IR2111 Gate drive

Discussion in 'The Projects Forum' started by Marie, Jan 28, 2008.

  1. Marie

    Thread Starter Member

    Nov 20, 2007
    24
    0
    Hi,

    I'm currently trying to build an active power filter. Part of the project consist of mounting a three phase inverter. I'm using MOSFET as switch and to ease implementation, i have chosen IR2111 half bridge driver from International Rectifier to drive the MOSFETs.

    Unfortunately, I've up to now fail to understand its operation. So, right now, i just trying to get an output without bothering about my system.

    Thus, I've tried to connect the circuit using the specification from IR-AN978. However, it did not work when I connect both the upper and lower switch.

    I've tried to use only the low side switch, but then again no output.

    I've attached a schematic of the circuit i tested. I'ld especially like to know if i'm connecting the power and control circuit ground correctly.

    Thanks n best wishes
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    In Figure 1, as soon as you get the MOSFET M1 turned on, it will place a direct short across the battery. You will likely see smoke and hear a loud "POP" as the MOSFET self-destructs, or the battery explodes.

    The load resistor needs to be between the drain of the N-channel MOSFET and the + supply, not across the MOSFET.

    As it is, the load resistor only conducts current until the MOSFET is turned on.

    Correction to my post - in figure 2, the load SHOULD BE connected between the two MOSFETS. In this case, at no time will both MOSFETs be conducting. However, you will also need a capacitor across the load to absorb transients (spikes) from inductive loads.

    I'm not familiar with the IR2111. Does it have charge pump for Cbs? 330uF seems offhand like a rather large cap; is that what the application note recommends?

    Added: from the datasheet; recommended connection for the IR2111.
     
  3. Marie

    Thread Starter Member

    Nov 20, 2007
    24
    0
    Hi,

    Sorry, I wrongly stated the value of the bootstrap capacitor. In fact, it is 330nF.

    I've posted a revised schematic. Did I get it correctly?

    However, I don't understand why there will be a short circuit. From what i've understand, both Mosfets are not supposed to turn on at the same time. Why then will the supply be shorted?


    I've also posted an abridged version of the recommendation of the bootstrap circuit from IRF.

    Thanks n Regards
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    OK, in your original post, the attached document had "Figure 1" and "Figure 2". In Figure 1, there was only one MOSFET, and the resistor was connected from the high side of the MOSFET to ground. In effect, you had replaced the high-side MOSFET with a short.

    In that situation, the lone low-side MOSFET should never be turned on.

    I see you've moved the batteries around. What is it exactly that you're intending to do? Active power filter? Well, power MOSFETS basically have two states; ON and OFF. They can be made to run in the partially conducting state, but it is not good for them.
     
  5. Marie

    Thread Starter Member

    Nov 20, 2007
    24
    0
    Ok I get what u meant.

    Basically, the active power filter will be used to eliminate harmonics from power line.

    I'm going to sense the harmonics load current and after some signal conditioning, generate an equal but negative polarity harmonics current. I then compare the current with a triangular waveform to generate a PWM signal.

    The PWM signal is used to drive the MOSFETs of the inverter.

    The inverter is a three phase PWM voltage source inverter which is connected to an inductor to generate the current source. This current is then injected into the power line to eliminate harmonics.

    The circuit of the gate drive I attached before, was only a way for me to test whether the driver works since no one I knew had ever used it n i couldn't understand the datasheet.

    Hope i managed to explain myself well.

    I've posted the design of the inverter.

    Thanks
     
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