon

    BlackMelon

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    Aug 7, 2012
    Thanks a lot Harald, you help me much! ^_^
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    As Harald says, the main problem is that when the optocoupler transistor is OFF, its collector will not rise all the way to +V_DRIVE because of the voltage divider effect caused by R1, R2 and R3, and QA will not turn OFF.

    You can test this out using a simulator such as the free LTSpice (http://www.linear.com) using a transistor for the optocoupler, driven with a square wave voltage generator through a resistor.

    You will see that the optocoupler's collector voltage does not swing over the whole voltage range. Then you can try changing resistor values and topologies and see the effects before you build it up. Of course simulation is not perfect, but it's surprisingly good. It certainly would have shown you this particular problem, and it will show you many other kinds of problems too.

    There are various ways to fix this problem. If +V_DRIVE is a fixed, known voltage, changes to resistor values and the addition of a base-emitter resistor on QA will be enough.

    You could use a two-transistor "bipolar emitter follower" buffer between the optocoupler and the base resistors - the circuit is like the QA/QB circuit but with NPN/PNP swapped, collector/emitter swapped, both bases connected together, and both emitters connected together (no need for series resistors like R4 and R5). You would also need base-emitter resistors on QA and QB because the bipolar emitter follower buffer doesn't pull fully up or down.

    Or of course, you could use a logic gate as a buffer.

    Generally though, I would use base-emitter resistors on both QA and QB. With base-emitter resistors, they will switch at a voltage closer to mid-way between +V_DRIVE and 0V where the input signal will be slewing more quickly, and they will turn off more cleanly, so the "stair steps" in the gate drive waveform will be shorter.

    Finally, why have you made R4 and R5 so high? R6 is meaningless when the driver has such a high output resistance. Cross-conduction in QA and QB will only be very brief. I would use something like 22 ohms for R4 and R5, and delete R6.
     
  5. BlackMelon

    BlackMelon

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    Aug 7, 2012
    In order to make the voltage swing at the optocoupler be full range (+V_DRIVE and GND), I'll have to increase its gain by increasing collector resistance, right? (I'll check the bandwidth too because increasing the collector resistance will narrow the bandwidth... I want this circuit to work on 20kHz PWM)

    If we use base-emitter resistors, will the gain of BJT be decreased? I'm still curious in your sentence "they will switch at a voltage closer to mid-way between +V_DRIVE and 0V", Did you mean the output voltage? If so, the MOSFET will be unable to turn off by PWM because the "low" state is higher than 0V so the 0% duty cycle will not shut the MOSFET down

    Oops! I forgot to delete R6 from the diagram lol...... You're right about putting R4 and R5.... If the cross conduction is very brief, I'll short them out :)
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

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    A simple Zener diode can cure a lot of symptoms :)
    [​IMG]
    Without the zener, Q3 will never be completely off. With zener it is.

    Note that Q1 is only a simplified simulation replacement for the optocoupler.
    Note too, that I have used your resistor values. Heed Kris' comments on the high values of the output resistors, use lower values in your circuit.
     

    Attached Files:

    BlackMelon likes this.
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    No. The optocoupler can only pull down to 0V. R1 is there so the output will pull up. So look at the circuit with the optocoupler completely removed. You have R1 from +V, R2 from QA's base, and R3 from QB's base. QA's base is never going to be more than 0.7V less than +V. QB's base is never going to be more than 0.7V higher than 0V. So you immediately have a voltage divider formed by R2 and R3 that will bias the junction (the collector of the optocoupler) at around half of +V. R1 will pull that junction towards +V, but not far enough for it to be less than 0.7V below +V, which is what you need in order for QA to turn OFF.

    Read my description a few more times. Also, Harald's suggestion of a zener diode in series with R2 is a good idea. I would also use base-emitter resistors on both transistors though.
    Yes, but that's not a problem. How much current are you expecting to get out of them? With R4 and R5 at 500 ohms, the transistors will saturate and the output current will be limited by R4 and R5.
    No. The transistors switch in response to voltage changes at the collector of the optocoupler. For cleanest switching, you want them to switch when that voltage is close to half of +_V, or lower. As that voltage approaches +V, its slew rate drops. So you want to keep the switching point below half of +V. Somewhere between half of +V and around 1.5~2.0V would be best. You'll see that the optocoupler output slews fastest around that range.

    That's why I suggest using base-emitter resistors, and why Harald suggested using a zener diode. In fact if your +V is fixed and known, you can add speedup capacitors to make the transistors switch extremely quickly.
    You've missed my point again. Don't short them out. Just reduce them.

    What's your +V_DRIVE voltage?

    Have you considered using an optocoupler with a digital output such as MOC5007/H11L1?
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I meant to post this earlier:

    mosfetdriver.png

    This is a fairly simple mosfet gate driver. Note that it is inverting.

    The gate current for turn-on is defined by R1, the gain of Q1, and the voltage that the gate plateaus at whilst turning on.

    The turn off current is similarly defined by R2 and the gain of Q2, and the input voltage.

    The power supply should not exceed VGS(max), assuming the mosfet's source is connected to ground.

    This circuit draws current when the mosfet is off, both from the power supply and from the signal source.

    A good starting point for R1 is 1k, and for R2 470 ohms (this will give you half an amp or so of base current for a typical mosfet).

    There are a number of improvements you can make, but they tend to reduce the simplicity of the circuit.
     
