MOSFET Switch issues

Discussion in 'The Projects Forum' started by mariust, Aug 31, 2009.

  1. mariust

    Thread Starter New Member

    Aug 19, 2009
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    I am trying to command the charging of a battery using a microcontroller and I use a P-type Power Mosfet to discontinue the charging supply after the battery has reached a certain voltage.

    I use a N-mosfet (Q2 - SI1012x) to drive the power mosfet (Q1) and the uController commands the N-mosfet (see the picture attached). The R2 resistor keeps the battery charging even if the uController loses power.

    The design is pretty straight forward and from my limeted experience should work. However when the Q2 is opened the voltage across R1 is about 2.4 V and that is not enough to close Q1 and disconnect the battery.
    I tried changing R1 with no efect (from 50 ohms to 100k). I also removed Q1 and inspected the board for any shortcircuit but I haven't been able to find an explanation for the problem.

    In my project the load is connected to the battery and I plan on using a similar design to disconect the load from the battery.

    PS: probably I should add a diode to block reverse current, but that is secondary at this moment.
     
  2. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    Your problem may be that Q2 is configured "high". As it turns on, the voltage across R1 increases toward 9V. Since its gate is limited to 5V or 3.3V by the MCU, Vgs will decrease and tend to turn Q2 off.

    Try something like shown here:

    [​IMG]

    Of course, with this "low" configuration, when Q2 is on, Q1 will be on. That is, the logic is inverted. NB: I included R3 by analogy to your diagram. It is probably not needed and if used, should be very low value, like 10 to 22 ohm.

    John
     
  3. mariust

    Thread Starter New Member

    Aug 19, 2009
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    Thank you for your response. The explanation was so simple but I never thought of it. I never thought of the potential at the source once the FET opened.

    Anyway the reson why I was reluctant to an inverter is that I want to use the same circuit to switch of the load. I also can't keep the MCU powered all the time and I want this switch to be opened by default so in an inverting topology I need the N-mosfet always open.

    In the new attached schematic the Gate of the N-fet (Q4) is default high so Q3 is always open. Since Vth ~ 0.7V and V_bat~9 V I think the switching off would work if I chose the R5/R6 ratio off 50 (R6=100Ω and R5=5k).

    Total gate charge of Q4 is ~750pC so for Vgs~10V => the equivalent capacitor of 75 pC and t ~ R6 * C = 7.5 ns for switching off.

    For switching on t ~ R5 * C = 375 ns.

    This is more than enough for my requirements if the new design and my calculations are correct.
     
  4. Tahmid

    Active Member

    Jul 2, 2008
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    For the switching transistor (Q3 in the latter diagram), you can use a pnp transistor. I attached the circuit.
     
  5. mariust

    Thread Starter New Member

    Aug 19, 2009
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    Why replace the mosfet with a transistor? I encountered several transistor schematics and the only benefit I could see is the faster switch time. However in my design a switch time of about 1 second is enough so a mosfet would do fine.
    Are there other benefits in using transistors to drive power mosfets?
     
  6. CDRIVE

    Senior Member

    Jul 1, 2008
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    Because it will work comes to mind and it's simple.
     
  7. Tahmid

    Active Member

    Jul 2, 2008
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    Transistor is simple and fast. One more thing, they have no gate charge issues.
     
  8. eblc1388

    Senior Member

    Nov 28, 2008
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    No, you cannot.

    The PNP will turn ON the MOSFET via the emitter-base forward bias junction on its own, without signal from the MCU.
     
  9. mariust

    Thread Starter New Member

    Aug 19, 2009
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    I think the PNP design would work only if the MCU can drive the Base close to V_bat so the Base - Emittor junction stops current flow to the ground. Therefore V_gate is V_bat and the Mosfet connects the load to the battery.

    Replacing the transistor with an NPN would seem to work the way I want it. When MCU signal is absent the transistor is closed therefore the Mosfet is ON. The 3V3 of the MCU would drive the Base high enough so V_gate of Q3 is low and the load is disconnected. I also think moving R4 between the Emitor and GND would limit the current when Q3's V_gate discharges to ground trough the transistor. It would also make the Base potential equal to output potential of the MCU.

    Also I would appreciate it if someone would confirm that my second design (the one with two mosfets) would operate as I described in that post.

    Best regards and lots of thanks.
     
  10. spinnaker

    AAC Fanatic!

    Oct 29, 2009
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    mariust,


    Sorry to drag out such an old thread but I am working on a very similar project. Did you get your battery to switch off? Are you willing to share?
     
  11. mariust

    Thread Starter New Member

    Aug 19, 2009
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    I used a schematic similar with the one in my second post. I will look at the exact circuit I used and reply to you later.
    I had no time to test it extensively but the initial results show it works. If your switch timing is an issue I suggest using a transistor to drive the Mosfet Gate. However I can do even with 1 sec switch time so no problem there.

    edit: The battery is connected to the load when a low signal from the MCU is present at the N-Mosfet gate.
     
  12. spinnaker

    AAC Fanatic!

    Oct 29, 2009
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    Thanks a lot.

    I was going to go with one of these MOSFETS. What do you think?
     
  13. mariust

    Thread Starter New Member

    Aug 19, 2009
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    If your application is similar with mine, you will need to use a P type as the main switch and an N type for the gate driver. The main issue when selecting the P mosfet is the power. For the driver N-mosfet you can select the cheapest as long as you have no other requirements.
     
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