Excessive current draw by strike lock

I have just replaced a power supply.

Output is 12VDC with continuous current of 2A and peak 5A.

The 12VDC is used to power :

1. a small radio transmitter, then
2. a GSM unit, then
3. an external 3 channel remote control receiver, then
4. 12V devices, including 2 strike locks for exterior gates, then
5. voltage regulator to 8VDC for powering the Arduino, then
6. voltage regulator to 5VDC for powering relay output boards, temperature sensors, ACS712 Hall Effect Sensor board, etc

The ACS712 is connected to the 12V line, between '2' and '3' above.

Under normal conditions, the sensor registers current draw between 1.25 A and 1.35 A.

If I trigger a relay to activate the first 12V strike lock on Gate 1 ( for 2 seconds ) , current increases to 2.51 A, and then back down to the 1.3A range. This is the 'heavy duty' lock.

If I trigger another relay ( same relay board ) to activate the 12V strike lock on Gate 2 ( also for 2 seconds ) , current increases to more than 4.7 A.

The cabling to both gates is the same type - 12 core 'house alarm' type cable.
Gate 1 is further in distance than Gate 1.
Gate 1's cable has data wires used for other items in the same cable.
Gate 2's cable is used only for the Gate lock.

The specs on the second strike lock is that it uses less than 500mA.

At this point, the system resets.

My logic is that the power supply is protecting itself and as soon as the current exceeds the rated 5A Peak, it shuts down. This causes the current to drop, and it turns itself back on again.

I am reasonably certain that the reset is not just the Arduino side, as the items mentioned in 1, 2, 3 and 4 ( before the Arduino system ) are also affected by the reset.

So in summary, Gate 1 ( heavy duty ) is drawing about 1.2A

Gate 2 draws in excess of 3.7A.

Any ideas on where my current is going ?

OK. Having done some additional searching, it would appear that a short is the most likely cause of the problem.

Would I be correct in thinking that if the 12VDC+ line ( controlled by the relay ) is somewhere damaged and touching a metal object, that this would be considered a short, even if the object being shorted to is not connected to the 12V circuit in any way ?

And that the circuit resistance would drop, current flow would increase ( just draining away as fast as it can through the short ), current draw would exceed the power supply 5A peak rating, and this would cause the reset ( which then disables the relay which disconnects the shorted wire from the 12VDC supply ) ?

Sounds like an excellent discussion of what might be occurring.

Have you tried measuring the resistance of each circuit? I'm guessing your 'strike lock' has a coil? If it binds it might draw more current.

Possible also that a relay closure is unintentionally wired in parallel with coil.

Many Thanks to you both for the replies.

I think that my first ( easiest ) option is going to be to inspect the wiring at each end for any damage. I do have some spare wires in the cable, so can connect a second wire to each 12+ and 12- . That way I should be able to measure resistance over each line to see if there is a problem ?

Second step would be to examine the lock unit for any problems.

Will give feedback once done.

I wonder if there isn't a back emf diode in each striker solenoid and that they are reversed. An inductor can be expected to draw a heavy current for several hundred uS but not continually as Op's explanation would seem to indicate. OTOH a reversed diode would do just exactly that and with any appreciable length of wire attached could well draw the currents mentioned by the OP.

Doc

I have just replaced a power supply.

Why?

Did everything work before you replaced the power supply?

Excellent question. The previous supply died the day before, and I only noticed 6 hours later when the battery shut down.

The exact cause - not sure. I have not had the time to spend on the old supply PCB, but since it has some components that I do not recognise, and a couple of really large capacitor looking thingy-ma-goodies ( read : 'i don't' really understand what I am looking at' ) , I will most likely err on the side of caution and just discard it.

Also, living at the coast, we do get a large amount of corrosion and have grown to accept that things do deteriorate faster here.

Did it all work before ? Yes. But I think that the previous supply may have had a higher peak rating, so it may have just managed to supply the short drain for the 2 seconds that the relay was on, before shutting itself down.

