Minimum power in coil to activate relay

[Rusttree]

I'm looking at using this relay for an automotive circuit. I want to make sure I understand the minimum power in the coil that will still activate the relay.

The datasheet specifies a pick-up voltage of 5.5V and a coil resistance of 225Ohm. Is it correct to assume, then, that the minimum power required to activate the relay is:
P = I^2*R = (5.5V/225Ohm)^2 * 225Ohm = 134mW

Or are there other factors I need to consider? I ask because I will be using an N-channel MOSFET to activate the coil and the FETs I'm looking at have fairly high Rdson. During an ignition start, a bad battery might dip pretty low, so I want to calculate how low it can go before the circuit won't work anymore.

Thanks!

 

[Mike odom]

during a start a good battery will dip down to about 3v. For automotive circuits, I use a buck/boost to supply a rock solid 5V from the battery.

 

[Rusttree]

I don't want to derail my own thread, but this is a point of confusion. I've looked at a couple of after-market circuits that control the start and ignition on a car (pushbutton start systems, etc) and they all just used a simple linear regulator to get to 5V or 3.3V. And from experience, they all seemed to work fine - even with well-used batteries. I would expect 3V from a nearly-dead battery, not a healthy one.

 

[Mike odom]

BTW, you can also do E^2/R to find power when you have voltage and resistance so you don't have to find the Intermediate I --> (E/R * E/R * R). The R cancels one 1/R to yield E*E/R for power.

P = E*I = I*I*R (replace E with I*R)
P = E*I = E*E/R (replace the I with E/R)

 

[Mike odom]

I don't want to derail my own thread, but this is a point of confusion. I've looked at a couple of after-market circuits that control the start and ignition on a car (pushbutton start systems, etc) and they all just used a simple linear regulator to get to 5V or 3.3V. And from experience, they all seemed to work fine - even with well-used batteries. I would expect 3V from a nearly-dead battery, not a healthy one.

I worked on an autostart circuit for diesel generators years ago, and this is what I remember. So what do these circuits do for cars with near dead batteries? Maybe they are tolerant of low voltage spikes so a simple linear regulator will do... like an ATMEGA can run down to 1.8V (linear regulators will follow the voltage down if they go below their dropout voltages). We were using an ASIC that needed a sharp 5V. If your relay needs a 5.5V minimum hold voltage, it certainly won't stand up to a low battery supply either.

 

[Mike odom]

Also, if your relay has a coil resistance of 250 ohm, then you'll need 5.5V to obtain that power. It's not like you can force it in with 2.2V and doubling the current...

 

[Mike odom]

If you are trying to determine how low the battery gets before the relay drops out, this is something you may just have to test to obtain. Also note that the Ron is characterized with Vds = Vgs, and if you are driving the gate with 3.3V but have a higher Vds, then you won't be able to calculate Ron from data given.

 

[Rusttree]

Also, if your relay has a coil resistance of 250 ohm, then you'll need 5.5V to obtain that power. It's not like you can force it in with 2.2V and doubling the current...

Oh I know. My intent was actually to calculate how much higher than 5.5V my minimum voltage would have to be. If, say, the FET had an Rdson of 10Ohm, then the minimum voltage to get 134mW would be 5.6V.

 

[dr pepper]

I've had some experience with engine management systems for vehicle engines, reasonable batteries tend to drop to 6v while the engine is cranking, however if its well cold or the battery is getting old then 3v is quite likely.
With vehicles relays are usually engaged before the engine is cranked, so the concern isnt so much the pulling in volatge of the relay, its more the holding voltage, as the relay(s) will be in when the starter engages.

It wouldnt be a good situation to have to pull in a relay while the engine is cranking, if you have to do that then as mentioned a buck/.boost reg capable of runing down to 3v would be a reliable way, better still use a tranny to switch the load if its not so high.

 

[Mike odom]

ok, for that you don't need to figure relay power, just current through the relay, that would yield the current through the mosfet, and give voltage drop across the mosfet. Add the two and you get total voltage. Again, not sure if this can be calculated... you will probably have to test it.

 

[Diver300]

On one engine that I was looking at some time ago, the engine management system stopped working below 5 V.

The maximum power is taken from the battery when the voltage is approximately halved by the load, so there is little advantage of going below 6 V.

The relay in the datasheet has a drop-out voltage that is less than 3 V, so as long as 3 V can be kept on it, it will stay energised.