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Poor Man's Hi Torque Servo, circuit

Discussion in 'Electronic Components' started by [email protected], Oct 2, 2007.

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  1. Guest

    Hello Electronically Superior Beings,

    I am electron challenged and I need a simple circuit to control a
    wiper motor that I'm using as a high torque servo.
    The motor has hi and low speed circuits and a "park " position.
    What I want to happen is this:
    S1 (NO) is closed and sends signal to some kind of device(D) that
    powers up the wiper motor. When motor cycles, S2 (NC) in series
    downstream of D, opens and stops wiper motor at desired location for
    30 secs. When S2 closes again, the wiper motor continues rotation back
    to "Park" and shuts off. D re-sets for new signal.

    I can't use S1 as switch for motor, because it necessarily moves away
    from activation force. I located a timer for the 30 second pause.

    What is D called and how do I get one?

    Thanks for your help,
  2. wrote in
    For the active switch in 'D' you could use a power MOSFET to let a positive
    signal voltage switch its resistance from several hundred megohms to a few
    milliohms. They are often used as motor switches for that reason. A good
    example is the IRF630, good for 9 amps, more for very short surges, and
    found cheaply on eBay, often in quantity.

    That leaves your switching able to work with small signals, which increases
    your options and lowers costs and raises reliability.
  3. Read slowly...

    Ok, here's a basic circuit:

    That was meant to be driven by a fast pulse train for speed control, but
    it's still just a switch, as you can work out from that diagram. Part
    numbers and voltages can vary but the idea is right.

    This doesn't solve your switching logic but it will make it easier to apply

    (I answer at the bottom so it's easier for others to follow. Top posting
    makes sense in private emails but not so well here).

    I'm too tired to go further but they're not all in bed in the US and
    Canada yet.
  4. John B

    John B Guest

    Thanks for your quick response, but could you speak a bit more slowly. My
    electronics vocabulary is in its infancy.

  5. ehsjr

    ehsjr Guest

    I'm assuming S1 is a normally open momentary switch.
    Device D is the relay (Relay1) shown below:

    + ---+---o o---+------+-------+
    | S1 | | |
    | | | |> | <-Relay
    | / | | | Contact
    +---o o---+ [Relay1] [Motor]
    Park | |
    | S2 |
    Gnd -------------------+---o->o---+

    You have not stated how S2 moves back to the closed
    condition, but in a reply stated that you have a
    30 second timer. If we assume that S2 is cam operated
    and stays in the open position until the motor is
    energized again, you can use a second relay that is
    held energized for 30 seconds after S2 opens. The
    normally closed contact of the second relay would be
    wired in parallel with S2. Here's the diagram, modified
    to show that:

    + ---+---o o---+------+----------+
    | S1 | | |
    | | [Relay1] |> | <-Relay
    | | | |
    | | | +-------+
    | / | | | | Contact
    +---o o---+--|<--+ | [Motor]
    Park D1 | | |
    | +---|<--+
    | D2 |
    | S2 |
    Gnd -------------------+---o->o---+--+---> Trigger
    | | Signal
    Timer Relay Contact-> '-> | | to timer
    | |

    How it works: When in the park position, the park switch
    and S1 are both open. Pressing S1 energizes Relay 1, its
    contact closes, the motor moves and the park switch closes,
    keeping Relay1 energized until the mechanism moves back
    to park. When the mechanism reaches the S2 position, S2
    opens and removes the (-) trigger signal to the timer.
    After 30 seconds, the timer relay de-energizes and the
    relay contact closes. That causes the motor to run again,
    moving away from the S2 position and closing S2. That
    sends the (-) trigger signal to the timer which energizes
    the timer relay. The mechanism returns to the park position,
    which opens the park switch and de-energizes Relay1, and the
    motor stops. Diode D1 protects the switches against the
    inductive spike created when power is removed from Relay1;
    D2 protects against the inductive spike from the motor
    when power is removed from it. The timer relay coil should
    also have a diode placed across it, wired the same polarity
    as the other diodes (banded end connected to the + side).

  6. John B

    John B Guest

    OK, a picture worth a 1000 words and good advise for forum posting/replys.
    But I have some homework to do. Thanks for your help.
  7. Guest

    I can't thank you enough for the hand-holding and spelling it out for
    me. I understand
    your clearly described circuit perfectly. I assume that there are some
    values for the diodes
    that I need to figure out, but I'm sure that I can handle that. Again,
    1000 thanks. You've made my week.
  8. ehsjr

    ehsjr Guest

    The diodes can be anything in the 1N400x "family".
    If your timer triggers on (+) instead of (-), you
    can move S2 to the + side. I would use a delayed
    dropout circuit instead of a timer (assuming the
    30 seconds does not have to be exact) like this:

    +12 ----------------+---[Relay]---+
    | |
    D3 |
    Trigger(+) ---[1K]---+-----+----| TIP120
    | |+ \e
    [POT] [470uF] |
    | | |
    Gnd -----------------+-----+------+

    The TIP120 is an NPN darlington power transistor. You set how
    long it keeps the relay energized with the 100K potientiometer
    (which you configure as a variable resistor by shorting one
    outside contact to the center contact).

    The + trigger signal from the circuit is interrupted when
    S2 (which is now on the + side of the motor) opens. The
    charge on the 470uF capacitor keeps the transistor conducting
    until the cap discharges through the pot. How long that takes
    depends on the resistance the pot is set to. As long as the
    transistor conducts, the relay remains energized, and its
    normally closed contact remains open, so the mechanism does not
    move. Once the cap discharges sufficiently, the relay drops out,
    the normally closed contact shorts out S2 and the motor moves
    again. The 1K resistor limits the current to the base of the

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