looking for simple way to drive 24VAC contactor coil from 3.3VDC & 2mA

Thread Starter

htroberts

Joined Jun 22, 2012
5
I have a 3.3VDC output from a developer board that I need to use to control the 24VAC coil of a definite-purpose contactor (like one you'd have in an air conditioner).

I could use something like this:
http://pcbheaven.com/wikipages/Transistor_Circuits/
but then I have another relay, and need to supply another DC voltage.

Any elegant way you can think of to get from 2mA @ 3.3VDC to ~7VA @ 24 VAC?

A latching or flip-flop arrangement would be even better...

Thanks,
Heath
 

MikeML

Joined Oct 2, 2009
5,444
...
Any elegant way you can think of to get from 2mA @ 3.3VDC to ~7VA @ 24 VAC?
...
You are asking us to violate the laws of Thermodynamics.

The power required to pull in the coil of said relay is several Watts. You have 3.3χ0.002 = 6.6mW available...:D If you can figure out how to do this without using an external power supply, I will buy stock in your company...
 

Thread Starter

htroberts

Joined Jun 22, 2012
5
You cant drive an AC relay with DC, you have to make a DC to AC inverter, or better still use a dc coil relay.
DC coil contractors are not commonly available, and the ones that are won't pull in on 2mA at 3.3V.

My question must have been worded poorly. I did'nt intend to drive the contractor coil _directly_. I need help designing an appropriate circuit to interface the input I have to the output I need.
 

cork_ie

Joined Oct 8, 2011
428
There are 3.3V solid state relays available but you would need about 50 ma to drive one. Is there anyway you could get that 50ma from your controller?
You would then use a small mains to 24V transformer and attach it next to your contactor using the same mains power supply to both. The solid state relay will do the switching for you without an extra power supply.
You may find a suitable SS Relay here
 

KMoffett

Joined Dec 19, 2007
2,918
So the development board is powered by 3.3VDC and the signal output pin can only supply 2mA.

Driver a logic level MOSFET with the 3.3v signal output of the board. Have the MOSFET switch the LED in an SSR from the 3.3VDC from the boards power source. The SSR then controls your AC contactor coil.

Actually most SSR's can be switched with currents as low as a mA. Look for the ones that spec an "input current", rather than "3-32VDC input" voltage. Though the 3-32VDV ones will probably also work. They have a 1500Ω internal current limiting resistor that is set to limit the LED current to a safe value at the highest input voltages. At low input voltage they don't draw much: (3v-1.7v)/1500Ω=0.86mA you might even be able to drive one directly from the board.

Ken
 

Thread Starter

htroberts

Joined Jun 22, 2012
5
So the development board is powered by 3.3VDC and the signal output pin can only supply 2mA.

Driver a logic level MOSFET with the 3.3v signal output of the board. Have the MOSFET switch the LED in an SSR from the 3.3VDC from the boards power source. The SSR then controls your AC contactor coil.

Actually most SSR's can be switched with currents as low as a mA. Look for the ones that spec an "input current", rather than "3-32VDC input" voltage. Though the 3-32VDV ones will probably also work. They have a 1500Ω internal current limiting resistor that is set to limit the LED current to a safe value at the highest input voltages. At low input voltage they don't draw much: (3v-1.7v)/1500Ω=0.86mA you might even be able to drive one directly from the board.

Ken
Thanks, this is a really good approach, I think. Now I just need to find an SSR that meets my i/o specs. I like the potential for optical isolation as well--don't want my $10 24VAC transformer to fry the $300 dev board--although it's not obvious how much isolation there really will be.
 

crutschow

Joined Mar 14, 2008
26,439
Actually, you can. Start with about 6 to 8Vdc applied to the contactor coil.

The DC drive voltage is typically about 1/4 to 1/3 of the AC rating...
You can determine the needed drive voltage by looking at the relay spec to determine the nominal AC coil current rating and also measuring the coil resistance with an ohmmeter. The required DC coil voltage then equals I*R.
 

cork_ie

Joined Oct 8, 2011
428
You can determine the needed drive voltage by looking at the relay spec to determine the nominal AC coil current rating and also measuring the coil resistance with an ohmmeter. The required DC coil voltage then equals I*R.
Not quite as simple as that unfortunately.
Two things need to be born in mind the power required to PULL IN the contactor is much greater than to HOLD IN the contactor and this is going to be reflected in the back EMF that will exist in the contactor coil. If the DC voltage is too low contactor pull in will be sluggish at best or at worst not pull in at all - result burnt contacts. If it is too high coil will heat. That is why PWM is so great.

I think he should stick with using a 24V AC transformer to operate the contactor coil.
 
Top