Hi All,

I have some basic electrical knowledge, can use a multi-meter and very logical thinker.... However I've reached the edge of my knowledge and in search of some help with the problem below... Be gentle with me!!

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 model, different supplier) and the same thing happens.

 

First thought: The solenoid is 3.4Ω and the supply voltage is 12V so the current through the solenoid will b about 3.5A. The supply is rated at 3A. Putting a 1Ω bulb in series would reduce the current to 2.7A, now within the capabilities of the supply.

 

First thought: The solenoid is 3.4Ω and the supply voltage is 12V so the current through the solenoid will b about 3.5A. The supply is rated at 3A. Putting a 1Ω bulb in series would reduce the current to 2.7A, now within the capabilities of the supply.

In my head, logically that theory makes sense to me! - Please excuse the ignorance here, however, I was thinking that instead of a bulb, if I added in a 1 ohms resistor would that give the same result!?

I have no idea if that is the "right" use of a resistor, looking online I would also need to choose a watt value!? Would adding a resistor solve this? If so, what rating resistor should I buy? If not a resistor, is there another way of lowering the Amp to below 3A?

 

A 1Ω resistor seems like a sensible solution. It would dissipate about 7.5W so it should be rated at least 10W, better 15W.

 

A 1Ω resistor seems like a sensible solution. It would dissipate about 7.5W so it should be rated at least 10W, better 15W.

Thanks for your help AlbertHall!! - Just so I understand the maths, how did you get to 7.5W ?

(Googling I read Power = Volts / Amps so that would be 11.8V / 1.57A = 7.5W but I don't know where the 1.57A comes from?)

 

You have the 1Ω resistor in series with the 3.4Ω solenoid making a total of 4.4Ω. The current is then 12V/4.4Ω = 2.73A. The power in the 1Ω resistor is then I^2 * R = 2.73A ^2 * 1Ω = 7.5W

BTW Power is voltage times current.

 

Fixed. I couldn't find a 15W resistor locally so used a 10W ceramic one which has cured the tripping, really appreciate the explanations AlbertHall as I like to understand the cause when finding solutions!!

 

Make sure there is a clear airflow around the resistor as it will get warm.