H bridge problem

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

  1. thatoneguy


    Feb 19, 2009
    Isn't that the function for the Vcc voltage input pin? I mean for the high voltage on the low side driver? It would switch between Vcc and ground, from looking at the block diagram.

    Assuming that, it could be any voltage that falls within the limits of the undervoltage lockout.

    COM would be -25V, Lo would go to the gate of the MOSFET, and Vcc could be Driver side Ground. Couldn't it?
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
    Actually the circuit is not weird at all. When used the way I drew it in the last two pictures of my previous posts it's just like a halfbridge where the other side of the load is held at middle potential between the bus bar voltages. 25V in this case.

    The -25V looks confusing at first and would be wrong if it were referenced to the same ground potential Vcc has.
  3. trimesh

    New Member

    Dec 20, 2011
    The thing is this isn't actually a H-bridge - it's half of one. The whole point of an H-bridge is that you only need a single supply to drive the load in either direction.

    The more conventional way of doing it would be to have basically two of the circuits you have described - but with the low side and high side drives going to opposite sides of the H for each. Then you just common the LD and HD pins on each chip and connect that to the SD pin on the other chip. This way, you don't get any shoot-through because as soon as you turn the drive on either chip on it disables the other and the turn-on delay in the low-side driver will mean that the gate capacitance in the upper FET has had time to discharge before the lower FET turns on.

    Of course, this is a lot more complicated and expensive - but looking at the parameters of your circuit (25V supply, 40R load) I would probably have chosen a part like the Unitrode / TI L293 that has 4 half-H-bridges (I.E. two complete bridges) in a single package and is rated at 1A / 2A peak. It also requires far less external components than the IR2110 does.
    strantor likes this.
  4. raychar

    Thread Starter Member

    Nov 8, 2011

    Sorry that I didn't mention about my circuit. Actually, it needs to run at +/-45VDC for the motor(as loading). I now test it with lower voltage and resistanc loading first.
    I understand that there are other H-bridge ICs like SN754410 to do it but can't meet these requirements. I understand that this can also be
    achieved by using single voltage and two IR2110 but it requires more components.
    For the splitted voltage, I heard form IR reply that this IC can do only in single voltage. But, I still hesitate as I found the schematic (see attachment) from the application note. Can someone help to explain if this schematic can also be applied to DC motor and in this voltage? The splitted voltage ground should not be connected to original circuit ground as someone already advised in this forum.

  5. praondevou

    AAC Fanatic!

    Jul 9, 2011
    Yes. the main problem in your initial circuit was the connection from load to the 2110 ground and the missing one from COM to the source of the lower MOSFET.
    Note that a net should have only one name, this is where the confusion started. -25V is actually Gnd, +25V is actually +50V and the other side of the load is not Gnd but the middle point between +50V and Gnd, i.e. +25V.

    It's then a simple halfbridge and should work.
  6. SgtWookie


    Jul 17, 2007
    Funny, I thought of that late last night, and was going to reply but I was simply too tired.

    Let's say that in the image you attached, HV1 and HV2 were both 25v.
    The point below HV2 would be approximately ground, and the point where HV1 and HV2 join would be approximately 25v, and the top of HV1 would be 50v. That would work.

    What you were trying to do is to go negative with the low side of HV2. That won't work.