Ohm's Law, revisited;

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Old 01-18-17, 06:48 PM
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Ohm's Law, revisited;

OK you guys that went to Electrician's College and learned more than Ohm's Law. This one has me baffled:

Double commercial entry doors with electronic unlocking using 24 VDC solenoids to operate the latches, with power supply designed for the purpose.
Solenoids are dual-coil: high-current, to snap the latching rods into unlock position, and a low current, to hold them in place for momentary entrance or all day, i.e., they are Continuous Duty so the doors can remain unlocked all day. A solid state potted circuit breaker adjacent the solenoid with 3 output wires (common, hi current coil, lo current coil) allows the hi current coil to fire for 1/2 to 1 second then disconnect to allow the lo current coil to continue operating. HI current coil is 1.2 to 1.5 Ohms, LO current is 115 to 120 Ohms.

Factory specifies 14 Gauge wire between power supply and locks for runs up to 50 feet or so, 12 gauge (stranded) beyond 50 feet.

A "sequencer" circuit in the power supply causes, upon a dry-contact input closure, to fire one lock, then 1 to 1.5 seconds later, the other lock, so they don't both experience the high current at the same time.

(con't)
 
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Old 01-18-17, 06:58 PM
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The Assa Abloy locks should use the voltage reduction method.
They fry with anything over 22vdc on them continuously.
 
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Old 01-18-17, 07:02 PM
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This locking design has existed for 25 years plus, with the only major design changes being the power supplies, that used to be with fused output, and are now with electronic overload circuitry.

This particular installation has 18 gauge wire because the power supply is above the doors, i.e., 10 feet from the locks.

After 8 months of trouble-free operation, the power supply now "overloads" each time the access controls call for the doors to unlock; within 1 or 2 seconds after the second door unlocks sequentially, the output cuts off, the doors lock back, and after about 10 seconds the power supply attempts to re-apply power, with the same result.

Troubleshooting for a short in the wiring and a check of the coil resistance check out fine.

Called tech support at factory, expecting a warrenty call for a new power supply, but when asking about wire size, said NO, NO, you must use 12 or 14 gauge wire! No more advice 'till you re-wire!

(cont)
 
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Old 01-18-17, 07:06 PM
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You're talking about the little solenoids that are actually inside the door lock ?
 

Last edited by PJmax; 01-18-17 at 07:20 PM. Reason: incorrect
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Old 01-18-17, 07:09 PM
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Thinking that an INCREASE in wire gauge would REDUCE the resistance and thereby INCREASE the current experienced at the power supply, I questioned the factory guy how that could prevent an overload, but he said "your voltage is dropping too much during the momentary surge.

I thought, ok, wise guy, I'll re-wire it....well I did and so far (a week) it's worked fine.

So with egg on my face, I'm trying to understand how this fixes it.
 
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Old 01-18-17, 07:10 PM
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The opening coil draws about 18.5 amperes at 24 volts. Using #18 wire the slightest bit of corrosion, even invisible, will cause an excessive voltage drop. The holding coil will draw about 0.2 amperes so is not a concern.

I agree with the factory, you need to use the larger wire.
 
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Old 01-18-17, 07:11 PM
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My thoughts..... it's not fixed and will happen again.

Are these the solenoids in the actual door lock or something on the latch/in the jamb ?
 
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Old 01-18-17, 07:13 PM
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Thanx PJ, no these are not your garden variety 60-100 Ohm coils used in electric strikes and electric locksets. These are big solenoids designed to operate Panic-Exit Devices with vertical rods to top and bottom of door; they require lots of force. Solenoids are 1-1/2 to 1-3/4" diameter.
 
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Old 01-18-17, 07:17 PM
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Solenoids are housed in the panic bars themselves, and operate the same linkage that the push-pad operate.
 
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Old 01-18-17, 07:19 PM
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Oh ok. I know exactly what you are working on now. Used a lot in hospitals.
 
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Old 01-18-17, 07:20 PM
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R, posts can be approximately 5000 words so no need to break it down to chapters/installments.
 
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Old 01-18-17, 07:21 PM
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@furd: but an excessive voltage drop would result in a corresponding current drop, no? At any rate, the voltage drop was not sufficient to cause the solenoids to fail to unlock...they would both (sequentially) seat, but a second later the power supply would quit (overload) and then reset in 10 seconds and attempt again and again.
 
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Old 01-18-17, 07:24 PM
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Sorry, I'm not the fastest typist in the world, and if I take too long to compose a post, I get kicked off and loose the document... very frustrating.
 
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Old 01-18-17, 07:24 PM
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Most likely the auto-reset circuit breaker simply tripped a bit late, not uncommon.
 
