Any way to use a capacitor to provide power-on and power-off pulses to a solenoid?

Thread Starter

huron11

Joined Oct 18, 2023
11
power diagram1.pngI have a solenoid that needs to be momentarily activated when power is applied and momemtarily activated again when power is cut off but with reverse polarity. My basic idea is to use a large capacitor in the circuit - the charging current will provide a positive pulse while the discharge current will provide a negative pulse.

Charging current is easy, however, I need this "black box" that could direct the energy in the capacitor back to the solenoid when power is cut off. I was thinking of using a depletion mode FET which will conduct when no input voltage exists but depletion mode FET needs a negative voltage at the gate to keep the FET off while power is on.

Keep in mind that there is no DC power to run the black box components when the AC switch is open!
I am not looking for a relay solution nor an IC solution, some discrete component solution is preferred.

Thanks for any suggestions.
 

AnalogKid

Joined Aug 1, 2013
10,944
I can see how to do this, but I can't do a schematic right now. How quickly does the 5 V supply output decrease when AC is removed?The basic plan is one p-channel and one n-channel logic-level power MOSFET, and a 1000 uF cap to power the timing circuit after the 5 V goes away.

5 V comes on, charges up 1000 uF local power store
Short delay
P-FET snaps on rapidly to activate the solenoid, and stays on continuously.
5 V goes off
P-Fet is turned off
Short delay
N-FET snaps on rapidly to activate the solenoid.
N-FET turns off when the 1000 uF local power runs down.

ak
 

Thread Starter

huron11

Joined Oct 18, 2023
11
I can see how to do this, but I can't do a schematic right now. How quickly does the 5 V supply output decrease when AC is removed?The basic plan is one p-channel and one n-channel logic-level power MOSFET, and a 1000 uF cap to power the timing circuit after the 5 V goes away.

5 V comes on, charges up 1000 uF local power store
Short delay
P-FET snaps on rapidly to activate the solenoid, and stays on continuously.
5 V goes off
P-Fet is turned off
Short delay
N-FET snaps on rapidly to activate the solenoid.
N-FET turns off when the 1000 uF local power runs down.

ak
It's just using those regular 5V USB charger cubes, not sure the decay rate.
 

crutschow

Joined Mar 14, 2008
34,047
AK's idea got me thinking, so here's the LTspice sim of a circuit that should do what you want:
When power is applied (green trace and yellow trace) the solenoid (red trace) is positively pulsed through D3 and C1.
This also charges up the bias supply C2 capacitor (blue trace).

When power is turned off, the power supply Cusb capacitor voltage (10μF typical) rapidly discharges through the bleed resistor R1, which turns on Q1 and thus Q2 from the bias capacitor voltage, to rapidly drop the C1 capacitor voltage to ground (yellow trace), giving a negative pulse to the solenoid.

1697684042070.png
 
Last edited:

AnalogKid

Joined Aug 1, 2013
10,944
A differentiator that turns an edge into a pulse is sometimes called a boxcar circuit. The faster the input edge is, the wider and more consistent the output pulse is. #6 is pretty much what I had in mind, except we don't know if the 5 V output snaps up fast enough to boxcar C1. That's why I suggested a p-channel MOSFET after D3.

Overall, I was thinking that a more heavy-handed approach to timing might be needed. A short R-C delay after the 5 V appears in case it doesn't snap up fast enough, followed by a Schmitt trigger gate to make a clean fast edge for a p-channel FET, and a short delay after it decreases, in case it wanders down so slowly that the gain of a transistor does not speed up the edge enough. followed by a Schmitt trigger gate to make a clean fast edge for an n-channel FET. A bit more complex, maybe not needed.

What happens if you change V1-SW into a pulsed voltage source with 100 ms rise time and 1 s fall time?

AND - as always, thanks for whipping up another great simulation.

ak
 

Thread Starter

huron11

Joined Oct 18, 2023
11
What is the minimum time the power will be on?
AK's idea got me thinking, so here's the LTspice sim of a circuit that should do what you want:
When power is applied (green trace and yellow trace) the solenoid (red trace) is positively pulsed through D3 and C1.
This also charges up the bias supply C2 capacitor (blue trace).

