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Turning a bipolar supply on/off with MOSFETs

Discussion in 'Electronic Design' started by Walter Harley, Nov 1, 2003.

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  1. I need to switch two 9v batteries (forming a bipolar supply) on and off,
    with the following requirements:
    - very low current drain, particularly when OFF.
    - control by whether a sense line is grounded or open. (I've shown a
    switch, but this could also be the open collector of a micropower

    Here's what I came up with. The MOSFETs are by Supertex, little TO92 things
    with around 0.5 ohm Rds(on) that cost less than $1. In simulation this
    works nicely (except that I had to use a VP2206 model instead of VP3203,
    since Supertex don't provide a VP3203 model). I haven't actually built it

    Am I missing something? Is it really this simple?

    It scares me a little that when it's on, the N-channel MOSFET is seeing Vgs
    = 19v (for fresh, 9.5v batteries). The rated max is 20v. Should I put a
    Zener diode in series with the collector of the 2N3906?

    ----------o-------+^+--- LOAD+
    V+ | |||
    | ===
    | ___ |
    POWER | 10MEG |
    _/ | ___ |
    .-o/ o-----|___|-o
    | | 10k |
    === | .-.
    GND | | |10MEG
    | | |
    | '-'
    2N3906 |-----'
    ___ |
    | 10MEG |
    | ===
    V- | |^| VN3205
    -o--------+|+------------ LOAD-

    created by Andy´s ASCII-Circuit v1.22.310103 Beta
  2. James Meyer

    James Meyer Guest

    Another 10Meg resistor in series with the collector would do the job and
    be cheaper.

  3. Mac

    Mac Guest

    I second that motion.

    Except I am a little worried about the high value resistor in series with the
    collector of the bipolar. If you (the OP) even have a micro-amp going
    through it, the N-channel FET will be fully turned on. With 100 nano-amps,
    there might even be some danger.

    So I would use a smaller resisor (or two as Jim suggests) to hold the
    N-channel FET off. I usually shoot for a 10:1 ratio of Ic:Ib.

    But I'm not real sure about this. I guess the thing to do is see how much
    leakage current you can expect from the bipolar.

  4. James Meyer

    James Meyer Guest

    The datasheet at shows 50
    nanoamps at 30 volts C-E as a max leakage current. That would only make 0.5
    volts across a 10 Meg resistor.

    The OP needs to make sure he cleans the pc board after soldering. :cool:

  5. Mac

    Mac Guest

    Yeah, but that's with VBE at -3 Volts, and at room temperature. I'm not
    sure this totally placates my nervousness. It could be that I'm just being
    irrational, though ;-)
  6. Duly noted. (I already do, but it hasn't been quite so important before.)

    What kind of current leakages can I expect from a not-cleaned board? Is the
    main problem flux that has absorbed moisture, or fingerprints, or ...?

  7. Look great, but I modified it a bit:

    The 'power on' consumption of the resistors is probably
    nothing compared to the load.

    Not tested.
  8. James Arthur

    James Arthur Guest

    Frank Bemelman posted:

    Doesn't turn on the VN3205 though, true?

    Like Mac and James Meyer, reliance on super low currents
    makes me uncomfortable too.

    Here's my entry...

    The following is less sensitive to leakage currents and
    doesn't expose either FET to excessive Vgs:

    ---------o--------+^+------ LOAD+
    V+ | |||
    | ===
    | ___ |
    220k |
    ___ |
    POWER | 220K |
    | |
    _/ | |<
    .-o/ o---o-----| 2N3906
    | |\
    === |
    GND ___ |
    | 220K |
    | ===
    V- | |^| VN3205
    --------o--------+|+------ LOAD-

    James Arthur
    "" -- delete it. (incoming e-mail subject to brutal filtering)

  9. Thanks! (Another plus is that I could use 1MEG instead of 220K, and that's
    a value that I already stock for this device, so one less resistor value

    The "naked" base of the 2N3906 scares me a little - I would have put a
    resistor there. But I can't prove why: after all, the only current paths
    are through 220k resistors or through the gate capacitance of the MOSFETs
    (~300pF, or ~4nC gate charge at 9v). It probably takes more than 4nC to fry
    a bipolar junction - so maybe this is a non-issue.

    What does the 220k across the B-E junction of the 2N3906 do? Is that to
    keep the base from floating, thus reducing leakage?

  10. James Arthur

    James Arthur Guest

    It might be exposed -- depends on your switch. When using a
    dome-and-membrane style switch, for example, it would be wise to ground the
    dome to divert static zaps from the user. An ungrounded toggle switch might
    conceivably have the same vulnerability. As drawn the circuit is safe provided
    the exposed pole and contact are grounded, which I assumed was your intent.

    If circumstances require it you could always clamp said base to the supply
    rails with a pair of diodes.
    Yes. If the board's super clean you could forgo it, but with it you needn't
    worry a'tall. Its value isn't critical.

    -- James
    "" -- delete it. (incoming e-mail subject to brutal filtering)
  11. Mac

    Mac Guest

    Good idea, except it won't turn on the VN3205. If you reverse the lower
    diode, then it will turn both FET's on all the time. But you gave me an

    Or replace the power switch with another FET. If the VN3205 is too
    expensive, use a cheaper FET. You can pull the gate down to V- to turn it
    off by default, then connect it to V+ to turn it on. When it turns on, the
    source terminal will rise up with the gate, so Vgs shouldn't ever see the
    full rail-to-rail voltage.

    Just a thought.


  12. Oops! ;-)


    That looks nice.
  13. You can also use a npn transistor, if that matches your stock better. Just
    flip the transistor and resistor.
  14. Good point. I use both, but I'd rather push up the NPN count than the PNP
    count. They're a little cheaper; and maybe they've got lower leakage? Just
    guessing, I'd have to look at the data sheets.

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