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Old 01-18-17, 07:28 PM
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...if I take too long to compose a post, I get kicked off and loose the document... very frustrating.
When you log in check the "Remember Me" box and you won't be logged out. Also, you can click on the "Restore previous text" (or something similar) line in the lower corner of the text box to restore most of your previous text.
 
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Old 01-18-17, 07:46 PM
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A little solenoid theory: the ones designed for DC current operate on an exponential power curve; for each unit of travel of the plunger toward the seated position, the force increases exponentially. The shape of the plunger's end has a lot to do with this characteristic. But that's why the potted circuit breaker in the lock allows about a half second at hi current to jerk the solenoid into it's seated position, at which point the current can be reduced greatly, because of the inherent high holding force in this position.

I just can't figure out why bigger wire has (so far) kept the power supply from overloading. BTW, this is not a switching power supply; it has a huge transformer and filter capacitor.
 
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Old 01-18-17, 07:55 PM
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Thanx, Furd...I wondered what "remember me" meant. I'll try that!!

And, late to trip or not, it was presumably tripping from overload. So why would bigger wire fix it?? And remember, it worked fine (on time clock, unlocked at 8AM, locked at 5PM) for 8 months.
 
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Old 01-18-17, 07:56 PM
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A switching supply would cut out when it saw a load that low.
That's why a linear supply is used. Probably with a good sized cap too.

I'm still not convinced the problems are gone. It worked for many months without a problem originally.

You're talking about a large load being switched by a time clock.
Possibly that switch is failing ?
 
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Old 01-18-17, 09:49 PM
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@PJ: Switching the output on/off is done with a "trigger" input; a dry contact closure across two input terminals on the power supply board. Completely isolated from the output load, for that very reason. The output board has two rather large relays that do the switching.
 
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Old 02-01-17, 07:35 PM
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Update: (3) weeks now and the problem appears to be solved. Still can't figure out why smaller wire was causing power supply to overload and trip, but bigger wire doesn't.
 
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Old 02-01-17, 07:48 PM
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That's certainly some good news.
 
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Old 02-01-17, 08:18 PM
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DC coils coupled to moving plungers experience a more complex current waveform than a simple resistive load. Current is ultimately limited to E/R but experience a slow waveform with a L/R time constant. As the L increases due to iron movement into the coil center, current will actually decrease, for just a while. If you have a resistive connection to the power source, that plunger movement will slow down. L stays lower and current could actually be higher than if you drove the coil with a higher voltage. You sped up the plunger with the larger wire size.
 
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Old 02-01-17, 08:21 PM
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...it was presumably tripping from overload. So why would bigger wire fix it?
I had to think about this for a while but it is the same as the electric water heater running on reduced voltage I posed to an apprentice electrician a year or two ago.

Neither amperage or voltage alone can answer the question, it is POWER that is important. Remember, power is the product of voltage multiplied by amperage. Remember also that the resistance of the solenoid coil does not change appreciably. Without showing the arithmetic simply stated if the interconnecting wiring has a voltage drop of one volt, the solenoid receiving only 23 volts the amperage will rise to 19.26 to supply the same power to the coil. If the voltage applied at the coil drops to 22 volts the current rises to 20.14 amperes.

This is a concept that a huge number of people do not understand, many of them professional electricians. If you need to see the arithmetic just ask.
 
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Old 03-05-17, 08:43 PM
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Reviving this thread, an update: Locks still working fine, concluding the bigger wire did the trick, but still trying to understand why.

@Furd: since I'm not a professional electrician, I don't feel too bad not understanding your explanation. Power (watts) is the result of a given voltage applied thru a given resistance, which results in a certain current and power.
Since I can't adjust the output voltage, the bigger wire would result in lower resistance, permitting higher voltage to the coil. The RESULT would be higher current demanded from the power supply. The higher voltage at the coil would RESULT in higher work done (watts). Is my interpretation off here?

The power supply can't feel the difference between the line resistance and the coil resistance, all it feels is the total resistance, and it's total resistance decreases a little with a bigger line, so more current is supplied to maintain it's output voltage.
 
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Old 03-05-17, 09:12 PM
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@telecom guy: while I don't know what the terms E, R & L mean, I'm assuming the current fluctuations you refer to take place while the plunger is in motion, (in my case, traveling about 7/8"). I'm assuming also, that once seated, current fluctuations stop. While I don't doubt your statement in the least, I'm thinking it's not a significant factor in my case, due to the time spans: when voltage is applied to the first solenoid, it takes only about 1/10 second to seat, then, by design, about 1 second later voltage is applied to the second solenoid, again taking 1/10 second to seat. The control circuit at each solenoid allows current to the low resistance coil for about .4 or .5 seconds before disconnecting and leaving the high resistance coil to keep the plunger seated.

My power supply was overloading a good full second after the second solenoid seated, ie., a half second after the high current demand ended.

