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Power Supply for Electromagnetic Door Lock

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Gabe Billings

New Member
I have an electromagnetic door lock and I don't quite understand how different power input will affect it. The instructions that came with it note that it can be used at either 12V or 24V, drawing 500mA or 250mA respectively.

First off, I don't understand the dual voltages. There's nothing in the instructions or a visible way to switch from one to the other. Is it just simply a question of using one input power or the other, without having to actually 'do' anything to the magnet?

If that's the case, is there anything wrong with using a voltage between those values? I have an old laptop power supply that's 15V/5A that I'd like to recycle, assuming I can use it without anything catching on fire.

I hooked it up and checked it with my multimeter and it's drawing 360mA. Seems to be working as it should.

So, anything wrong with using this power supply?
 

dougy83

Well-Known Member
I have an electromagnetic door lock and I don't quite understand how different power input will affect it. The instructions that came with it note that it can be used at either 12V or 24V, drawing 500mA or 250mA respectively.
I think you mean 12V or 24V, drawing 250mA or 500mA respectively. An electromagnet is just a coil of wire (+ flux guides), so increasing the voltage will increase the current it passes (due to wire resistance), in addition to its pulling power.

In answer to you question, any voltage 12-24 will be fine.
 

Gabe Billings

New Member
Misinformation

It looks as thought we're both right. I double checked the spec page from the place I got the lock, and it says "Current Draw : DC12V/500mA(default) or DC24V/250mA".

Certainly, though, you seem to be correct, since when testing I saw 210mA at 12V and 360mA at 15.25V, the current rising with the voltage.

Whatever the case, knowing that nothing bad will happen using a voltage in the middle is reassuring. Thanks.
 

Boncuk

New Member
24V/500mA and 12V/250mA would make sense to me.

Applying DC to a coil it is just like a resistor. Use Ohm's law to calculate for current flow at any voltage.

Boncuk
 

smanches

New Member
Post the link to the spec sheet. I would disagree about the suggestions so far. It's most likely the model of the lock that denotes what voltage it uses.
 

Hero999

Banned
Does it have two coils?

You might need to connect them in parallel for 12V operation and series for 24V operation.
 

Ross Craney

New Member
Does it have two coils?

You might need to connect them in parallel for 12V operation and series for 24V operation.

Hero gets the prize. They do indeed have two coils running them at an intermediate voltage will result in it either being under driven or overdriven depending which winding you use. An important point is to ensure the armature plate is mounted with the supplied rubbers so it can self align to the magnet. There only needs to be a poofteenth amount of misalignment for the magnet not to hold properly
 

dougy83

Well-Known Member
Or maybe the manual refers to 2 different models. Running the 12V lock from 15.25V is going to dissipate 60% more heat - which may not be a problem for a door strike due to the short duty (but may be a problem for a electromagnetic lock).

You can put a 10 ohm 2 watt resistor in series with the coil if you're worried about the extra heat. The resistor will reduce the coil voltage to 12.3V and will dissipate 0.85 watts.
 
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