Electric lock tripping PSU

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Hi All,

I have a 240v Access Control Power Supply which outputs 12VDC / 3A via a built in relay, connected to a 12VDC electric lock (which has two solenoids & a 1n5408 diode in parallel across the terminals) and a 12VDC RFID keypad. I purchased these as a project to help me learn about RFID and pros / cons of installing in a domestic environment.

The Problem
Whenever I trigger the lock to open via the keypad (or just using a push switch) the PSU appears to trigger a protection and after engaging the lock for a split second, the PSU and keypad power down for a second then back on. If I replace the electric lock with a 12v AC bulb, it works correctly, the bulb stays lit for 3-4 seconds and the PSU doesn't power cycle.

After much trial and error, I stumbled upon if I connect the 12v AC bulb in series with the lock, everything works okay, the bulb lights, the lock opens and the PSU doesn't power cycle. After further trial and error with the components I had to hand, if I replace the bulb with another 1n5408 diode in series, connected with two wires using crocodile clips, the lock sometime works without the bulb. If I remove one of the wires with clips, the PSU power cycles again when triggered.

I measured various things with my multi-meter and this is what I found:

Relay voltage output = 11.8v DC
Solenoid resistance = 3.4 ohms
Crocodile clip wire resistance = 0.7 ohms
Bulb resistance = 1.0 ohms (its box states AC12V 50Hz 4.5W - MR16 275MD)

My Questions
1) What could be causing the PSU to trip?
2) Why would adding a bulb change that?
3) How can I solve this without using a bulb ?

Any questions or want me to test anything?

Help!!!

P.s. I tried another PSU (same type, different supplier) which resulted in the exact same problem.
 
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What's the current rating of the PSU? What's the pull-in current of the solenoid?

The only info I have for the PSU is as above, 12VDC / 3A output - Using an online calc, I worked that out to mean the PSU can output 35W / 3.93 Ohms ?

A couple of extra bits of info on the lock...

1) The "manual" (piece of paper) says 10W 8-12V
2) There two solenoids in the lock, I tried connecting only one of them and the same thing happens.
 
I purchased these as a project to help me learn about RFID and pros / cons of installing in a domestic environment.
Strange thing to do - I would have thought that initially you would benefit a great deal more if you learnt about basic electrical and electronic theory.


My Questions
1) What could be causing the PSU to trip?
2) Why would adding a bulb change that?
What's the pull-in current of the solenoid?
What's the pull-in current of the solenoid?
What's the pull-in current of the solenoid?
What's the pull-in current of the solenoid?
What's the pull-in current of the solenoid?
What's the pull-in current of the solenoid?


3) How can I solve this without using a bulb ?
Knowing what you are doing, and how the things which you have decided you should fiddle with actually work, would be a good start.

http://www.google.co.uk/search?q=electronics+101


Using an online calc, I worked that out to mean the PSU can output 35W / 3.93 Ohms ?
  1. Power supplies do not output resistance.
  2. Does the fact that you had to use an online calculator to see what a spec of 12V 3A meant show that you don't even know about Ohm's Law?

Seriously - the words before, run, and walk come to mind.
 
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Strange thing to do - I would have thought that initially you would benefit a great deal more if you learnt about basic electrical and electronic theory.

[snip]

Seriously - the words before, run, and walk come to mind.

Really? - Does everyone have to know everything about every subject!? You can't ask questions and get sensible replies??

BAS, you sound like a grumpy old man looking to find fault rather than actually help people which is a shame as I suspect you may actually be rather intelligent!?

Let's test that theory... since posting the original message, I've found the mathematical answer via someone rather more interested in helping. So BAS, can you enlighten the other members on the cause / solution for this problem?
 
Last edited:
Power supply 3 amp, lock 11.8/3.4 = 3.47 amp, add one ohm for light 12/4.4 = 2.73 amp. If the current is 3 amp and resistance is 3.4Ω then 3 x 3.4 = 10.2 volt and volt drop across a diode is approx 0.6 volt as 1.8/0.6 = 3 so you would need three diodes to drop the voltage enough. It is simple ohms law.
 
Really? - Does everyone have to know everything about every subject!?
Not everything.

But if you do not understand the basics upon which everything else is built then you will not be able to build anything stable.
 
Provide the details ( make and type numbers ) of the lock and the power supply.

In-rush current is a likely cause but not the only one.

One assumes you have connected the correct polarity of voltage to the coil, reverse polarity and the diode across the coil will conduct and put a short circuit across the PSU. The end of the diode with the white band must be connected to the POSITIVE from the PSU.. ( the diode's function is to reduce the back EMF voltage spike that is created when the solenoid is de-energised ).

Why are there two solenoids. ?
 
Power supply 3 amp, lock 11.8/3.4 = 3.47 amp, add one ohm for light 12/4.4 = 2.73 amp. If the current is 3 amp and resistance is 3.4Ω then 3 x 3.4 = 10.2 volt and volt drop across a diode is approx 0.6 volt as 1.8/0.6 = 3 so you would need three diodes to drop the voltage enough. It is simple ohms law.

Correct, that's more or less the cause/answer I was given else where. The solution chosen was to add a 1Ω 10W resistor in series (instead of diodes above). Which then gives the 11.8/4.4 = 2.68A and the lock works without tripping the PSU overload protection! :)

BAS, so you think the meaning of life is explained by Ohms law too? I thought it was 42!

Bernard, not sure why there are two solenoids, they are connected in series to pull an internal lever. (From the outside the lock looks like one you would find on electric gates)
 

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