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Simple plan - will it work?

Discussion in 'Electronic Basics' started by Bdyne, Nov 11, 2003.

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  1. Bdyne

    Bdyne Guest

    I am installing a 115VAC cooling fan for my computer. The fan will be
    mounted in an enclosure feeding 3" diameter flex hose to the computer
    case. I will be using a 5 VDC signal from a fan connection on the
    motherboard to close a small radio shack relay (5VDC - 120VAC) and
    start the fan. Is it this simple or am I forgetting something? I
    don't want to burn the house down!
     
  2. ?

    Generally speaking the idea will work. But I can imagine some things you'll
    have to check.
    - Check the current rating of the contacts of the relay. (Just to stay on
    the safe side.)
    - Check the 5V control voltage you want to use. If it is directly connected
    to the 5V power you will have no problem but when it is switched somewhere,
    the switch (a transistor for instance) may not be able to provide the extra
    current for the relay.

    When you connected the fan and the relay, check the circuit, particularly
    the relay when it switches the fan off. If there's a heavy sparking at the
    contacts you may suppress them by placing a capacitor parallel to the
    contacts. I guess a 1nF/250VAC will do.

    petrus
     
  3. Don't forget the free-wheeling diode across the relay coil (at the
    driver end of the cable).
     
  4. Don Bruder

    Don Bruder Guest

    Ummm... Why?

    Perhaps I'm daft, but isn't a freewheel diode only needed in cases where
    a permanent magnet type (which almost always means DC-powered) motor is
    being controlled? Unless I've misunderstood things really badly, your
    typical AC "fan motor" type doesn't act as a generator/alternator when
    freewheeling down after power-off, since there's no juice on the stator,
    therefore there's no magnetic field to be cutting/cut by the windings of
    the rotor, which renders it functionally nothing more or less than a
    flywheel. (Albeit it fairly poor one, since they're usually pretty light)

    Conversely, cutting the juice to a PM-type motor DOESN'T "turn off" the
    magnets that make up the "stator", so the magnetic lines of force are
    cutting/being cut by the windings of the rotor, producing juice as long
    as the rotor continues to spin after power-off.
     
  5. It's not the motor that's in question.
    Unless I've missed the whole point here, he wants to pick a 5VDC relay
    coil to control a 115V fan. Since the relay coil (and the wire between
    the the driver and the coil) has significant inductance, a free
    wheeling diode is needed to clip the transient when the driver turns
    off. ...otherwise the driver goes to driver heaven and he's out a
    motherboard.
     
  6. Lord Garth

    Lord Garth Guest

    The poster said to put the diode across the RELAY COIL not the motor.
    When the relay de-energizes, you will create a reverse EMF. The diode
    will shunt that so as not to possibly damage the driver transistor.
     
  7. Don Bruder

    Don Bruder Guest

    D'oh! Musta missed that part. Somehow, for some reason, I was thinking
    that the poster was saying put the diode across the MOTOR side of
    things, not the relay coil side. In that case, I withdraw the question.
     
  8. Don Bruder

    Don Bruder Guest

    Now that it's been pointed out to me that you were saying "put it across
    the relay coil", you're 100% correct. I must have missed that when I
    read the message I replied to. Had I seen/understood that as your
    intent, I would have been completely in agreement, and wouldn't have
    bothered to post my response. Chalk it up to a brain-fart, I guess.
     
  9. Bill Bowden

    Bill Bowden Guest

    The diode goes across the relay coil. The coil is a inductor
    and produces back emf of L*di/dt which is a bad thing without
    the diode.

    -Bill
     
  10. You're right, I should have mentioned it. The 5V power will not be harmed,
    but if the control voltage is switched by a transistor you can easily blow
    it (the transistor) if you don't use a free-wheeling-diode.

    petrus
     
  11. Bdyne

    Bdyne Guest

    You guys kind of lost me (noob)- but I'm sure glad I asked here before
    cooking my new mobo! Thank you all very much.

    I understand what a diode is - like a check valve. If I look at the
    bottom of the relay I have 2 contacts for the DC and 3 for the AC.
    Physically where in the circuit are you putting the diode? And is the
    diode one of those resistor-looking ones or the three wire ones (I
    would guess the 2 wire).

    Again thanks for helping me NOT fry my new board.
     
  12. Since the current in an inductor (the relay coil and long wire) cannot
    change instantaneously, when the driver transistor turns off the
    current will try to go somewhere. Without the freewheeling diode there
    is no other path so the voltage rises until it can go somewhere. Zap,
    your off transistor is no longer off (if it's anywhere to be found. ;-)
    With the freewheeling diode, the current flows through it into the 5V
    supply, effectively shorting out the inductor. Of course when the
    driver is ON the diode is backwards biased, so it's not in the picture.

    +5V
    | |
    C| o /
    relay C| /
    C| ./
    | o
    | |
    |
    +5V |
    | |
    - Freewheeling diode |
    ^ |
    | |
    +-------------------------+
    | long wire
    |/
    -| driver
    |>
    |

    created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de
    You'd better check the current rating of the driver too. ...and please
    be careful with the 120V circuit! It'll do some damage to your board
    too (and you).
     
  13. Rich Grise

    Rich Grise Guest

    Just a nitpick, but it should be:

    +5V
    |
    +----------+
    | | |
    | C| o /
    Freewheeling diode - relay C| /
    ^ C| ./
    | | o
    +----------+ |
    |
    |
    |
    |
    |
    |
    +----------------------+
    | long wire
    |/
    -| driver
    |>
    |

    Cheers!
    Rich
     
  14. Since we are picking nits, you should keep in mind that wiring has
    inductance, too. The cleanest way to use the diode is at the switch
    as follows:

    +5V
    |
    +--------+
    | | |
    | C| o /
    | relay C| /
    | C| ./
    | two | o
    +5v | long | |
    | | wires |
    +-------------+ |
    | |
    - |
    ^ |
    | |
    +----------------------+
    |
    |/
    -| driver
    |>
    |
    0v
     
  15. Rich Grise

    Rich Grise Guest

    Yeah - you've got all that inductance in the wire in series with the
    inductance in the relay coil, so you get an even bigger spike! (and
    more concomitant radiated noise)

    :)
    Cheers!
    Rich
     
  16. It's better than frying the driver! We had several IBM 3033s
    catch fire because of exactly this problem (inrush limiters stuck
    on). Customers weren't impressed and management was *not* amused
    at this little slight. Especially so, since the drivers had the
    diodes built=in, but the designer didn't want to wire 48V on the
    card and instead mounted diodes on the contactors. No, the free-
    wheeling diode goes on the driver. After all, it's not there to
    protect the relay coil. ;-)
     
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