# How to make a 220 V AC solenoid lock

I need to make a powerful solenoid lock using 220 V AC.

Let say I will use 21 gauge wire and get a length of 24 Ω. This will draw 9 A from the 220 V, so 1980 W.

This will be great force electromagnetic but the problem is this: the 21 gauge can only withstand 1.2 A maximum; to avoid getting maximum current I need to make the resistance 183 Ω, which means about 3 km of wire!

And this will cause less force because the current now is 1.2 A; calculating other gauges in the same manner will result in this too-long length.

Is my calculation wrong, or is the reality not like the theory, or what?

How to make a solenoid coil that can connect to 220 V safely, and the gauge?

• Are you planning on wasting ~250 W continuously on a solenoid? Your average solenoid uses an iron core to increase the inductance and hence lower the current. Jul 21, 2022 at 23:18
• How 240w will be wasted?
– Tito
Jul 21, 2022 at 23:20
• This looks like a duplicate of this question, are you the same person who posted that one? You both seem to be asking a lot of questions about solenoids and coilguns. Jul 22, 2022 at 0:06
• I see this post before and is same what iam looking for but no answer till now
– Tito
Jul 22, 2022 at 0:26
• I am not sure whether the solenoid inductance will become significant at 50 Hz or 60 Hz. But if so, the inductance might reduce the current slightly when you are using AC. If necessary, you could use a large area core for your wire in order to get high inductance, or put a resistor in series to reduce current. Most solenoids have lots and lots of turns of fine wire. Jul 22, 2022 at 2:43

The 24Ω is the DC resistance of the wire. It is only relevant at AC if the solenoid has no core - because its inductance is so low that it has no effect at 50-60Hz mains frequencies.

You'll need a core in the solenoid to increase its inductance, and thus also its AC reactance at 50/60Hz. That way the magnetizing current will be manageable and you won't need lot of wire.

Solenoid locks used for doors are an item you just buy: it's a coil on an armature, usually running off 12V or 24V, and a metal plate that sticks to the armature when the doors are closed and the current is applied. Those work fine from DC, and their DC current is impedance limited.

An example is this Amazon listing, where the picture below comes from. The coil takes 0.3A at 12VDC. The holding force is about 180kg. You can buy similar locks that are much more powerful, yet consume just as little power. The idea is not to go brute force at the problem, but use magnetic circuits to your advantage. An electromagnetic lock is, after all a combination of an electrical and magnetic circuits. Yes, magnetic circuits, you read it right. Interesting stuff and worth learning about.

Is my calculation wrong, or is the reality not like the theory, or what?

The magnetic field strength is directly proportional to the MMF (magnetomotive force) expressed in ampere turns.

It's not the current alone that decides the field strength but also the number of turns.