Docedison:
I wonder if there isn't a back emf diode in each striker solenoid and that they are reversed. An inductor can be expected to draw a heavy current for several hundred uS but not continually as Op's explanation would seem to indicate. OTOH a reversed diode would do just exactly that and with any appreciable length of wire attached could well draw the currents mentioned by the OP.

Doc

Inductors resist change in current, so you would not expect current spikes at all. Voltage spikes yes, but not current.

Actually the striker solenoids are possibly rated for AC, not DC? In that case the DC current flow could be higher
than expected.

The original description says that data lines are sharing a cable with the heavy current wires - this could
explain the resets perhaps. Never wise to put delicate signal wires alongside high power wiring.

OK. the plot thickens.

Just purchased a new strike lock and connected - same problem of drawing too much current.

The specs on the new lock are that it uses 12VDC 350 mA.

Removed the strike lock and connected with new wires direct to the power supply, and again same result.

So this tells me the probability of wiring problems to the gate are slim, and the likelihood of the new lock being faulty the same as the old lock also not probable.

Next I stripped down the old lock and found that the contacts go directly to a large solenoid coil that pulls a plate that allows the lock to open.

However, I don't see any other circuitry or components whatsoever. Now I know that a 12V relay works in the same manner, and that requires a diode over the contacts.

Am I correct to assume that the high current that I am registering when energising this lock, is actually reverse current from the coil shutting down when it is de-energised, and this is in effect shorting the power supply and causing the reset ?

If so, should a diode over the contacts on the lock correct the problem, and what size diode should I use ?

In my box-of-bits, I have some 1N4001 and MBR340 ( 3A ) diodes. If either of these would do the job, which way should it be orientated - silver band on the diode to the positive terminal, or Gnd terminal ?

For any relay operated from DC add a free-wheel diode (rating same as current of coil). For both AC and DC a snubber
circuit can be used.

But before going further measure the winding resistance of the coil with a multimeter - if it really is 12V 0.35A, it should have
a resistance of 34 ohms or so. Measure both directions - if there's a difference that implies a diode is present.

Am I correct to assume that the high current that I am registering when energising this lock, is actually reverse current from the coil shutting down when it is de-energised, and this is in effect shorting the power supply and causing the reset ?

No, the kickback from a coil when switched off is momentary - it might generate a voltage high enough to damage the power supply
of course - have you checked the supply is working normally?

MarkT:
But before going further measure the winding resistance of the coil with a multimeter - if it really is 12V 0.35A, it should have
a resistance of 34 ohms or so. Measure both directions - if there's a difference that implies a diode is present.

Thanks for the reply MarkT

As suggested, I have measured the coil in both directions, and reading 24 ohms both ways.

I assume then that this indicates no diode present ?

The power supply does appear to be working ( I assume correctly ) as all other components connected to the supply are operating perfectly.

Would you suggest adding a diode to the lock contacts should solve the problem, and if so, which diode ?

24ohms means they take 0.5A. Any diode capable of 0.5A peak.

OK. Have added the 1N4001 to the strike lock over the contacts, with the 'band' on the positive lead side.

Peak Current when energized has reduced from +5 A ( which caused a momentary power supply reset ) to around 4.5 A, so it is not causing the power supply to reset anymore.

Unfortunately, it is still too high and is causing the armed response radio transmitter to register and transmit a 'power low' signal.

In essence, I think that this new power supply is not capable of supplying enough current for everything that I have connected to it.

Maybe I need to re-think the setup and remove the 2 gates which have coil relays in them, to a separate power supply. The fact that these gates will not open electronically when we do have a power outage, is not a major concern.

Since I am supplying power to them via a relay board connected to the Arduino, it should be a simple task to move the COM from the existing supply to a different supply, as no common ground would be required.

