Needed: 5 seconds of AC
Needed: 5 seconds of AC
(OP)
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
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





RE: Needed: 5 seconds of AC
RE: Needed: 5 seconds of AC
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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RE: Needed: 5 seconds of AC
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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"
RE: Needed: 5 seconds of AC
RE: Needed: 5 seconds of AC
They make some cheap pneumatic timers that should work with in +/- 1 sec.
Barry1961
RE: Needed: 5 seconds of AC
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RE: Needed: 5 seconds of AC
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
RE: Needed: 5 seconds of AC
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.
RE: Needed: 5 seconds of AC
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
RE: Needed: 5 seconds of AC
Dude
RE: Needed: 5 seconds of AC
RE: Needed: 5 seconds of AC
RE: Needed: 5 seconds of AC
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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RE: Needed: 5 seconds of AC
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RE: Needed: 5 seconds of AC
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RE: Needed: 5 seconds of AC
thread797-121652
RE: Needed: 5 seconds of AC
#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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RE: Needed: 5 seconds of AC
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RE: Needed: 5 seconds of AC
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.
RE: Needed: 5 seconds of AC
RE: Needed: 5 seconds of AC
When you disconnect power you will have oscillations, which leads to two questions
1 - Barry mentioned a dc cap - won't it cause violent damage to the dc cap when the polarity reverses?
2 - Even if you use an ac cap (my preference), how will the dc relay respond in presence of oscillating ac? I would think you are in danger of dropping out.
To my thinking, if a cap is an option it would be with the existing ac coil. If you are going to dc coil why don't you just use the reverse biased diode?
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RE: Needed: 5 seconds of AC
You also need a second diode that will keep the capacitor charge from backfeeding into the control circuit.
RE: Needed: 5 seconds of AC
With a $10 cost limit you are going to run into problems with any electrical solution. Have you considered a mechanical change? I am thinking of backspin ratchets on pumps. They work in a similar manner, but what they do is have a sloped side to the end of lock pin and a sawtooth arangement on the tops of the shaft cam lobes. So even though the pin technically drops when the power fails, it "rides" over the tops of the sawtooth cam lobes until the velocity of the shaft is slow enough for it to drop down all the way into the slot (past the slope) and lock the shaft.
"Venditori de oleum-vipera non vigere excordis populi"
RE: Needed: 5 seconds of AC
I'd say a diode would be almost the same but with more loss may not hold in as long. The cap alternative sounds like it may have problems. The size of the cap to store a reasonable amount of energy would be quite large (and expensive).