Needed: 5 seconds of AC

[controldude]

Hello all, I hope you can help me with this.

I'm running a motor, and when I lose power I have a mechanical lock with a solenoid design. If I lose power, the solenoid drops and hits a cam on the motor's shaft. Normally this doesn't happen while the motor will be running. But if the motor is running and I lose power, I don't want the plunger to slam into the cam if it was running. So basically, I want just about 5 seconds of 120 AC to the solenoid before it drops out, enough time for the motor drift to a stop. Is there an inexpensive circuit design for this? I'll gladly take pointers in the right direction.

-Dude

 

[benta]

How about placing some kind of hydraulic damper on the solenoid plunger?

 

[electricpete]

Some other thoughts.

There are a wide variety of time delay solenoids available off the shelf.

You could get tricky and try to put a cap in parallel with the solenoid to help keep current flow after power cutoff but I don't think that would be reliable. For one thing it may depend on the point in the cycle where power is interuppted.

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[jraef]

The easiest one-off solution to this would be to just go out and buy a small inexpensive UPS, like the kind they sell too cheaply to be of any real use with a full PC, but probaly good enough to hold in 1 solendoid for 5 seconds (depending on the size of the solenoid!). Something like this:

http://www.circuitcity.com/ccd/productDetail.do?oid=57438&WT.mc_n=4&WT.mc_t=U

Then simply wire a circuit in parallel that runs power to the solenoid trough the UPS only if the motor was running, by using an off-delay timer. That way you will not have the delay every time you want to set the plunger (which would be the problem with a hydraulic or pneumatic dampener).

If this is an ongoing OEM thing, you may want to consider a big cap as E-pete suggested, but 5 seconds can be an eternity if your soledoid coil power is high. If so, consider using a DC solenoid and placing a small battery and a floating charger in the circuit. Porbably cheaper than that cheap UPS.

"Venditori de oleum-vipera non vigere excordis populi"

 

[buzzp]

There is a device, talked about in another thread somewhere, that provides ride through power to contactors for up to a minute, I think. This device looks ok but never used it. Looks to be exactly what your looking for. It is in an ice-cube relay package (8 pin). If I remember the link Ill post it, otherwise you might search the threads for it.

 

[Barry1961]

An AC cap is not going to work since you will not know at what point of the cycle power will cut off. Positive charge, negative charge, zero charge or in between. If you do manage to ensure the power cut off at a full positive or negative charge you will then be supplying a DC voltage to the coil.

They make some cheap pneumatic timers that should work with in +/- 1 sec.

Barry1961

 

[ScottyUK]

I think benta probably nailed it first time. A pneumatic or hydraulic damper - a dashpot - would slug the response for a suitable period. How powerful is the return spring on the solenoid?



----------------------------------

If we learn from our mistakes,
I'm getting a great education!

 

[Marke]

Hi Controldude

I agree with the comments above except for the capacitor. While this would work on DC, I do not think it will work so well on AC.
You could slow down the operation of the solenoid by electrically dampining it by putting a short across it's terminals when the power drops out. This can be done with a changeover relay. I do not think that this will be enough (5 seconds) but it will certainly slow it down.

I would look at the UPS option.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[itsmoked]

Cap no way I agree.

I take it we are refering to a dashpot that only works in one direction? Because you will certainly *smoke* that solenoid if its energized direction is slowed by more than about one second.

 

[Barry1961]

Just a thought............

If you could change the solenoid coil to DC you could could use DC caps. I am GUESSING you would get one time constant before the voltage got low enough to drop the coil out.

For a rough guess using a 90VDC coil that draws .5 amps.

V/A=R, 90/.5=180 Ohms

One time constant is R*C (voltage drops to about 37% of full)

Time/R=C 5/180=.027 Farad

Now .027 Farad is a lot of capacitor but you can stack some in parallel. This is not uf (micro farad).

Just a thought. I would wait for one of the pros in here to make a better estimate of what you would need before buying anything.

Barry1961

 

[controldude]

Wow, you all act fast! Thank you. Well, I will explain a couple more things, because I don't think the dampers would quite fit well. First, the solenoid drops out due to gravity, it is mounted vertically. There are no springs. Next, the solenoid in my application must retract immediately when power is reapplied. But if power goes out, the solenoid should stay energized for at least 2 seconds (5 would have worked too, but 2 is ok also). Also, the UPS is also too large for our product, and I was aiming for a $10 solution or something interchangeable. I can see why the cap wouldn't work, because of the low frequency, ac operation of the solenoid, and instability of having the peak of the wave. The time delay solenoid may work, but I've had difficulty finding them on the web. Don't forget, this is a total loss of power. The solution needs some type of storage element. Would an AC inductor work? Lastly, the DC solenoid may work also. I might try this. Oh, the solenoid has a coil resistance of about 88 ohm, and full load amps of 1.22 A. Hope this helps!

