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.
snipped-for-privacy@gmail.com wrote in news: snipped-for-privacy@22g2000hsm.googlegroups.com:
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.
"John B" wrote in news:8OidnfzZI4OmApzanZ2dnUVZ snipped-for-privacy@netlojix.com:
Read slowly... Ò^O
Ok, here's a basic circuit:
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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 it.
(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.
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:
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).
Ed, 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,
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]---+ | | +-----|
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