Excessive current draw by strike lock

Other Hardware General Electronics
21

Yes. The sensor is connected to an Arduino Mega analog pin.

The 12VDC supply is used for the strikes, strobe, etc.
From that, an 8V voltage regulator supplies power to the Arduino, and to a 5V regulator.
The 5V regulator powers the relay boards, temperature sensors, hall effect sensor, and LCD display.

I can appreciate the possibility of what you are suggesting, and agree that a voltage drop could affect the sensor reading.

However, the fault causes the main power supply to reset - this is indicated by the sudden phone call from the security company. The radio transmitter for the security company is connected directly to the main power supply.

That power supply has a 5A peak rating, so under normal conditions, the 12V power board registers 1.25A ( excluding the current pulled by the security radio which is located before the power board ). As soon as the strike gets power, it draws sufficient power to exceed the peak ( not sure of how much above 5A it tries to pull ) and the reset occurs.

If it was a case of only the Arduino that was reset, then I could agree with your scenario.

22

outofoptions:
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?

I am reasonably confidant in the reading. I also have a second strike lock ( a different type, but contains a coil all the same ) and when I activate that one, the current increases from 1.25A to around 1.9A. Other components that also use the 12V supply are also drawing around the expected / rated current for each of the items.

Yes, I tried connecting the strike directly to my power board ( 12V after the fuse, diode and hall effect sensor ) using much heavier ( and shorter ) cable than what is used to the gate. Result was the same, so wire gauge / wire short can be discarded as a direct cause.

23

Again, have you tried powering the system directly from a 12V battery? How much current does the strike in question draw when measured with a multimeter?

24

It is a 12V system, and the resistance over the coil is 24 ohms ( yes, measured with a multimeter ). This means the current should be 500mA. So how could it possibly consume more than 6 times that ?

25

My concern to do that would be the damage to the battery if the coil did draw 5A ( or more ? ). I am not certain that a simple 9Ah battery is designed to withstand such a sudden draw.

The 24 ohm was measured across the coil contacts while it was disconnected.

26

DaveO:
It is a 12V system, and the resistance over the coil is 24 ohms ( yes, measured with a multimeter ). This means the current should be 500mA. So how could it possibly consume more than 6 times that ?

So, you have measured the actual current consumption with a multimeter? As in multimeter in series with load. Have you substituted a hefty 12V battery for the 12V power supply? Do you get the same current printouts with the battery substituted for the power supply?

I wouldn't worry about damaging the battery for a short test.

27

No, not the consumption. I measured the disconnected coil over the 2 terminals and got a reading of 24 ohms.

You are suggesting powering the coil with something like a car battery ( the strike is rated, I think, from 8 to 18V ) with an amp meter in series and seeing the reading ?

I am off to the local auto electrician first thing in the morning to get the car aircon repaired, and will take the lock with me. I am sure he would have a decent amp meter ( which I do not - yet - have ).

28

How about a schematic of what you have wired up? I am not convinced from your various descriptions that things are wired correctly.

29

CrossRoads:
How about a schematic of what you have wired up? I am not convinced from your various descriptions that things are wired correctly.

Sounds like a reasonable idea - give me an hour ( hopefully ) to get the kids into bed and I will sketch it out.

30

DaveO:

[quote author=Papa G link=topic=163753.msg1225712#msg1225712 date=1367513690]

DaveO:
It is a 12V system, and the resistance over the coil is 24 ohms ( yes, measured with a multimeter ). This means the current should be 500mA. So how could it possibly consume more than 6 times that ?

So, you have measured the actual current consumption with a multimeter? As in multimeter in series with load. Have you substituted a hefty 12V battery for the 12V power supply? Do you get the same current printouts with the battery substituted for the power supply?

I wouldn't worry about damaging the battery for a short test.

No, not the consumption. I measured the disconnected coil over the 2 terminals and got a reading of 24 ohms.

You are suggesting powering the coil with something like a car battery ( the strike is rated, I think, from 8 to 18V ) with an amp meter in series and seeing the reading ?

I am off to the local auto electrician first thing in the morning to get the car aircon repaired, and will take the lock with me. I am sure he would have a decent amp meter ( which I do not - yet - have ).
[/quote]

Yes, that is what I'm suggesting. Get a cheap Chinese multimeter as well. I paid $10 for the one I use when I'm working on my vehicle.

DaveO:

CrossRoads:
How about a schematic of what you have wired up? I am not convinced from your various descriptions that things are wired correctly.

