How to choose which component to use

Discussion in 'General Electronics Chat' started by Questioner1.0, Jul 22, 2013.

  1. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    When one is working on a project that requires the use of an IGBT, how does one choose which driver to use?
    What information is needed?
    Where does one look to find good drivers?
    Does it have to be an IGBT driver or will a MOSFET driver work? This question is more along the lines of curiosity but I'm having a little bit of difficulty with understanding the exact differences between a power MOSFET and an IGBT besides the first needed a certain current to turn on and the other requiring a specific voltage level
    Thanks in advance for any answers
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Both MOSFETs and IGBTs have have a similar high impedance input that looks like a capacitor and require a specific voltage level to turn on. It's a bipolar transistor (BJT) that requires a current (as well as a low voltage) to turn on.

    The driver required depends upon the how fast you want the IGBT to turn on and off, and the input capacitance of the IGBT. The driver must have the current drive capability to charge and discharge the input capacitance at the speed you need.

    Edit: Google "gate driver IC" for appropriate driver circuits.
     
  3. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    I'm probably completely missing it, but where do you find the current drive capability? I'm not completely sure that I know what it is that I'm looking for.
    I attached the datasheet for the driver that I'm looking at, so maybe you could point me in the right direction?
    Thanks
     
  4. BobTPH

    Active Member

    Jun 5, 2013
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    Does this help (first feature bullet in the datasheet)

    • High Current Output Stage: 1.0 A Source/2.0 A Sink


    Bob
     
  5. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    so a current output of 1A source/2A sink should be enough to drive an IGBT with input capacitance of 2650 pf at 50 to 200 Hz right?
     
  6. BobTPH

    Active Member

    Jun 5, 2013
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    At 50 to 200Hz the switching time is not that critical. But do the math.

    1 A will charge 2650p to 10V in 26.5 nSec.

    Bob
     
  7. crutschow

    Expert

    Mar 14, 2008
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    The formula is [t = (V*C) / i] where t is the time, V is the change in voltage on the capacitor, C is the capacitance, and i is the current charging (or discharging) the capacitor.
     
  8. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    where did that formula come from? It looks like I need to go back and review all my basics
     
  9. Papabravo

    Expert

    Feb 24, 2006
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    It comes from computing the voltage across a capacitor of a given value. The charge(voltage) on a capacitor is proportional to the integral of the charging current.

    http://lmgtfy.com/?q=Voltage+across+a+capacitor

    Follow the top result and scroll down to the section titled Current-voltage relation
     
  10. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    what does the input current values mean? Do they have an impact on how the chip works if you exceed the values listed?
     
  11. Potato Pudding

    Well-Known Member

    Jun 11, 2010
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  12. Papabravo

    Expert

    Feb 24, 2006
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    If you exceed the input current specifications for a driver chip it will stop working. Give us an example of such a specification, symbol and any text that goes with it. Better yet provide a link to the datasheet and tell us which page you're looking at.
     
  13. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    page 2 near the bottom, under Logic Inputs. It states values for the input current in the high state and the low state.
     
  14. Papabravo

    Expert

    Feb 24, 2006
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    Sorry - missed the original link to the datasheet

    Those are currents are required by the input stage. As long as the driving source can supply those currents there should be no problem establishing the correct logic levels. The amount of current required by an input is not something you can control externally; it is an intrinsic feature of the input.

    Where this might be a problem is if you tried to put a 1 Megohm resistor in series with the input. In that case only a fraction of a microamp would flow through the resistor and you would not be able to quickly establish the desired logic level. The other thing to look for is the presence of protection diodes on inputs. If you connect the input so say a 12V car battery, or a negative voltage source, the protection diodes will try to clamp the input to Vcc or Vee minus a diode drop, and a significant amount of current will flow. If enough current is pulled through the protection diodes you will fry the chip.

    The presence of these protection diodes is indicated by the voltage specification under Maximum Ratings
    Logic Input Vee - 0.3 to Vcc
     
  15. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    so if on the low state of the input stage there is no current then it won't be able to get the correct logic level?
     
  16. #12

    Expert

    Nov 30, 2010
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    It is much more difficult to answer, "What happens if I do not follow the instructions?" than it is to answer, "What do I need to do to follow the instructions correctly?".
     
  17. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    That is a good point. So how do I go about making sure that the input is getting the right current and neither not enough or too much?
    As it stands right now I'm getting no load in the low state and about 1mA in the high state.
     
  18. BobTPH

    Active Member

    Jun 5, 2013
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    By putting the correct voltage on it. All that spec is saying is how much current will be drawn by the input. The inputs are basically voltage driven, not current. These can be ignored unless you are trying to drive many inputs from a not very capable output.

    Bob
     
  19. Questioner1.0

    Thread Starter New Member

    Jul 20, 2013
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    Oh ok thank you
     
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