High Side MOSFET Drivers

Discussion in 'General Electronics Chat' started by Wendy, Jan 31, 2010.

  1. Wendy

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    [​IMG]

    OK, how would you drive the two top MOSFETs in this design?

    I redrew the picture in my PWM in the book article, but I'm curious how it is done, and am pretty sure there is a simple solution.

    Here is the redrawn schematic I used to solve the problem for the illustration.

    [​IMG]

    I know this design has problems with shoot through, but I'm not worried about it at the moment.

    Thanks.
     
  2. MikeML

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    You cannot connect the PFets as you show. The body diode is always forward biased. PFet source has to be tied to +100V. Gate drive must be applied relative to the source, so must swing from +100V to +90V to control the gate.
     
    Last edited: Jan 31, 2010
  3. SgtWookie

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    Bill,
    With P-ch power MOSFETs, you connect the source terminal to the more positive supply. With N-ch power MOSFETs, you connect the source terminal to the more negative supply.

    High-side drivers require bootstrap caps; otherwise you won't be able to get a high enough Vgs. If the total voltage from +V to Vdd or GND will exceed about 15v, you have to start worrying about Vgs limitations; usually around +/-20v. An easy fix for that is to use a Zener diode, but it'll start bleeding off Vgs pretty rapidly once the gate is turned on; so you can't keep it turned on very long before bringing it back low again and recharging the cap. Otherwise, you'll need a supply higher than +V to get the gate high enough. Many gate driver ICs have built-in capacitive charge pumps for a supply.

    One easy way to connect the low-side logic to the high side is to use optocouplers/optoisolators. However, they have an additional problem; slow rise and fall times.

    Gate driver ICs are certainly the low parts-count way to go.
     
  4. Wendy

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    Opps. :rolleyes: :D It will be fixed.

    [​IMG]

    If the Source is +V or near to it on the top MOSFET, where will the voltage come from to turn it on? I believe you will need 10V across the Gate Source to do this, I don't see where the drive voltage is, or how you can make it short of doubling the Vcc voltage somewhere. For the sake of argument, Vcc is +12VDC.


    As for the offending schematic...

    [​IMG]

    Better?
     
    Last edited: Jan 31, 2010
  5. hgmjr

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    Bill_M,

    I like the alternate p-channel/n-channel solution for its simplicity. It may be worth mentioning in the description of such an implementation that care should be exercised in generating the control signals to avoid the potential for having both n-channel and p-channel in either side of the h-bridge from being turned on at the same time.

    hgmjr
     
  6. Wendy

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    OK, I was thinking something similar, but I didn't want to wander too far astray. I'll add something about shoot through.

    It's your fault though, I was fat, dumb, and happy until you mentioned high side drivers. I'm beginning to suspect it isn't as easy as I thought it would be. :D

    I have a different solution for BJTs that seems to work.

    [​IMG]

    I don't know how well it will handle higher frequencies, but the few times I've used it it seems to work. I developed the idea after I joined these forums, and figured out the down side to 555 darlington outputs.
     
  7. hgmjr

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    MIA CULPA.

    hgmjr
     
  8. MikeML

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    Bill,

    Stare at this. It shows some of the ideas you should consider as essential parts of an H-Bridge. I can send you the .asc file for it if you want to play with it, replacing transistors with fets, for example.
     
  9. SgtWookie

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    You're gonna hate this, because it isn't exactly simple ;)

    See the attached; it's just a model of a half-H bridge driver. The other half would simply be a repeat of the side that's shown, only reversed.

    There's a discussion of this type driver on this page:
    http://www.innovatia.com/Design_Center/High-Side Drivers.htm

    Note the difference between the low-side and high-side driver in my example; D3 and C3 have been added.

    Note that C3 is not charged until the low-side MOSFET has been turned on for a brief period of time. When that occurs, C3 charges through D3.