I might be more inclined to think along your lines if the power supply kicked off immediately upon high current demand from second solenoid, but even if your assertion had significant effect in my case, the bigger wire would allow higher voltage at the coils and increase the current spike you describe, making even more likely an overload.
 
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Old 03-05-17, 09:40 PM
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I'm sort of left wondering if the fourth component of this circuit is the cause of the unexplained behavior, said circuit being the power supply, the wire, the solenoids and the potted circuit breakers, as described in post #1.

Since they're potted, I can't tell how they work, but my first theory is they work by heat; when voltage is applied, an internal component heats up and in .4 to .5 seconds breaks the high-current contact, and as long as voltage is present this contact stays off. When supply voltage is discontinued, it apparently takes only a fraction of a second to reset and be ready for the next application.

SO, if the previous set-up with smaller wire caused more voltage drop at the PCB input, it might result in a slightly longer time span to heat up before breaking circuit....this would result in the power supply having to provide the higher current for a fraction of a second longer.

STILL, this fraction of time longer should be balanced by the lower current being demanded due to the smaller wire.

So if y'all were not confused before, you probably are now. I guess I'm one of the few who want to know why something works, rather than not work.
 
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Old 03-06-17, 08:21 PM
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Maybe the key question is what turns off the high current solenoid. At 1.3Ohm and 24V source, that pull-in coil will draw some 18Amps. We don't know the power supply rating, but it very likely not 444 Watt rated. That means, it can deliver that 18 Amps for only a fraction of a second before it overloads. So, the shorter the travel time, the less time the power supply needs to power it. 18AWG wire of a dozen round trip feet will have a significant fraction of the total load resistance. That means a sluggish plunger, and the travel time will be significantly longer. Yes, the current drain is less, but that could be offset by the current on time. Total power from the power supply is higher. Be aware that a fair amount of power will be wasted in the 18Awg wire; the larger wire will not heat as much.

So, going from 18 to 14 AWG does this: Current spike is higher, current pulse is shorter, plunger travel time is shorter, and wire heating is less. It's the area under the current/time curve that represents power supply load. I'm submitting that the area is less for the larger wire. We need measured details to work the exact math. Current waveforms, in particular.

At work, I'm doing very similar work. I'm doing a monitor for very large circuit breakers. They use a remote 125V dc control voltage to move a solenoid. I monitor coil current. You can look at the waveform and tell exactly when the trigger point is reached and when the plunger is no longer moving. The coil will burn out in about 5 seconds if left on. There is a contact set that removes power to the coil when, and only when, the trip has occured. If it's jammed, it will burn out.
 
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Old 03-07-17, 07:49 AM
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Try this:
  1. Draw your circuit showing each device; Power supply, Exit device, Wire and Connections.
  2. Make a label for each device that includes: Resistance of the device, Current flowing through the device, Voltage drop across the device and Power consumed by the device.
  3. The label for the power supply will be the maximum Power supplied at the specified Voltage (from the specs). If not stated, current can be figured with Joule's Law. Internal resistance of the power supply is not important. The label for the coil will use the inrush current (around 16A iirc).
  4. Using Ohm's Law and Joule's Law, fill out your labels based on the current required by the coil at its operating voltage. Using 18 gauge wire, and the current drawn by the coil, figure the voltage drop. Don't forget to either show and figure the resistance for each conductor (out and back) or double the resistance.
  5. Now fill one out using 14 gauge wire.
Remember that in a series circuit: Current is the same at all points, Resistance, Voltage drops and Power are cumulative. Becuase the coil needs x Amps @ y volts and E = IR, the voltage drop is greater in 18 gauge than in 14 gauge and the coil isn't getting the same voltage. Since P = IE and E has dropped, P has to drop. (I think not considering Power may have obscured what you were looking for.)

You were probably close to the limit with 18 gauge wire and increased resistance at a splice sent it over the inrush cliff.

If you set up some formulas in a spreadsheet, you can play with the numbers and see the results. Set up each "device" separately and see what effect a change has on a device and on those "downstream".
 
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Old 03-07-17, 08:33 PM
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Thank you, gentlemen. I think I'll conclude the possibility of cumulative small factors such as a), power supply slightly below spec on overload threshold, b), one or both solenoids slightly lower resistance than spec, c), potted circuit breakers (PCB) disengaging from high-current contact slightly longer than spec. and d), a weak splice or 2 in the wiring.

But perhaps the likeliest explanation might be the PCB's sensitivity to an expected amount amount of current, (therefore, heat) to switch off high current circuit in a spec'd time, and with less current, even if only a reduction from 18A to 16A, the resulting extended ON time may not be offset enough by the reduced current sensed by the power supply. The doggone PCB's list for nearly $200 so I won't destroy one to see how it works unless I acquire a defective one in the future. If so, I'll let y'all know.

Cheers!
 
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