# How can you determine the current needed to turn on a relay coil?

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#### Zanderist

##### Member level 3
I've encountered a problem with a relay (a small one you could pick up from radio shack).

I had it set up so an osicalling infrared led would switch on a transistor, which in turn would turn on a relay.

It didn't work, I was using a 9 volt wall wort at 500mA that ran through a 7805, giving me about 4.98 volt as the supply.

So I got rid of the transistor and used a mosfet, that didn't work either.

So I got rid of the power supply all tower and went straight for a 9volt battery, which then was working perfectly through the mosfet.

My problem is if I am using a pot (which I am monitoring, with a meter, the resistance of.) the moment it turns on the resistance changes.

I believe the threshold voltage required to turn-on the Relay is different if rising on a ramp waveform or on a pulse waveform.

+++

• Zanderist

### Zanderist

Points: 2
There should be a datasheet that describes the operating current of your relay.
An alternative is to measure the coil resistance , if you also know the relay nominal operating voltage you can calculate the current.
I didn't get the resistance part, maybe a schematic would help,
My problem is if I am using a pot the moment it turns on the resistance changes. :?:

Alex

P.S. relays turn on with lower voltage than the rated one, is that what you are asking, the minimum voltage/current that your relay turns on or the relay consumption?

• Zanderist

### Zanderist

Points: 2
check whether the IR driving the mosfet perfectly.
what are you using to sense IR, TSOP or IR photodiode ?

• Zanderist

### Zanderist

Points: 2
it is very simple..
I=V/R
I = current needed to turn on a relay coil. (in ampere)
V= applied voltage across relay coil ( you can find this marked on relay. or see data sheet)(in volt)
R= resistance of relay coil ( you can measure it with multi meter.) (in ohm)

by this value (I) you can calculate resistance for triggering the device (like transistor )
and for success full operation you must check the datasheet of triggering devie( transistor) that is it capable of carrying current ,, voltage and gain

• Zanderist

### Zanderist

Points: 2
Just one thing I want to point out to everyone, was that point of this experiment was to see an electrical singal get converted into light, transmitted and then turned back into an electrical signal by means of a mosfet.

The experiment was successful I was able to control a relay through low frequency PWM (about 10hz anything higher made it hum) which would also cause a small motor to switch on with the relay clicks. It also raised some other questions, like the one being asked in OP.

@ alexan_e

I can't provide a schematic at the moment but to describe it like so might bring clarity.

I have the pot in series with a multimeter mearsuring current (I screwed up when I said resistance) which then runs into relay's coil.

When the relay's coil pulls in or I hear the click, then current drops. (Just one point, I'm just testing the relays coil, nothing is at the terminals.)

One thing I want to know is what is driving the relay's coil? Is it the voltage? or is it the Current? I'm thinking it's current because the greater the curren,t the stronger the electromagnetic becomes.

@ asimkumar

The mosfet is driving it as planned, I've connected the frequency generator directly to the gate to see if that was the problem. The only thing I've done when using the sensing diode (it looks like an led but emitts no light and is clear) is I have a pot to bais the gate to ground which allows me to adjust it's sensivity to the IR led connected to the frequency gen. (of course I have to hold the IR led relatively close because I don't have a focusing lens fixed to the sense diode)

However one thing to consider, is that it only works when I am using a 9 volt battery. When I use a wall wort that should give me half an amp of current (rated as half an amp) running through a 7805 to bring it down to 5 volts, it seems as if I lose all my current through the regulating process.

it is very simple..
I=V/R
I = current needed to turn on a relay coil. (in ampere)
V= applied voltage across relay coil ( you can find this marked on relay. or see data sheet)(in volt)
R= resistance of relay coil ( you can measure it with multi meter.) (in ohm)

by this value (I) you can calculate resistance for triggering the device (like transistor )

This is the way to calculate the current consumption of the coil at the rated voltage but the switching occurs at a lower voltage.
For example with
the 12v relay has a resistance of 720 ohm which gives a current of 12/720=16.5mA but can actually pick up with just 8.4v which means 8.4v/720=11.7mA

One thing I want to know is what is driving the relay's coil? Is it the voltage? or is it the Current?

The coil has a resistance, to increase the current you have to increase the voltage across it, you can't provide more current that the load demands and on the other hand if you apply a voltage and the power source can't provide the current that the load wants then the voltage drops as well, there is a balance between these two.

Alex

very simple and practical way to find current is drive the relay with DC supply and in the path place and ammeter in series , you'll know the current needed to drive the relay.I've used this method when no datasheet is associated with relay.To find the current through resistance method is not good because the current you will determine will significantly change after going into saturation.

V
Points: 2
To determine the pull-in voltage or current of relay (or a solenoid valve), you have to apply DC. In this case, the relation between V and I will be simply ruled by the coil resistance, as already discussed. But you apparently applied a PWM waveform (although the information is somewhat hidden in your post). In this case, the current will chance because the coil inductance changed. The details depend on the exact circuit, which hasn't been revealed. And I'm not motivated to discuss all possible variants in advance.

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V
Points: 2