H bridge problem

Discussion in 'General Electronics Chat' started by raychar, Dec 20, 2011.

  1. raychar

    Thread Starter Member

    Nov 8, 2011
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    Hello,

    Can someone help to see any problem in the attached schematic. It is supposed to drive a motor. But I connected a resistor for testing first. Before applying +/-25V, there are slight pulses in HO and LO, once they were connected, the gate driver blew out.

    Thanks,
     
  2. raychar

    Thread Starter Member

    Nov 8, 2011
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    Oh, here is the schematic,
     
  3. strantor

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    Oct 3, 2010
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    Are you sure the IR2110 is capable of using the split power supply? the datasheet shows the low side FET switching ground, not a negative supply. I have no idea if that's correct, just something I noticed.
     
  4. praondevou

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    Jul 9, 2011
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    The COM pin needs to be connected to the lower MOSFET source terminal, that's the gate signal return path.

    Why is there 10Ohm between the VS pin and the source of the upper MOSFET?
    Why is there a 10Ohm resistor at the anode of the 1N4148?

    Do your input signals have a deadtime?
     
  5. strantor

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    that won't create an shorts between -25V and GND within the IR2110?
     
  6. SgtWookie

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    Jul 17, 2007
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    No.

    Raychar,
    Besides the COM terminal not being connected to -25v/your lower MOSFET source terminal, praondevou mentioned dead-time.

    You can help to prevent shoot-through by using Schottky diodes across those 39 Ohm gate resistors, anode towards the MOSFETs. This makes the MOSFETs turn ON more slowly than they turn OFF.

    If you are trying to breadboard the circuit, forget it - that won't work.
    You need to follow the guidelines in the Application Note:
    http://www.irf.com/technical-info/appnotes/an-978.pdf
     
    strantor likes this.
  7. Audioguru

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    Dec 20, 2007
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    I added some contrast to your grey on grey image.
     
  8. trimesh

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    Dec 20, 2011
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    The fact that the external low-side FET is connected to a -25V supply will reverse bias the pulldown FET on the low-side driver in the chip, and the parasitic diode in there will effectively supply an effective 25V of gate drive to the external FET - so it will always be turned on.

    You could try putting a 25V zener in series with the gate - this would fix the "always turned on" problem - but you would likely still have problems with high-frequency operation because this wouldn't let the low-side FET in the driver discharge the gate capacitance of the output FET.
     
  9. praondevou

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    Mmh, the OP used the same ground symbol for the load as for the rest of the circuit. This would mean that the -25V voltage is measured against this ground. So it cannot be connected to the COM pin, that would be a short. The source of the lower MOSFET however needs to be connected to the COM pin of the 2110.

    I suppose the OP wants to invert the voltage at the load without using a Fullbridge.

    The only solution I see is using a halfbridge. (with the load connected to the middle point of two capacitors)

    So, what's the frequency you will be using?
     
  10. thatoneguy

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    Feb 19, 2009
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    From the Datasheet, Page 1

    [​IMG]

    [​IMG]

    [​IMG]

    I'm just posting these so all the info is up front. I'm not sure about the Zener on the negative side MOSFET, after looking at the internal construction.


    Shouldn't COM on the low side be tied to -25V? The ground on the output side is different from the logic ground - Lines on logic side, triangle on output side, so there should be isolation already.
     
    Last edited: Dec 20, 2011
  11. raychar

    Thread Starter Member

    Nov 8, 2011
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    Hello everybody and thanks,

    1) Adding diodes(1N4148) in parallel with gate resistors
    2) COM not tie to ground, instead to tie to souce of lower mosfet.
    3) delete the 10 ohm resistor which connects with VS.

    -Is it for sure the circuit will function then?
    -Can someone help to propose another type of circuit of H-bridge which IR2110 is applicable as I think +/-voltage may not be suitable in this gate driver??

    Thanks,
     
  12. SgtWookie

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    That's what has been said; he needs COM going to -25v.
     
  13. shortbus

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    Using a dual supply like shown, wouldn't the bottom mosfet also in effect be a high side switch? The gate would need to go 10V above the source. Its my understanding that a low side switch must go directly to common.
     
  14. SgtWookie

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    Jul 17, 2007
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    Interesting - it doesn't work with IR's own IR2110 model. I'm not certain if that is a fault with the model, or the way a real IR2110 would operate, as the model is a macromodel, not a component model.

    The IR2110 model is 540 lines long. I'm not anxious to try to "de-bug" it, as I don't have a real live IR2110 to compare it to.

    One thing that is not obvious is when the lower MOSFET turns on, the boost cap will charge to the supply voltage plus the lower rail voltage. If the lower MOSFET ever turns off, the upper MOSFET's gate will have more than enough of a charge to exceed the maximum Vgs.

    Adding a resistor in series with the diode and 15v Zener across the boost cap helps that out, but the lower MOSFET won't turn off in the simulation.
     
  15. strantor

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    yeah thats what I thought when I looked at it. it outputs 20V max to the gate, but that would be -5V. and -5v is below the low limit spec of -.3v. but i'm not sure what the "20V spec" is in reference to. maybe you could do it with 2 high side drivers. or maybe you can do it with just the one, as discussed. I don't know; I've never seen a high/low driver used with split supply like that before, so I am interested in how it turns out.
     
  16. praondevou

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    Jul 9, 2011
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    This is what the OP has right now. It can not work, since the -25V has the same reference as VCC. So he can not join COM and the lower source. Which is necessary for the MOSFET to function.
    [​IMG]

    IMO he needs either this:
    [​IMG]

    Or this:
    [​IMG]

    Did I think this through?

    EDIT: thinking again the last two pictures are the same thing...:rolleyes:
     
  17. SgtWookie

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    Jul 17, 2007
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    Look, guys - it's Vgs.

    You'll never want both MOSFETs turned on at the same time. That leaves the load for a current path to ground.

    There is no point in getting the lower MOSFET's source terminal above ground; that would in fact be undesirable.

    Since Vs will always be <= ground, Vgs will be able to go up to the output supply voltage.
     
  18. SgtWookie

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    Jul 17, 2007
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    praondevou,
    Your last two schematics are not the same thing.

    You have COM grounded in the middle schematic, and the source terminal is also grounded. That will work.

    In the last schematic, you have ground on COM, and -25v on the source, and the two are wired together, for infinite current flow between the two. Things would get very hot, very quickly.

    I tried putting -25v on COM and the source terminal in a simulation, and it didn't work - which somewhat surprised me.

    However, the model is a macromodel, not a component model - and I don't have an IR2110 here to test.
     
  19. praondevou

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    In the last picture I used a different ground symbol for the power part to indicate that these are completely isolated sources. There is no current flow possible between -25V and COM.

    The connections are exactly the same.:)
     
  20. praondevou

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    Jul 9, 2011
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    @raychar

    This is a weird application. ;) I don't know if it works, it should though. I don't have a 2110 here, I'm tempted to try it out.

    Can't you just use a full bridge?
     
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