Electronic – A robust way to supply power to a solenoid

power supplysolenoid

Solenoids typically heat up very quickly and it seems a few seconds of power supply is already adequate. Does anyone know of a robust/established way for supplying just a few seconds of power to a 12V/1A solenoid? How about cap discharge?

Thanks

EDIT 1 The solenoid pushes and pulls a lock for my rabbit house, so continuous operation isn't needed. In fact, my solenoid needs a fair bit of cooling time.

EDIT 2 My current plan is to drive it with 8051 and a transistor, but I'm open to alternatives. I could get the 8051 to give the pulses, but I fear that my code is buggy and frying my rabbits as a result. So I was hoping to come up with a relatively fail-safe method.

The solenoid is spring loaded, so that the pin retracts when the solenoid is engaged (spring is compressed) and releases when it's not (spring is released). With this motion, I'm planning to get it to unblock and block the gate (thus, a lock).

The rabbits will be locked up at night.

Best Answer

Yes. This is easily achieved with a switch and a battery. No need for a capacitor.

schematic

simulate this circuit – Schematic created using CircuitLab

Operation

  • Press the button.
  • Hold for required number of seconds.
  • Release the button.

This satisfies your requirements as stated in your original question.

As you suspected, a capacitor could be used to provide a pulse to operate the solenoid.

schematic

simulate this circuit

Here we'll charge up the capacitor through a resistor. If the switches are left on the resistor will limit the current to a safe value that won't overheat the solenoid. Let's figure out some values. First the solenoid.

$$R = \frac{V}{I} = \frac{12}{1} = 12 Ω$$

The power (heat) in the solenoid is given by

$$P = V·I = 12·1 = 12 W$$

Let's limit the current to 0.25 A. This will require a total resistance of 48 Ω (12 V / 0.25 A) so we'll need a 36 Ω resistor. We'll use a 39 Ω as this is a standard value.

Let's say we need a 0.5 s pulse to move the solenoid. We don't have a figure for the inductance of the solenoid so we'll just treat it as a resistor for now.

The time constant for an R-C circuit is simply R·C. Therefor

$$C = \frac{t}{R} = \frac{0.5}{12} = 0.041 F = 41,000 uF$$

This is a lot but not impossible. 4,700 uF capacitors are readily available. Assuming your rabbit is on a tight budget you might find capacitors in old power-supplies, etc. You need to make sure that they are rated for 12V or higher. Simply parallel all the capacitors making sure to keep the polarity correct.

I'm suggesting a 3-way, centre-off switch. This way the solenoid can be reversed for open and closed.

If finding the capacitors proves to be a problem we'll have to add some electronics to do the timing.

Edit after Update 2.

The requirements are simpler now because the solenoid is known to be spring return.

schematic

simulate this circuit

I tried to come up with an alternative but it's hard to beat the old 555 timer. This configuration is a monostable. When Q1 turns on it triggers the 555 which turns on Q2 and the relay. Time delay is set by \$R4 \cdot C2\$.

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