    After C3 is charged, when Q1 turns on, Q2 sources current to M1's gate via D1 & R19. Since the low side of C3 is connected to M1's drain, and C3 starts off being charged to V+ (12v)-Vf(D3), it acts like a simple charge pump.

    The problem with this idea is that C3's charge is rapidly drained via Q2's base through Q1's collector and via R3 to the drain of M1. Both are necessary. When Q1 turns off, R3 pulls the base of Q3 low to rapidly discharge the gate of Q1.

    You can't hold the high side MOSFET on indefinitely with this design; after about a millisecond, C3 is so discharged that the MOSFET gets into the linear region and rapid oscillations occur. Increasing the size of C3 is not a good solution, as the charge current via D3 will increase dramatically, and so will the ESR of the cap.

    Even with this relatively high parts count design and the aformentioned shortcoming of limited on-time, there are no protections against under-voltage, no shoot-thru protection, and a goodly area of the board will be taken up with all of these components. The addition of Zeners from the gate to source terminals will provide some protection against exceeding maximum Vgs, but will also increase the discharge rate of C3.
     
  10. SgtWookie

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    Variation on a theme - this one is a good bit more simple; see the attached.

    The disadvantage is the slow gate charge time due to them being charged via 1k resistors, which limits the circuit to fairly low switching speeds. I'm using a 1kHz square wave to drive the gates in this example.

    The Zener D2 really isn't necessary for 12v operation; I simply included it to show where it goes.

    It's necessary to use a pair of NPN transistors to drive M1's gate, as it's signal needs to be inverted from M2's gate. This causes a slight mismatch in propagation delay times between the two gates. One fix for that would be to use a 74x265 quad complementary output gate, but I don't think they make it anymore.
     
  11. hgmjr

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    Bill_M,

    You may want to avoid getting bogged down in the details and just provide a generic block and label it as h-bridge driver controller.

    hgmjr
     
  12. Wendy

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    I'm not too worried about how they drive it, that is fairly basic.

    [​IMG]

    If the power supply is 12VDC, and the load has 12VDC across it, where does the Vgs to turn on the transistor M1 come from? Your sim showed 24VDC pulses, where is the 24V coming from?

    There is probably a hole in my theory of operation, but I'm under the impression you need 10V between Gate and Source to turn the transistor on.

    I don't think I've seen any equivalents to emitter followers in MOSFETs.

    I'm off to read the other links.
     
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  13. Wendy

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    I've already finished the PWM circuit description by a simple disclaimer. It isn't about teaching H-Bridges, it is about PWM. I could get the design shown there to work if I had to.

    This thread is about my own education. There is something I don't understand, so I'm working on it.
     
  14. Wendy

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    OK, read through the article. If I understand it correctly there needs to be a section (that is not shown, labeled VGH) that is basically a voltage doubler (double the power supply voltage). It is the extra voltage that does the job. The zener prevents the gate from having excessive voltage on the Gate Source, and also prevents it from going too negative. Does this sound like a correct assessment?

    This doesn't explain the circuit shown in post #12 though.
     
  15. SgtWookie

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    The lower side of C1 is connected to the drain of M1. When M1's source terminal is at/near 0V due to M2 being on, C1 charges to nearly V+ potential.

    When Q2 turns off, the gate of M1 charges via R1, turning the MOSFET on.
    This causes the drain/source connection resistance to be very low; Vds becomes nearly 0 as Rds decreases. Thus, the source terminal's voltage increases to nearly that of V+

    Since the low side of C1 is connected to the source terminal, it's potential also becomes nearly V+. Remember, C1 was already charged to nearly V+ when the source terminal was at nearly 0v potential. Since Vd is now at nearly V+, and C1 was charged to nearly +V, the two add up to nearly 24v.