Sounds like a reasonable idea - give me an hour ( hopefully ) to get the kids into bed and I will sketch it out.

Excellent suggestion.

31

OK. Here's the diagram. Apologies if it is a bit crude - I have tried to keep it as simple as possible.

I have drawn the positive lines in red, and ground lines in blue. Added A, B, C .. in purple just in case anyone wants to reference them.

board1.jpg
board1.jpg786×551 46.3 KB

32

You are drawing 4+ amps and NOT blowing the fuse on the power board even though that is a 25% over current? It was over 5? I can see a fuse perhaps surviving one or two times but not repeatedly.

33

outofoptions:
You are drawing 4+ amps and NOT blowing the fuse on the power board even though that is a 25% over current?

That's what is so damn confusing. The sensor says 5A, then the power supply resets. And the strike is powered after the fuse and diode.

The power supply is a 5A Peak Current rating, so understandable that it resets at 5A.

34

This was a comment I made early in the discussion, clearly wrong as written. After having read all that follows. First a fuse unless otherwise marked/made as being a fast acting fuse requires about 10 seconds to blow.. at 200% of rated current. Fuses are solely for the prevention of fire.. So the problem with the strike coil seems to point to a defective coil... IMO something within the coil moves when activated (I used to use solenoids for all my work, they controlled the irrigation valves that I was making controllers for). Occasionally one wouldn't work except from a battery and grew noticeably warm after 30 seconds or so of being connected to a hard 12V supply (A fresh 12V 7A SLA battery) and these solenoids usually had shorted turns inside the coil and they would move when the coil was activated drawing more current than a static test with an ohm-meter would indicate possible.
In other words a defective striker coil that fails when activated. I might add that all my solenoids were activated by dumping a charged 4700 uF cap into the coil with a relay for polarity and a Mosfet for the switch that grounded one end of the coil.

I wonder if there isn't a back emf diode in each striker solenoid and that they are reversed or damaged. 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 diodeor damaged would do just exactly that and with any appreciable length of wire attached could well draw the currents mentioned by the OP because if the resistance of the wiring connecting striker coil to the controller... A quick test would be to clip one end of the diode and see if it changes and proof would be to measure the current drawn by the coil directly from the battery.

Doc

35

Docedison:
This was a comment I made early in the discussion, clearly wrong as written. After having read all that follows. First a fuse unless otherwise marked/made as being a fast acting fuse requires about 10 seconds to blow.. at 200% of rated current. Fuses are solely for the prevention of fire.. So the problem with the strike coil seems to point to a defective coil... IMO something within the coil moves when activated (I used to use solenoids for all my work, they controlled the irrigation valves that I was making controllers for). Occasionally one wouldn't work except from a battery and grew noticeably warm after 30 seconds or so of being connected to a hard 12V supply (A fresh 12V 7A SLA battery) and these solenoids usually had shorted turns inside the coil and they would move when the coil was activated drawing more current than a static test with an ohm-meter would indicate possible.
In other words a defective striker coil that fails when activated. I might add that all my solenoids were activated by dumping a charged 4700 uF cap into the coil with a relay for polarity and a Mosfet for the switch that grounded one end of the coil.

I wonder if there isn't a back emf diode in each striker solenoid and that they are reversed or damaged. 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 diodeor damaged would do just exactly that and with any appreciable length of wire attached could well draw the currents mentioned by the OP because if the resistance of the wiring connecting striker coil to the controller... A quick test would be to clip one end of the diode and see if it changes and proof would be to measure the current drawn by the coil directly from the battery.

Doc

Thanks for the reply doc.

And for the info about the fuse.

My original suspicion was a faulty coil ( in the strike lock ) so I changed the complete unit with a new one - same result.

This did give me an old lock to strip down, and all I found was the 2 contacts connected to a small coil, with a metal piston through the center, which did the unlock action. There are no other components what-so-ever in the lock unit or coil.

As far as time is concerned, the lock / coil is energized for 2 seconds ( in the code ) and resets after at least a full second has passed.

I have removed the lock and will be visiting the auto electrician in the morning.

36

Just been looking at the board, and measured between Ground and the relay's NO pin, and it shows 17 ohms. But if the disconnected coil was at 24 ohms by itself -- shouldn't the reading be higher -- 24 ohm coil + 100ft cable ?

