voltage divider or regulator?

Discussion in 'General Electronics Chat' started by LaurenceR, Jul 16, 2014.

  1. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    I have built a high current driver board that can handle 12 to 24 volts dc up to about 30 amps. To turn the mosfet on you have to place a voltage on the gate. At 12 volts I can simply take the same supply voltage, pass is through a 100 ohm resistor and turn the mosfet on. At 24 volts this would be too high a voltage for the gate as the spec sheet says max gate voltage should be +-20 vdc. I was thinking of using two 2000 ohm resistors to form a divider and cut the voltage in half giving 6 volts for the 12 volt supply and 12 volts for the 24 volt supply. I tested the turn on voltage with a 100Kpot and found that at 2 volts the 12 volt supply allowed max current and at 24 volt supply 4.1 volts delivered max current draw. there fore a minimum of 6 volts should turn on the mosfet completely with either supply. Since the gate doesn't theoretically draw any current I wasn't sure if I would be better off with a voltage divider or a linear voltage regulator, want to draw as little current as possible as this is a solar application that can't waste current.
    Thanks in advance for any insights.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Using a resistor divider to drive the gate is fine. You can even go to higher value resistors if you want to save a little more current. But you want to drive the transistor with a high gate-source voltage to fully turn it on, 10V being a typical value. That will dissipate the minimum power in the MOSFET when it is on.
     
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  3. Fibonacci

    New Member

    May 23, 2014
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    Usually the gate-source voltage is between 10 and 20 volts, conect a voltage divider to do this and maintain the 100 ohms resistor to prevent oscillations due to parasit inductances and capacitances in the circuit.
     
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    The Thevenin equivalent impedance of the divider can affect efficiency. The higher the impedance, the more slowly the MOSFET will turn on and off due to the gate capacitance. This is a simple R-C charge and discharge calculation.

    ak
     
  5. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    At what frequency is the FET being switched on/off? At high frequency the gate capacitance charge time is of greater concern.
     
  6. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    QFT.

    For a simple resistive divider the equivalent resistance is simply the value of the resistors in parallel.
     
  7. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    Thank you.
    Is the 100 ohm a max value or can I use much higher resistors in the voltage divider. Apparently no one sees the need for a linear voltage regulator?
    Thanks again.
     
  8. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    The frequency is usually around 12Hz and has never exceeded 100Hz. I wouldn't think this would cause the capacitance to act as a large current draw.
    Thanks.
     
  9. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    After giving this some more thought I guess the questions is whether to use a voltage divider with high resistance in order not to waste power in the off state, or to use a linear voltage regulator and a 100 ohm resistor to protect it from parasitic influences. Does anyone know at what resistance level the turn on time would start to cause over heating?
    Thanks again for your help.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    For 100Hz maximum frequency the turn-on and turn-off time can be fairly slow before significant power would be dissipated in the MOSFET. This time is related to (but not equal to) the RC input time-constant where R is the equivalent gate resistance and C is the gate capacitance. What MOSFET are you using?
     
  11. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    IRFP064N Pbf
     
  12. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    I understand the turn on/off time could be slow to satisfy the frequency requirements however I noticed when testing the turn on voltage necessary for full turn on that when I increase the voltage slowly by hand it caused a fast heat build up in the mosfet (12 amps @ 24 volts). So I am wondering what practical amount of maximum resistance can I use to cut down on wasted current and not cause the mosfet to heat up. the linear voltage regulator doesn't draw anything until voltage is drawn from it.
    Thanks again.
     
  13. crutschow

    Expert

    Mar 14, 2008
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    Changing the gate voltage by hand would generate a lot of heat in the transistor since you are staying in the active region for a long time.

    My simulations with a similar MOSFET at 24V, 30A and 100Hz square-wave switching frequency gave the total dissipation in the MOSFET (including the ON resistance losses) with a 10V gate drive pulse for the following equivalent source resistances:

    10kΩ 1.5W
    5kΩ 1.0W
    1kΩ 0.6W
    100Ω 0.5W

    So it would seem that 5kΩ equivalent divider resistance in series with the gate or less generates an acceptable power dissipation in the MOSFET. 1W should make the transistor warm but not excessively with no added heatsink.

    If you switch the input of the resistor divider then it will draw power only when the transistor is on.
     
    Last edited: Jul 17, 2014
  14. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    Thank you for all the information you provided. It must have taken some time to develop as it really tells the story of what is going on in the mosfet. I'll stick with your recommendation of 5K or less. The only question I still have is the current draw. You stated there wouldn't be any current draw when the mosfet is off, there is still a circuit to ground through the voltage divider resistors in series. Using 5k resistors wouldn't there be 10k resistance at 10 volts resulting in .001 amp draw in the off state?
    Thank you again for your information and the effort you took in providing it.
     
  15. crutschow

    Expert

    Mar 14, 2008
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    If you switch the input to the resistive divider then there is no current in the off state. So if you have 24V divided down to 10V by the divider then you would switch the 24V.

    For example, for 10V at the MOSFET gate with 24V source you could use use a 10kΩ resistor in series with a 7.15kΩ resistor to ground. That gives a Thevenin equivalent resistance to the gate of 4.17kΩ.

    Incidentally, the simulations I did to determine the power dissipation were from the free LTspice program from Linear Technology. It's well worth downloading and learning if you plan on doing any more electrical designs.
     
    Last edited: Jul 18, 2014
  16. LaurenceR

    Thread Starter Member

    Feb 7, 2013
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    Many thanks you answered all my questions. I really learned something today. That's a good day.
    I'll try to find the LTspice program. It would be very useful for my work.
    LaurenceR
     
  17. crutschow

    Expert

    Mar 14, 2008
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    Here's the LTspice location. The learning curve can be a little steep but it's well worth if if you do any analog or switching circuit design. There's a tutorial for it to help you get started along with Example Circuits and a Demo Circuit Collection to demonstrate various types of simulations.
     
    Last edited: Jul 18, 2014
  18. LaurenceR

    Thread Starter Member

    Feb 7, 2013
    97
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    Many thanks.
     
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