Mosfet in SMPS

Discussion in 'Homework Help' started by bug13, Sep 24, 2012.

  1. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    [​IMG]

    as seen in picture, the circuit on the top left is a part of a circuit from a switching mode power supply, the circuit on the low right hand side is a circuit I draw.

    So why do we need 3 transistors to switch the circuit, when I can use only 2, or there is something wrong with my design?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    We sometime use this additional transistor to speed-up turn-off time.
    This BJT provides additional current to quickly discharge MOSFET input capacitance.
    And we do it almost for free without decreasing Rc resistance.
     
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  3. crutschow

    Expert

    Mar 14, 2008
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    Another problem with your circuit is that there is nothing to limit the current through the zener diode if Vin is greater than the zener voltage, and that would likely zap the zener.
     
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  4. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    That's how I understand, is it because the larger current drive the MOSFET turn on/off faster?

    In my circuit, the turn off speed is faster than the turn on speed?

    Please let me know if I'm wrong.

    PS: can I add a small capacitor across with the pull up resistor to increase the turn on time??
     
    Last edited: Sep 24, 2012
  5. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Thanks for pointing that out crutschow.
     
  6. Jony130

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    Feb 17, 2009
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    Yes, because more current faster charge and discharge MOSFET parasitic input capacitance.

    I think that sill turn on time is smaller then turn off time.

    But how this capacitor increase the turn on time ?
     
  7. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    my idear is, a larger in rush current goes through the pull up resistor, and turn on the mosfet faster.

    so why do you think the turn on time is smaller then the turn off time in my circuit, might be I still don't understand the circuit, judging by your last repley
     
  8. Jony130

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    Feb 17, 2009
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    But which resistor in this diagram is pull up resistor ? R1?
    And R1 has no effect on turn on time.

    [​IMG]

    It seems to me that Q2 provides lower resistance path for charging current than Q1 for discharge current.
     
  9. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Sorry, I meant the pull up resistor in lower right circuit of my OP.
     
  10. Jony130

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    I'm still confused, and I don't know which resistor you call pull up resistor (next time label all components on the diagram).
    I see only one pull up resistor connect between BJT collector and input voltage supply.
    The BJT base resistor don't speed-up the charging time.
    Also MOSFET gate behaves like a capacitor. So adding additional capacitor won't help to improve the situation. The only thing that help is to increase charging/discharging capacitor current.
     
  11. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    Using a simple resistor to pull the gate up will switch the FET very slowly. FETs have a lot of gate capacitance and require high peak current drivers to switch them quickly which is essential to keeping switching losses low.
     
  12. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Thanks, your explanation with Jony130's make me understand it now :)
     
  13. bug13

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    Feb 13, 2012
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    Sorry about the confusion, the pull up resistor I'm talking about is the one in parallel with the zener diode.

    My idea is to add an capacitor in parallel with the pull up resistor, when pull up, the large in rush current through the capacitor speed up the turn off time.

    It properly doesn't work, but I just don't seem to understand why my ideait doesn't work.

     
    Last edited: Sep 26, 2012
  14. Jony130

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    Feb 17, 2009
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    Let me try to explain this puzzle.
    The "normal" circuit look like this
    [​IMG]

    Where Cgs represents MOSFET input capacitance.

    Now let as try to switch-on the Q2 and see what's going to happen.
    Switching Q2 to ON state start charging Cgs capacitor.
    Cgs current flow from +Vcc ---> Cgs---> collector - emitter Q2---> gnd.

    [​IMG]
    And this charging current is high because Q2 transistor is saturated region. So our MOSFET is quickly ON, we have small turn-on time.
    So after short period of time our Cgs is full charge to almost 10V.


    Now let's see what happens when we turn-off the Q2.

    [​IMG]

    So Q2 is in cut-off stage. And now the only path for Cgs discharge current is through R1. So after we quickly cutt-off Q2 the MOSFET remains in ON state due to Cgs is sill charged to 10V. And Cgs slowly discharged through R1. So we have a very long turn-off time.
    I hope that now it's clear that adding a capacitor in parallel with the pull up resistor R1 will not help at all. What worse this additional capacitance increase turn-off time.
     
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  15. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Hi Jony130,

    Thank you for taking the time and cutting all those graph to explain it, I am totally get it now.

    Thanks a lot :)
     
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