Is this evidence of a short waiting to happen when the relay connects the 12V on com to the NO pin ?

or is this caused by the diode on the coil contacts ?

ps .. re-measured with multimeter black lead on Ground = 12.1 ohms, and with red lead on Ground = 19.4 ohms. I assume this indicates that the diode is in place ?

37

with the striker coil disconnected?? If So inspect the cable/wired to the striker coil for sharp pinches or a screw or nail driven through it. One of my jobs in the far past was to service car phones and the major cause of failure on installation was either a cable laid over a sharp piece of car body or a screw driven through the cable. My quick test was to get another cable and lay it around the car from control head to radio and power it up via the new cable... When I had to go outside to the installation area (10 installation/service bay's wide) and prove to an installer that it was his error, I'd hand him his paycheck minus the time it took me to prove him wrong and tell him to collect his tools and let me inspect his toolbox before he left.. Only had to do that once..

Doc

38

OK. Update time.

Took the lock down to the auto electrician.

Connected the striker directly to a 12V battery, and energised as it should.

Also tried on a stand alone standard 12VDC alarm backup battery ( 7Ah ) and also operated fine.

Added an amp meter on the positive line in series, and registered 3.3A. So it looks like my home power supply and home system wiring is OK, and the hall effect sensor is registering the right current range that is actually being used. ( I was starting to wonder if I had some other serious error in my complete setup )

Went back to the security supply store. Chap there says that they usually connect these strikers to a wall plug-in AC adapter - input 230 VAC, output 16 VAC, 16 VA, 1A. He had a box full of these and says they never had any problems using them. He can't understand why it is drawing 3.3A. The rating is 350mA, 8-14 Vac / dc.

To solve the problem of not having a gate lock : if I want to connect the striker to one of these AC adapters, would it be safe to connect 1 of the AC lines directly and permanently to the striker ( like I currently have the 12VDC Ground connected ) and pass the other line through my relay ( com to no pins, controlled by the Arduino ) ?

Does it matter which line I use for the permanent connected line ?

Should the striker still have a diode over the contacts, or is that not applicable if I am using an AC supply ?

Is it possible that a coil that draws 3.3 A on a 12VDC circuit, would draw less than 1A on 16VAC ?

ps .. just found this statement on google :
DC coils need enough resistance in the windings to limit the current, and AC ones limit it with inductance.

So does this mean that the coil ( most likely designed for AC usage ), if used with AC would draw the rated 350mA, but has little resistance, so if used on a DC circuit, would be the same as a 'short' ( as suggested by the low 12 ohm reading between Ground and the positive side of the coil when in the off position ) and would rely on the DC power supply having a high enough peak rated to be able to 'over-power' the short for the duration that the coil is energised. If this is the case, wouldn't a resistor in series on the 12VDC supply to the coil limit the current, thereby preventing the total and system fatal short ? I think that if this is the case, the supplier should clearly indicate that although the coil is rated for AC and DC, the use of DC does require additional circuitry and / or a much more powerful ( and hardy ) power supply.

39

my 2 cents;-

Observation:

  1. Magnetic Field Interference of ACS712.
    The sensor is sensitive to external magnetic fields.
    App-note of allegromicro, Managing External Magnetic Field Interference When Using ACS71x Current Sensor ICs.
    App-note

  2. DC strike lock current changing rate.
    DC strike lock= Inductor
    "Ohm's Law" for an inductor
    v=L*(di/dt)
    Changing the rate of current is base on v/L, and current slowly top up to V/R.

Solution:

  1. confirm correct current by current meter.
  2. magnetic shielding box to block MFI (Magnetic Field Interference) for ACS712.
  3. Use separate power supply with DC-DC SSRs (Solid State Relay) for both strike locks or opto-isolator relay board to isolate between strike locks and rest of system. The key issue is floating ground.
  4. Use 24 V power supply with 24 V DC strike locks to prevent long and thin wire runs could lead to voltage drop at the product and further problems.
40

To solve the problem of not having a gate lock : if I want to connect the striker to one of these AC adapters, would it be safe to connect 1 of the AC lines directly and permanently to the striker ( like I currently have the 12VDC Ground connected ) and pass the other line through my relay ( com to no pins, controlled by the Arduino ) ?
Yes, obviously making sure to isolate both the AC leads from Ground.

Does it matter which line I use for the permanent connected line ?
No.

Should the striker still have a diode over the contacts, or is that not applicable if I am using an AC supply ?
No diode.

Is it possible that a coil that draws 3.3 A on a 12VDC circuit, would draw less than 1A on 16VAC ?

Yes, very likely.