    KrisBlueNZ likes this.
  9. BlackMelon

    BlackMelon

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    Aug 7, 2012
    Harald, I appreciate in your zener diode... As my understanding, it will dramatically reduce the current flowing through R1 when the diode is reversed bias (when we want the QA to be off which is the problem that I asked you) The reverse voltage of zener will do the job... however, how is method different from increasing the base resistor of QA that Kris was talking about? I just want to know the advantages of both solution...

    Steve, The picture below is your circuit which has been separated into two case according to turning D1 ON and OFF...
    the problem is when the voltage of the collector of Q2 go high (May be 10V) Can we guarantee that the diode D1 is OFF?
     

    Attached Files:

  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Your simplification is wrong.

    the diode provides a path from the gate to ground via the transistor. I'll draw you a simplified circuit when I'm in front of a real computer.
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    When the input is low, Q2 is turned off, so the circuit is effectively this:

    Mosfet on.png

    Current flows via Q1 to the gate.

    When the input is high, Q2 is turned on and Q1 is turned off. The circuit is effectively this:

    Mosfet off.png
    Current flows to ground from the mosfet gate via the diode and Q2.

    You might ask, "Can both Q1 and Q2 be turned on, shorting out the power supply?"

    Mosfet driver.png

    The answer is no. Here's why:
    • For Q1 to be on, point A must be a higher voltage than point B (it must be about 0,6V higher)
    • For D1 to conduct, Point B must be a higher voltage than point A (it must be about 0.6 V higher)
    • The above two cases cannot be both true at the same time.
    Q1 turns of slightly before D1 starts to conduct when the input signal rises.

    Q1 turns on slightly after D1 stops conducting when the input signal falls.

    In addition to that, because the current through Q1 and D1 fall to zero almost immediately, there is little stored charge, so they switch rapidly.
     
  12. BlackMelon

    BlackMelon

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    Aug 7, 2012
    Right now I'm doing a simulation about transient response... I wanna compare between signal transistor 2N2222A and power transistor TIP41C... The circuit below is simulated and I've found that the response of TIP41C is so ugly >_<

    (The graphs' color in the simulation plot are the same as the probes' color in the circuit)

    My questions are:
    1) Should we use 2N2222A to be a MOSFET gate driver? (Its maximum collector current rating is around 1A and I think in the gate current in practical might be around half of an ampere)
    2) Could you explain me about the baker clamp? (http://en.wikipedia.org/wiki/Baker_clamp) I want to know when the diode D1 in the baker clamp is turned on (I want to know when the voltage at its anode is greater than its cathode) so I'll be able to identify the time that D1 clear out those excessive charges which are the problems in fast switching Circuit.PNG Simulation.PNG
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Use a 2N2222, it will have a higher gain at the currents you're interested in (you can limit current using RG

    I see no Baker clamp in my circuit. I'm also not aware of any reason why it might help.
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    A baker clamp effectively places a limit on VBE, not allowing it to exceed VCE (or a voltage based on VCE).

    In the circuit I have given you, Q1 would not benefit from a Baker clamp, but Q2 probably would. You would have to consider the effect of a larger VCE on the turn off time for the mosfet, and ensure that the mosfet is in cutoff significantly before VGS of 1.2V (so no logic level mosfets).

    It would be interesting to compare the switch on latency as well as the rise and fall times with a baker clamp on Q2.

    Can you try it and report back?
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    This might be useful reading.
     
  16. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I agree, a Baker clamp on Q2 in Steve's circuit in post #11 is a good idea if you want Q2 to switch OFF quickly. The Baker clamp would prevent Q2 from saturating, and the associated delay on switch-off; that's what a Baker clamp does.

    As Steve pointed out, if you use a Baker clamp on Q2, when the control signal is high and the MOSFET is supposed to be OFF, the circuit will not pull the gate voltage down so close to the 0V rail. In fact it may be as high as 2.5V, approximately. So you need to make sure that the MOSFET's VGS(th) voltage (gate-source threshold voltage) is guaranteed to be at least 2.5V otherwise you risk not turning it OFF fully.

    To speed up the turn-off of Q2 I suggest adding a resistor between Q2's base and emitter, and possibly a small capacitor across the base diode if the switching is continuous. And make sure that whatever drives the driver can pull high and low fairly strongly.
     
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    A capacitor across R2 would also help, assuming the signal is driven from a low impedance source. Choosing the capacitor is a non-trivial problem
     
  18. BlackMelon

    BlackMelon

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    Aug 7, 2012
    The result of comparing all ideas here is very interesting!! As you can see, the green trace is a driving signal which is 5V square wave..... The result from baker clamp is very lousy! (blue waveform) so I'll be investigating the result further. Now, let's compare the yellow waveform (capacitor across the base resistor) and the red one (normal gate driver circuit).......

    As I zoom in to see rising edges of them, adding capacitor across the base resistor provides an undershoot while the original gate driver circuit has no problem to deal with....

    This is just a result from OrCAD Pspice.... Please try the other simulation programs so we can discuss further :)
     

    Attached Files:

  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I agree with the comments about a simple baker clamp in this instance. I do it without looking at your circuit or results, but on the basis that I have been simulating the circuit I posted. In my case the drive was via a comparator with a very weak pullup (22k) though.
     
    BlackMelon likes this.
  20. BlackMelon

    BlackMelon

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    Aug 7, 2012
    Steve, are you gonna say that capacitor across base resistor works on your case? (very weak pullup)

    PS: I'm still curious in your capacitor across base resistor.. I guess that if the driving frequency go higher, the impedance of capacitor will be lowered and the current will flow through capacitor much higher than through the base resistor. That will make the transistor to be able to switch fast. Am I correct?
     
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