Dude

 

[buzzp]

Still cant find the post I referred to but it is relatively new and the product I mentioned sounds like exactly what you need. Anyone else remember this thread?

 

[buzzp]

Ok, I found the thread. Here is the link to the product. Look for "coil lock relay" from

http://www.pqsi.com

. I think you will find this is exactly what your looking for, although I have never used it.

 

[electricpete]

I was the first to mention CAP and please notice I say it would be tricky and possibly not reliable. BUT not as bad as you guys made it out to be. In analysing the configuration we should remember we have TWO stored energy sources - an inductor and a cap.

The cap energy storage depends on cap V which is same as inductor V.
The inductor storage depends on inductor I.
Since it is an inductor, I and V are 90 degrees apart and there will never be a point when the stored energy is 0.

Since there is very little resistance in the circuit the energy dissipation rate will be small. The LC circuit current and voltage would oscillate at sqrt(LC).

I would think if you select C to put resonant frequency above line frequency by approximatley 10%.. That way the frequency of the oscillation does not cause 60hz relay to dropout (although the decaying current magnitude would cause dropout at some point). Also I believe in this case (LC resonant frequency not too far away from line frequency), the initial conditions of energy stored in IL and VC will tend to add, rather than subtract.

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[electricpete]

I just saw mention you would consider change to DC coil and I agree that sounds like much better option. All you need is a diode which is reverse biased during normal operation. It will allow inductor current to continue after shutoff.

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[electricpete]

To clarify, diode is in parallel with the inductor and reverse biased during normal operation.

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[buzzp]

see this thread for a similar discussion

thread797-121652

 

[electricpete]

Well, I have made some errors above that I would like to correct before someone else does.
#1 - the resonant frequency is not sqrt(LC). It is 1/sqrt(LC). (And that is in radians/sec, not hz)
#2 - I may not have been right about the need to select a resonant frequency close to (but not equal) to line frequency. I did a little analysis of an idealized LC circuit given arbitrary IL(0) and VC(0). It suggests that the solution IL(t) will have two terms 90 "degrees" apart (where degrees is defined on the basis of the resonant frequency). One of those terms depends only on IL(0) (not VC0) andone depends only on VC(0) (not IL0). Regardless of the initial values of IL(0) and VC(0) they cannot cancel each other out since they are separated by 90 degrees.

It would not be too hard to completely solve the problem using assumption of an ideal LC circuit with voltage removed abruptly. I'm sure there are some real-world issues to consider. One is that the power system may not go away abruptly and may interact with the L and C to remove their energy on the way down.

Buzz you made a comment in the other thread that you believe all ac relays (presumably plunger type) have an internal rectifier. Some do and some don't. It is not required since force only goes to zero briefly when powered by ac.

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[electricpete]

Umm. I guess my explanation for why an ac relay doesn't always need a rectifier should have mentioned something about shaded poles that help keep the flux from going to zero. Sorry for the incorrect info and for going off on a tangent.

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[mc5w]

If you use a UPS you would to power most of the control power off of the UPS so that your timing relays will work correctly.This would be the simplest solution. The UPS would need to use a true sine wave inverter such as sine coded pulse width modulation like with a variable frequency drive. A square wave or quasi sine wave inverter would mess up your solenoid and control relays.

You could convert the solenoid to direct current. This would be a lot more complicated than using a UPS. You would need to meaure how much alternating current the solenoid uses before the armature closes - you would have to block the armature in the dropped out position, briefly apply 120 volts AC, and read the meter. You would need to take another current measurement with the armature pulled in. You could then use a 24 vac class 2 control transformer, bridge rectfier, and a big capacitor rated at least 70 volts DC. The bridge rectifier would need to be rated 5 or 10 amps and about 100 volts top withstand inrush current - a 75 va class 2 transformer would limit inrush current to around around 10 amps. This would give you about 40 volts DC that you can run through a 7812 or 7824 regulator to give you enough voltage top pull in the solenoid. You would need a slow operate relay, say 1 second, that bypasses the economy resistor when the solenoid is supposed to be pulling in. The reason why you need an economy resistor to to avoid burning up the solenoid with the pull in current.

You are now wondering why I would post such a complicated alternative method. It is to keep you from making mistakes.

You could also get a replacement solenoid that is straight DC. This would avoid sticking in an economy resistor.

Turns out that Telemecanique makes a capacitor package that extend out the drop out time of a control relay or motor starter. One application is for use of a contactor at the load terminals of a soft start that is used for removing the soft start completely from the power circuit once a motor is running. There is a water utility in Florida that used 3 contactors with each soft start to isolate the soft start once the motor was running. This vastly cut lightning damage to the soft starts. Source: An issue of EC&M magazine from about 13 years ago.