24V Mosfet High Side Switch

Discussion in 'General Electronics Chat' started by Regulator, Nov 6, 2008.

  1. Regulator

    Thread Starter New Member

    Nov 1, 2008
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    Hi,

    This is my first message here and I have a doubt about a mosfet switch circuit.
    I have an inductice 24V load and want it to be grounded when it´s off, so I used a FDS4935A mosfet to switch it and a BYG10M diode to protect the circuit from reverse voltage (the load will be in parallel with the diode). The max. VGS of the mosfet is 20V so i used a resistor divider network with a BC817 activated by a 3.3V output of a PIC32. When the BC is on, VGS will be 12V, turning on the mosfet and when th BC is off VGS will be 0 turning it off (at least it is what i think).
    I don´t have the parts to make a prototype and i´m using electronics workbench to simulate it. I used an "ideal" mosfet of the library (the FDS4935 is not there) and the circuit didn´t worked well, the VDS is too high and the current on the load too low. Is there a problem with my circuit or with workbench?
    Thanks.
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    The problem is with your circuit. Vgs has to be about 10Volts for the MOS to fully conduct. If you apply 12V to its gate, initially it will fully turn on but then it will almost turn off because the voltage on its source will rise to some positive voltage and Vgs will not equal 12V (as you apply) but less (Vgs=Vg-Vs).
    You can use a special IC called gate driver to drive the MOS (it will make the voltage on the gate higher than the power supply voltage as to have the proper Vgs) or use a P-channel MOS which needs a negative Vgs to work and make your life easier. With a p-channel you will connect the source to Vdd and make its gate 12V less than Vdd to turn it on.
     
  3. Regulator

    Thread Starter New Member

    Nov 1, 2008
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    Thanks for the quick answer mik3.
    I think there is a confusion here, the mosfet i´m using already is a P-channel one (http://www.fairchildsemi.com/ds/FD/FDS4935A.pdf) and it´s source is connected to VDD as you said. This way I think the source will always be at 24V right?
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, the source will always be 24v.

    The problem is that you have a high impedance path to Vdd and to ground via the resistive divider.

    The gate has a fair amount of capacitance. It will take time to charge/discharge the gate across those resistors. The larger the resistors, the longer the RC time, and the longer your MOSFET will stay in the linear (high-resistance) region, getting warmed up. :eek:

    P-ch MOSFETs generally have considerably higher Rds(ON) values than their counterparts. For this reason, you're generally better off to use an N-ch MOSFET.

    But since you have a diode in the circuit, I'm thinking that what you really want is a synchronous buck driver. The diode is replaced with another N-ch MOSFET.

    Go to Microchip's website, and get the datasheet for a MCP14628 - 2A Synchronous Buck Power MOSFET Driver. That should give you some ideas.

    Synchronous rectification is all the rage nowadays. MOSFETs fit into the scheme quite nicely, as when they're properly controlled they act like near-ideal diodes.
     
  5. mik3

    Senior Member

    Feb 4, 2008
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    For a DC signal at the base of the BC it should work. However, if you apply a high frequency square wave to the BC then it wont work properly because the voltage divider resistors (22K) are too high and wont charge the gate of the MOS fast, thus it wont turn fully on and you will loose power on the output. If you are going to use it in a switching application then reduce the voltage divider resistors as more as you can. This is important as not to heat the MOS too.
     
  6. Regulator

    Thread Starter New Member

    Nov 1, 2008
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    Thanks for the help guys.
    My application don´t require high frequency and PWM modulation. I´ll just turn on and off a 24V solenoid about a dozen times in ah hour, anything less than a second is acceptable. Anyway I replaced the 22k resistors for 4k7.
    Thanks for the MCP14628 advice SgtWookie, i didn´t fully understood its application but I think is for PWM and high frequency modulation, right?
    But the main problem is why the circuit didn´t worked on workbench? I´m afraid to ignore the simulator and then when the board is ready, i realize that he was right and i was wrong.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    What is the resistance of your solenoid? When you modeled it, did you include the resistance? Your circuit should work in the application you described.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    I simulated a modification of your circuit using Linear Technology's free LTSpice/SwitcherCadIII; I changed the resistance values on the gate to speed switching, and added a 10v Zener to allow a lower value resistor to be used on the low side.

    Not knowing what the inductance or resistance of your solenoid's coil is, I gave it a value of 22mH and 390 Ohms (serial), which may be off considerably.

    The green trace is the high side of D1 (Vout of the MOSFET)
    The red trace is the MOSFET's gate voltage.
    The pink trace is the current through L1.
     
  9. Regulator

    Thread Starter New Member

    Nov 1, 2008
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    Oh man, what can I say?? Thanks a lot SGT!
    I´ll install the SwitcherCad right now, it seems to be a better simulator.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    Good deal. :)

    Note that text in BLUE is comments (such as 10v Zener and 390 Ohms)
    In order to change the values, you'll need to right-click on the components - particularly L1 to change it's resistance/inductance values.

    Even with the resistors much lower in value, current flowing through them is still minimal (10-15mA).

    Download the 2nd file I attached to your \Program Files\LTC\SWCadIII directory, then it'll appear in your File/Open dialog.
     
    Last edited: Nov 10, 2008
  11. vishal_injatkar

    New Member

    Sep 30, 2009
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    Hey SGTWoogie, Can you please explain how does the 10V zener allows lower resistor values on the low side?
    Thanks
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    MOSFETs generally have a maximum Vgs (gate voltage in respect to the source) generally have a limit of +/-20v. Since our OP had a 24v supply, without some kind of voltage clamp (like the Zener diode) they were forced to use resistors of the same size to get a Vgs of -12 when turning the MOSFET on.

    Adding D2, a 10v Zener, clamps the gate voltage at -10v. So, rather than having to use nearly equal-sized resistors to get the gate to -10v, R3 (the switched pull-down resistor) can be a much lower value than R2 (the static pull-up resistor).

    The circuit I posted isn't ideal, but would work fine for a low-speed circuit.
     
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