Before moving the power sent to the strike lock to a different ( separate ) power supply, I have one thing that is still baffling me :

The main 12VDC line from the battery backup power supply is used to power the armed response radio transmitter, and then goes to my 'power board'.

On the Power Board, I have a 3A glass fuse, and then a MBR340 ( 3A ) Diode. The output from there is through an ACS712 hall effect sensor breakout board ( used to measure and log the current used ). 12VDC Power for the strike locks is taken after that point. Ground is permanently connected to the common ground, and 12VDC to the control relay COM, with NO to the strike.

So if the strike lock is drawing around 3A, making the total system ( after the fuse and diode ) current around 4.5A, shouldn't this have blown the fuse, or would the 3A diode be protecting the fuse from blowing ?

The strike lock is rated 350mA, and I have a 1N4001 diode over the lock contact points, so how could it possible draw around 3A ?
Wiring is not a factor, nor is the relay board used to activate the strike, as connecting the strike direct to the power board causes the same result.

DaveO:

[quote author=Papa G link=topic=163753.msg1223276#msg1223276 date=1367362765]
Did everything work before you replaced the power supply?

Excellent question. The previous supply died the day before, and I only noticed 6 hours later when the battery shut down.

The exact cause - not sure. I have not had the time to spend on the old supply PCB, but since it has some components that I do not recognise, and a couple of really large capacitor looking thingy-ma-goodies ( read : 'i don't' really understand what I am looking at' ) , I will most likely err on the side of caution and just discard it.

Also, living at the coast, we do get a large amount of corrosion and have grown to accept that things do deteriorate faster here.

Did it all work before ? Yes. But I think that the previous supply may have had a higher peak rating, so it may have just managed to supply the short drain for the 2 seconds that the relay was on, before shutting itself down.

[/quote]
I may have missed it in another comment but have you tried running the system directly from a 12V battery, thereby eliminating the power supply? It's entirely possible that what I highlighted in red has been the answer all along and you just need to fit a heftier power supply.

Thanks Papa G

I have little doubt that a power supply with a higher peak rating ( like the previous supply ) would most likely survive the excessive current draw for the short period that the strike is energized.

But it still doesn't explain why a 350mA strike is drawing 3A which causes the existing supply to reset.

It is also quite possible that the 'expiry' of the old power supply could have been caused by this excessive draw, having placed strain on the components.

I was prepared to isolate the power to the strike by adding a separate power supply / transformer just for the strike locks, but I think that I first need to try to understand why it is happening.

So I am back to the question of why the fuse didn't blow, and is the 3A diode in line after the fuse protecting the fuse, but failing because of the current ?

DaveO:
Thanks Papa G

I have little doubt that a power supply with a higher peak rating ( like the previous supply ) would most likely survive the excessive current draw for the short period that the strike is energized.

But it still doesn't explain why a 350mA strike is drawing 3A which causes the existing supply to reset.

It is also quite possible that the 'expiry' of the old power supply could have been caused by this excessive draw, having placed strain on the components.

I was prepared to isolate the power to the strike by adding a separate power supply / transformer just for the strike locks, but I think that I first need to try to understand why it is happening.

So I am back to the question of why the fuse didn't blow, and is the 3A diode in line after the fuse protecting the fuse, but failing because of the current ?

You're measuring the current via the hall effect device, right? Can I presume that you are reading it via an analog in on the Arduino? If so, then imagine this:

the current demand on the new supply causes its voltage to drop below the dropout voltage of the Arduino regulator and the supply voltage to the microprocessor drops to say, 3.5V. If at that time your sketch reads the analog input for the hall effect (current sensor) it will use the 3.5V supply as the reference for the ADC but your calculation will be done assuming a 5V reference and the result will be a higher than actual current. Just a theory, mind you.

Are you sure about your current reading and the decimal place is not misplaced? Have you tried hooking the strike up directly to the board and not installed? Are you sure about the gauge of your wires to the latch?