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Simple solution to send a short pulse to Solenoid for n milliseconds?

Bobby2017

Oct 15, 2017
2
Imagine a Ding Ding doorbell scenario where we want to save our solenoid from burning if a visitor keep the button pressed for a long time.

For a 12v Solenoid If power button is turned on I want to send a short pulse to Solenoid i.e It activate it for n milliseconds and then deactivate immediately and then when power is turned off it again sends a similar short pulse

I am basically an Arduino guy and do not know much about basic electronics I can make this circuit using arduino but for this to work my arduino needs to remain working all the time whereas I am looking for a simple solution for it where I don't have to keep anything ON 24/7

Any help will be much appreciated

Thanks

kellys_eye

Jun 25, 2010
6,514
Monostable, non-retriggerable.

Can be programmed into an Arduino but there are a multitude of discrete circuit - and simple IC circuits like the 555 - that can do it.

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,834
A raterh simple circuit for the first requirement (turn-on):

The capacitor allows an inrush current into the solenoid (inductor) until the capacitor is charged. Then current no longer flows through this part of the circuit. When the pushbutton is release, the capacitor discharges through R1 and L1. However, this will not create the necessary current to activate the solenoid unless the value of R1 is very small which in turn will increase power dissipation as long as the pushbutton is pressed.

This improved circuit will force a current through the indctor at turn-on and at turn-off. It requires a changeover pushbutton to do so:

Since no resistor is required, power dissipation is greatly reduced. Note that the current at turn-off is in the opposite direction of the current at turn-on. Consequently the magnetic filed in the solenoid will be reversed, too. Depending on the application this may or may not be an issue (e.g for a doorbell).
If it turns out to be an issue, a more complex circuit e.g. using 555 timer circuits may be needed.

The size of the capacitor depends on the current required to operate the solenoid and the time you want the solenoid to be 'on' when the pushbutton is pressed. As a first approximation use t ~ sqrt(L*C) for selecting C when t and L are given. You may have to experiment with different values for C to get satisfying results.

Bobby2017

Oct 15, 2017
2
Thanks for the rely but either way something needs to be
A raterh simple circuit for the first requirement (turn-on):
View attachment 36817
The capacitor allows an inrush current into the solenoid (inductor) until the capacitor is charged. Then current no longer flows through this part of the circuit. When the pushbutton is release, the capacitor discharges through R1 and L1. However, this will not create the necessary current to activate the solenoid unless the value of R1 is very small which in turn will increase power dissipation as long as the pushbutton is pressed.

This improved circuit will force a current through the indctor at turn-on and at turn-off. It requires a changeover pushbutton to do so:
View attachment 36818
Since no resistor is required, power dissipation is greatly reduced. Note that the current at turn-off is in the opposite direction of the current at turn-on. Consequently the magnetic filed in the solenoid will be reversed, too. Depending on the application this may or may not be an issue (e.g for a doorbell).
If it turns out to be an issue, a more complex circuit e.g. using 555 timer circuits may be needed.

The size of the capacitor depends on the current required to operate the solenoid and the time you want the solenoid to be 'on' when the pushbutton is pressed. As a first approximation use t ~ sqrt(L*C) for selecting C when t and L are given. You may have to experiment with different values for C to get satisfying results.

Thank you for looking into this and providing a simple solution with details

Your first solution is workable for me and half of the solution to what I am looking for, but if sending a pulse on button release is not possible then I think the first half is good enough (as long as I hear a full ding ding on button pressed)

Can you please suggest the values of capacitor and resistor in case I use 12v power supply and 12v Push/pull type Solenoid?
Also if power button remain on for long time like 30 minutes or in case when button gets stuck would this cause to burn Capacitor?

Many Thanks

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,834
Can you please suggest the values of capacitor and resistor in case I use 12v power supply and 12v Push/pull type Solenoid?
Not possible without knowing the value of the inductor. If you don'tt know this value, you need to try different values for the capacitor. Start e.g. with an electrolytic capacitor of 1000 µF / 16 V (observe polarity).
The resistor is not critical. Assuming an interval of at least 1 second between pushes of the button, a value of e.g. R = 300 kΩ/C (C in µF, for example 1000 µF -> 0.3 kΩ = 300 Ω) will suit.The smaller the resistor, the faster the discharge of the capacitor, but the higher the power dissipation. The resistor needs to be rated at P=144 V²/R (e.g. 300 Ω -> P = 0.5 W).

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