IGBT flash driver

Discussion in 'The Projects Forum' started by auvie, Oct 9, 2013.

  1. auvie

    Thread Starter New Member

    Oct 8, 2013
    25
    0
    Hi all,

    I'm designing a camera flash unit that will use an IGBT to control the pulse-width of the flash. I came across a part from Toshiba (GT8G151, data sheet at http://www.semicon.toshiba.co.jp/info/docget.jsp?type=datasheet&lang=en&pid=GT8G151) that appears to be specifically designed for the task, and I'm currently trying to make sense of the example gate drive circuit they provide (see attached). I'm new to IGBTs and FETs, and so I'm hoping someone here might be willing to shed some light.

    First off, I'd like to control the IGBT with an AVR (probably an AT90PWM3B), which isn't able to source more than 20mA from any one of its outputs. The gate resistor that the GT8G151 data sheet seems to like is 62ohms, which would source close to 50mA at 3V -- it seems the AVR really shouldn't be used to drive the IGBT's gate directly. Given the large amount of current the IGBT is handling (pulses of 50A or more), there's probably a lot that can go wrong, so I have a feeling this is not the only reason.

    And so we have the example gate driver circuit, which has two FETs and a BJT. I think I understand the basic operation: A high signal activates the BJT (with the 1.2k resistor causing just a small amount of current to be sourced from the signal provider, i.e. AVR) which in turn deactivates the N-ch FET and activates the P-ch FET, allowing the 3.3V supply to reach the IGBT (through an unnamed resistor that I assume is the gate resistor and should be 62ohms); A low signal does the opposite, deactivating the BJT and subsequently activating the N-ch FET, deactivating the P-ch FET, causing the IGBT to see ground at the gate and break the xenon lamp circuit.

    There are several things I don't understand, including the function of both the 20k resistor and the 470ohm resistor... why are they there, some sort of protection?. I also don't get the choice for 910ohm and 91ohm resistors... some kind of voltage divider? And finally, I can't figure out why the FETs are even there in the first place, instead of just connecting the BJT collector to 3.3V, and the emitter to the IGBT, through the gate resistor... perhaps the FETs are also added protection?

    Any help would be appreciated, as I want to understand the circuit fully before I incorporate it into my design!
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    An IGBT like a MOSFET has a near infinite gate input resistance since it looks like a capacitor with significant capacitance (as stated in the data sheet). Thus you need to charge and discharge this gate capacitance quickly to obtain rapid switching of the IGBT. That's the purpose of the two complementary MOSFET push-pull drivers. They can generate the large current pulse needed to charge and discharge the IGBT gate capacitance rapidly.
     
  3. iimagine

    Active Member

    Dec 20, 2010
    129
    9
    The 470 resistor helps turn off the transistor faster. Same as the 20k
     
  4. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    For fast switching you should use a fast saturated-switching transistor such as the 2N2369 for the NPN bipolar on the right.
     
  5. auvie

    Thread Starter New Member

    Oct 8, 2013
    25
    0
    Thanks guys for your input! It all makes more sense now. The 910ohm and 91ohm resistors are starting to make sense to me as well:

    1)when the trigger input is high, the N-ch FET's gate is pulled to ground to disable it, and the P-ch FET has a 91ohm connection to ground, and a 910ohm connection to 3.3V -- the link to ground is "stronger", causing it to enable.
    2)when the trigger input is low, the P-ch FET sees 3.3V through 910ohm and disables, however the N-ch FET only seems to see 0.3V, but still manages to enable? That part is still a little fuzzy to me.

    crutschow, I'll certainly look into the 2N2369. Do you have any additional recommendations for the n-channel and p-channel FETs? Does the 910ohm/91ohm resistor combo influence the decision?
     
  6. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    2) Why do you say the N-MOSFET only has 0.3V? :confused: Both transistor gates are at 3.3V when the BJT is off so the N-MOSFET is ON.

    The MOSFETS can be small devices (say 1A-2A current rating) but they need to be logic-level types that can fully turn on at 3.3V (that's not the threshold voltage which will be much lower). International Rectifier, Infineon, and Fairchild are good sources of such devices.
     
  7. iimagine

    Active Member

    Dec 20, 2010
    129
    9
    2) when the trigger input is low, the gate driven transistor is cutoff so there is no current flow from its collector to emitter therefore it's collector will be at 3.3V

    As for the 910/91 voltage divider I have absolutely no idea why it is necessary, You can remove the 91ohm resistor and the circuit will still run the same .
     
    Last edited: Oct 10, 2013
  8. auvie

    Thread Starter New Member

    Oct 8, 2013
    25
    0
    okay, I get it now. I was a little confused as to the voltage seen at the collector when the BJT has no current flow. It all makes sense now.

    Indeed, it does seem like there is no need for that 91ohm resistor.

    Thanks again to you both!
     
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