Solenoid drop out delay 1

[memepe]

I have a 24 VDC solenoid (resistance 60 ohms, # turns = 2000).
How do I calculate the drop-out time and voltage?

there is a diode wired in paraell (Sp) to protect the supply circuit from the back-eMF. I believe that this slows the circuit down on drop-out but it does not slow it down enough. How do I calculate the response time and what effect would putting a resistor in series with this diode have? Would it slow the drop-out time down?

What about a capacitor? How would I calculate the time for this?

Are there any good sources for electrical solenoid design references that could help.

 

[itsmoked]

You cannot 'calculate' the drop out time. It is an imperical value that results from the solenoid's design. You must check with the manufacturer.

Yes diodes can extend the drop out time some. Yes you can help them drop out quicker if you use an external resistance to help dissapate the previously stored energy. With a diode/resistor or better(ie faster) with a snubber circuit; resistor/capacitor.

Your statement is a bit confusing you state:[green]"I believe that this slows the circuit down on drop-out but it does not slow it down enough"[/green]

Why are we talking snubbers and diodes if you want the relay to take LONGER to drop out?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Skogsgurra]

A capacitor will always work. But you need to know the drop out voltage. And that can be measured with simple means.

Just connect a large enough capacitor parallel to the coil. But not directly - connect via a normally open contact of the relay. Apply voltage. The relay pulls in. Then remove voltage.

The capacitor voltage starts falling and when it has reached drop out voltage, the relay does exactly that. The contact opens and capacitor discharge is stopped. If you have a voltmeter connected to the capacitor, it will show the drop out voltage.

For DC relays, the drop out voltage is surprisingly small. A 24 V coil may stayed locked even at 5 V, sometimes even lower. The voltage falls down to about 9 V in one time constant (coil resistance times capacitor value) and to just above 3 V in two time constants. So your relay will probably drop out somewhere between one and two time constants.

If, for instance, you need a 1 second delay, the time constant shall be between 0.5 and 1 second. That means that you will need a capacitor being C = 1000000/R microfarads (using microseconds to get microfarads) or 1000000/60 = 16 700 microfarads or, if relay drops out at lower value, a bit less.

Test a 10 000 uF. They are small and cheap. Do not forget that the charging of the capacitor will be a problem with some contacts. But I do not think that it will be a problem in your case.



Gunnar Englund

http://www.gke.org

 

[OperaHouse]

A big capacitor can be rough on some low current contacts and even cause welding. A 10 ohm resistor in series with the capacitor can be enough to sole this problem.

 

[nbucska]

how much delay do you want?
I suggest adding an RC delay followed by a two ransistor
buffer -- needs only 6 parts.

Plesae read FAQ240-1032
WEB: <

http://geocities.com/nbucska/>

 

[memepe]

Thanks for the info so far.

More details on the application

The solenoid activates a spring load piston. When deactivated the plunger is mechanically pushed away from the pole face. That is what causes a spike when the solenoid is deactivated.
The current diode protection or equivalent needs to remain in place so that the control circuitry does not experience the voltage surge.
I am looking for a delay of between 0.5 and 1.0 seconds on the drop out and minimal delay on the pull in (currently about 50ms) An RC circuit might be the answer if I can get it to work in one direction only.
Size is limited.
Where can I find out how to calculate the time delay by an RC circuit and how to design one that works in one direction only.
I am a mechanical engineer by training and have only designed basic circuits using diodes, resistors, and a few H-bridges.

 

[itsmoked]

That is a longgggggg time in relay land.. Some sort or RC is probably the wrong way to do this. Why can you're controls just wait a while longer or something.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[waross]

Hello Kieth and Gunnar.
What do you think about a capacitor in series with a charging resistor. The capacitor resistor combination in parallel with the solenoid. With a diode across the resistor the capacitor will charge through the resistor and discharge through the diode and the coil. The solenoid should be able to pull in quickly without too much current or voltage drop as a result of capacitor charging. The capacitor can be sized to provide the proper time constant for the drop out. If that results in a ridiculously sized capacitor an interposing relay could be used with the capacitor delay on the relay.
Respectfully

 

[Skogsgurra]

With all respect, guys. I think you are overcomplicating things a bit. I have used the technique that I described in my first post - and it works. The trick with the contact was only if you want to know the drop-out voltage. A capacitor across the coil is the simplest and most reliable way to go. Especially if the time value isn't critical.

The delay is only on drop out. That's when the coil resistance is part of the equation. Switching on will not be delayed since the path then is directly from source to coil. There will probably be a current surge when switching on. But a circuit with a 9.6 W coil always has very sturdy contacts, so that will not be a problem. Enough talk and speculation. Just go an buy some capacitors - I think that something like 10 000 microfarads will give the desired delay. But do get some other values as well to get some latitude in your test. Buy 35, 50 or 63 V units. They will be physically smaller than your coil.

Of course, they shall be eletrolytic. And do observe polarity when you connect them. They will get destroyed if you feed them the wrong way. And, if this is a mobile application, see to it that the capacitors are fastened with oil resistant strips. The leads will break from the vibrations if you don't.

Gunnar Englund

http://www.gke.org