When power is turned off, the power supply Cusb capacitor voltage (10μF typical) rapidly discharges through the bleed resistor R1, which turns on Q1 and thus Q2 from the bias capacitor voltage, to rapidly drop the C1 capacitor voltage to ground (yellow trace), giving a negative pulse to the solenoid.

View attachment 305297
Wow! Thanks for the effort of the drawing and testing out the circuit. The output puses are exactly what I wanted. If I understand correctly, the inital pulse is just from charging the capacitor/solenoid through D3. When 5V is lost, Q1 gets forward biased through R3 and R1, this in turn forward biases Q2 which provides the return path for the capacitor to discharge. Great! Now is there a way to reduce the component count??
 

Thread Starter

huron11

Joined Oct 18, 2023
11
A differentiator that turns an edge into a pulse is sometimes called a boxcar circuit. The faster the input edge is, the wider and more consistent the output pulse is. #6 is pretty much what I had in mind, except we don't know if the 5 V output snaps up fast enough to boxcar C1. That's why I suggested a p-channel MOSFET after D3.

Overall, I was thinking that a more heavy-handed approach to timing might be needed. A short R-C delay after the 5 V appears in case it doesn't snap up fast enough, followed by a Schmitt trigger gate to make a clean fast edge for a p-channel FET, and a short delay after it decreases, in case it wanders down so slowly that the gain of a transistor does not speed up the edge enough. followed by a Schmitt trigger gate to make a clean fast edge for an n-channel FET. A bit more complex, maybe not needed.

What happens if you change V1-SW into a pulsed voltage source with 100 ms rise time and 1 s fall time?

AND - as always, thanks for whipping up another great simulation.

ak
Yes, I was thinking about a Schmitt trigger myself also, a simple R-C might put the FETs into resistive range and drain the energy in the capacitor into heat instead of to the solenoid.
 

crutschow

Joined Mar 14, 2008
34,047
Now is there a way to reduce the component count?
Not likely.
My design goal is always KISS, so I don't offhand see how parts can be eliminated and the circuit still work.

If anything, depending upon how fast the 5V power comes up, additional parts may be needed.
 

eetech00

Joined Jun 8, 2013
3,823
AK's idea got me thinking, so here's the LTspice sim of a circuit that should do what you want:
When power is applied (green trace and yellow trace) the solenoid (red trace) is positively pulsed through D3 and C1.
This also charges up the bias supply C2 capacitor (blue trace).

When power is turned off, the power supply Cusb capacitor voltage (10μF typical) rapidly discharges through the bleed resistor R1, which turns on Q1 and thus Q2 from the bias capacitor voltage, to rapidly drop the C1 capacitor voltage to ground (yellow trace), giving a negative pulse to the solenoid.

View attachment 305297
Hi

Can you re-run this sim, but with a duration of, say, 60 minutes, between 3 switch pulses, then post the results?
 

eetech00

Joined Jun 8, 2013
3,823
[
The polarity of the pulse is determined by whether the supply is going on or going off.
Any potential storage component (Cap(s)) is discharged after a period of time.
So the state of the circuit is, potentially, always the same on power up. So I don't understand how it can determine what the polarity of pulse should be.

How could that be affected by the length of time it is on or off?
Just to ensure the caps are fully discharged in your sim.
 

LesJones

Joined Jan 8, 2017
4,172
Here is a suggestion using a relay and a diode.
Connect the relay coil directly to the supply.
Connect the supply negative to one side of the solenoid. and to the NC contact on the relay.
Connect the other end of the solenoid to the common contact on the relay via the capacitor.
Connect a diode between the positive supply and the NO contact on the relay. (Cathode to the NO contact.)
When the power is removed the relay the relay will drop out connecting the end of the capacior to the negative supply which will generate a negative pulse to the solenoid.

Les.
 
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