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Author Topic: Excessive current draw by strike lock  (Read 2264 times)
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South Africa
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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.
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Dallas, TX
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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.

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.
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South Africa
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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 ?

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Dallas, TX
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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. smiley
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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? 
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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. smiley

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.
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South Africa
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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.
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Dallas, TX
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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?
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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 ?
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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?

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

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.

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.
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