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remote device blowing fuse/power supply

Discussion in 'Electronic Basics' started by thomas, Jul 23, 2004.

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  1. thomas

    thomas Guest

    I have a remote device at the end of 100' of 22 AWG wire. The 5V remote
    device has a 2200uf capacitor and draws 30mA and up to 1A intermittently.
    My local 9V power supply is rated at 2.4A and I fused it with a 1.5A fast

    When I connected the remote device to the power wire it blew both the fuse
    and power supply. The long wire and large remote capacitor were just too
    much for it.(?) Was my mistake in making the connection with the power
    supply hot? If I make the connection with the power supply off won't it
    cause the same problem upon powering up? What can I do next time?
  2. The long wir should not be the problem, it just increases the load
    resistance, thus decreasing the load current. The capacitor of course
    will cause a high current peak, which could cause trouble.
    Fuses don't really protect electronics - the electronic usually is
    dead looong before the wire in the fuse melts.
    Check your wire for a shortcut and provide some soft - start at your
  3. JeffM

    JeffM Guest

    I have a remote device at the end of 100' of 22 AWG wire. The 5V remote
    You seem to be saying that you're running a 5V device on a 9V supply.
    Is there something you're not telling us?
  4. thomas

    thomas Guest

    Yes, I get a couple volt drop from the wire length, then I run it through a
    regulator to get 5V.
  5. I'm not sure what the problem is, but if there isn't any soft start,
    the current to charge that 2200uF cap will be about 3.3A.

    Here is one simple-minded soft-start circuit:

    ie 10 ohms/10W
    IN V ___
    | | |
    | | |
    | | P-MOSFET |
    | | |
    | ||| |
    | === |
    --- | |
    C --- | |
    | | |
    '------o |
    | OUT V
    | | R
    | |

    If the Vgs(th) of the P-MOSFET is Vx, and
    the supply voltage is Vcc, then the P-MOSFET
    will turn on after approximately

    T = R*C*ln(1 - Vx/Vcc)

    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    The idea is that the RC circuit is pulled up to Vin initially, and
    then relaxes down to GND. On the way, when the voltage passes
    12+Vgs(th) of the P-MOSFET, it'll turn on the mosfet, bypassing the 10
    ohm resistor.

    Bob Monsen
  6. thomas

    thomas Guest

    I'm not sure what the problem is, but if there isn't any soft start,
    This simple (minded) soft-start is probably just what I need. The web is
    full of soft-start MOSFET references but practically nothing on the simple

    I found the IRF9Z34 MOSFET which looks like it will work. It is rated at
    55V, .1ohm RDS, 17A ID Cont. and has a Vgs(th) of 2-4V.
    With a Vgs(th) of 2-4V does this mean I'll be slowing down just the initial
    ~3V charging of the 2200uf cap?

    If I want to slow by .5 seconds the initial charging of the 2200uf cap then
    RC=~1.4, giving something like 10uf and 140R, or 2.2uf and 630R. Am I going
    about this right?

    Do you think .5s delay is enough?
  7. No, the fact that the resistor is switched in is only dependent on
    what the resistor/cap combination connected to the gate are doing, not
    what the right side connected tot he load is doing.
    You dropped a factor of 1000. For 1/2 second, you need

    1/2 = -R*C*ln(1-4/9)

    So if you pick a 10uF cap, your resistor should be something like
    82,000 (to pick a standard value)
    Yes, in fact, I bet 22 ms is enough. You just need to charge up that
    2200uF cap on the other side of the wire. The time constant for a 10
    ohm resistor and a 2200uF cap is 22ms. OTOH, .5s won't hurt. The cap
    should be completely charged when the circuit switches the 10 ohm
    resistor out.

    Note that the power dissipated in that 10 ohm resistor will be I^2 *
    10, so if the initial voltage across the cap is 0, you have an
    instantaneous power of about 8 watts dissipated by the resistor. That
    number will drop quickly, as the cap charges, but it means that you
    probably shouldn't use a 1/4W resistor; a 1W resistor will probably be
    ok, a 2W would be better.

    Bob Monsen
  8. JTM

    JTM Guest

    With a Vgs(th) of 2-4V does this mean I'll be slowing down just the
    I see
    oops...uF is 1/1,000,000F
    Using the formula: .5 = R*C*ln(1--3/7) with Vx=-3V (midpoint of the Vgs(th)
    range) and Vcc=7V since there is a couple volt drop across the wire.
    This time I calculate R*C=1.4, giving 10uF and 140K.
    Yes, I understand. Maybe I'll go with 1/4s.
    Ok. I have some 2W 6.8R around. With 7V and 2200uF the time constant is
    15ms, @1A.
    Thanks, Bob
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