    Note that the voltage across C1 doesn't change much once it's charged (except for leakage through the Zener); just the voltage on the low side of C1.
    Take a look at the red plot of V(M1gate,Load) - that is actually the Vgs of M1. V(Load) in this case is equal to V(M1 source). In the lower plot, the grey-blue plot of V(M1gate) goes from 0v to 24v because it's referenced to ground, not the source terminal of M1.

    Think of it as a switched-capacitor voltage doubler.
    Does this make sense?

    For standard MOSFETs, that is true. Look at the red plot; you'll see Vgs vary from 0v to around 11v.

    Well, power MOSFETs are really best used as switches. However, I suppose you could use a power MOSFET with a very low threshold voltage as a voltage follower; but it would probably be much more noisy than a bjt.
     
    Last edited: Jan 31, 2010
  16. Wendy

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    Blew my mind there, you picked the other schematic to explain.

    [​IMG]

    Off to the side, you show V+ as being the power supply plus something (V2?). I think this is where I'm having problems.

    Where does the extra voltage come from if you have a +12VDC power supply?

    C3 pumps the Q2 collector to the +24V level, which is what turns on the gate of M1. True statement?

    I think I see it now.

    If I read this correctly this circuit depends on being switched, if it is on any length of time C3 will discharge, defeating the purpose.

    /
     
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  17. SgtWookie

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    Sorry about the mix-up - just substitute C1 for C3, and the explanation is the same.

    OK, here's a relay switching the ground side of a cap. Same thing as what's happening with the MOSFET:

    [​IMG]

    Initially, the cap isn't charged; +12v is on both sides of the cap. When the relay's coil gets energized, the low side of the cap gets switched to ground. I added R1 so you can see the RC charge curve.

    When the relay's coil gets de-energized, the low side of the cap gets switched to 12v; and the cap is still charged to 12v. So, the potential across the cap hasn't changed, but since the low side of the cap is now referenced to 12v, the high side of the cap measures nearly 24v to ground (less the Vf of the diode, of course)
     
  18. SgtWookie

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    (image omitted)
    Added image:
    [​IMG]

    I've corrected my prior post: C3 -> C1

    "V2" is simply the default reference designator for the signal source. "V1" is the V+ 12v supply, V2 supplies a 1kHz 0v-5v square wave that I simply called "toggle" for no particular reason, other than it toggles between 0v and 5v.

    I really didn't mean for it to be confusing - I should've labeled "V2" as "555" instead, and "Toggle" as "555out". Then you would've known exactly what I was simulating. Sorry about that.

    The low side of C1 is tied to the source terminal of M1; the high side of C1 follows the source terminal voltage, offset by V(C1). Since V(C1) is about 12v, and Vs(M1) changes from 0v to nearly 12v, the voltage on the high side of C1 referenced to ground becomes nearly 24v.

    Ah, no. :)

    By George, I think you're getting it! :D

    Yes, the high side needs to be cycled in order to recharge C1. Even if the Zener weren't in the circuit, you'd have leakage in the transistor and C1 itself, along with some very minor gate leakage current. Eventually, the MOSFET gate voltage would decrease, Rds would increase, and then you'd have lots of problems. The "eventually" won't be very long, BTW.

    As I mentioned before, many gate driver ICs have built-in charge pumps that keep the equivalent of C1 charged continually. If you really wanted to, you could build a 555 charge pump to keep the cap charged up (and I know you really want to... ;) )
     
    Last edited: Feb 1, 2010
  19. Wendy

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    I do a lot of thinking whilst I drive, I had the same thought. The zeners keep the voltage from exceeding the safe ranges (in both directions) of the MOSFET gate. A 555 voltage doubler or tripler would drive a MOSFET mostly OK, though current surge could be a problem.

    Answers another question I've been mulling over, how to drive a N-Channel MOSFET as a series pass element for a positive power supply, only way I could see it before was with a P-Channel.

    Thanks.
     
  20. MikeML

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    I think its called "driving while distracted" :D

    To use an NFet as the pass device, put it the negative lead of the supply...
     
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