RELAY SWITCH!!!!

[nero1985]

im starting to use relay switches and i wanna know how long would a 9v battery last with a 5v relay and what value of resistor i would need to get 5v from a 9v battery

[R.G.]

Relay coils have a specified coil resistance. That coil resistance determines how much current flows through the relay coil when a voltage is placed on it.

Although relays are specified by voltages, they're actually current activated devices. A "5V" relay may have coil resistances of 50 to several hundred ohms, depending on the size and physical design. My personal favorite, the NEC EA2 relay series has a 5V relay with a 178 ohm coil. What that tells you is that it takes 5v/178 = 28.1ma to run that coil.

The EA2 will operate (pull in) at voltages of 80% of nominal, or 4V: that means it will operate on 4v/178 = 22.5ma.

How that relates to your question is this:
- If you hook a 9V battery directly to a 5V EA2 (which is a representative low power relay) the coil current will be 9V /178 = 50.5ma. A 9V battery is a 160ma-Hr to 320 ma-hr device, so the battery would last 160/50.5 = 3 to 6 hours. However, the relay coil would probably burn out before then.
- if you hook a resistor in series to keep the current down to 28ma, then the coil won't burn out, but the battery will last 5.6 - 11 hours before being exhausted.

You can't pick a resistor to get 5V from a 9V battery; you have to pick a resistor to fit the current that flows in the 5V relay coil when there is 5V across it and to soak up the excess voltage. A 9V battery is usually 8.8 to 9.3V, but sags to as low as 7V when getting low. If you had a perfect 9.000V battery, and a 5V relay coil operating on 5V across the coil, you could figure the necessary resistor by dividing the 5V across the relay coil by the coil resistance to get the coil current, then using that current and the difference between 9V and 5V (that is, 4V) and dividing *that* by the relay coil current to get the correct resistor value. As a hint, it will be 4/5 of the resistance of the relay coil.

[Paul Perry (Frostwave)]

You could save current drain wiht a latching relay, this 'holds on' without continuous current (a trick involving two coils & an internal magnet I think. Mouser has little PCB mount ones.)

[Peter Snowberg]

Latching relays come in two styles.... one coil and two coil. The two coil units have a coil to set either state while the single coil units use swapping the coil polarity to set the state.

[nero1985]

i found some relays at RS, i was thinking about using those relays that are always on one position unless u give it some current, so i think im going to got for those and im going to use a 100k trim pot to drain current to the relay so that the relay doesnt break!!!!!!!!!!

[Hal]

i would think you'd wanna use an adapter....

[Athin]

if battery life is a concern - don't use relays - trannies or 4066's are the way 2 go...

[R.G.]

i found some relays at RS, i was thinking about using those relays that are always on one position unless u give it some current, so i think im going to got for those and im going to use a 100k trim pot to drain current to the relay so that the relay doesnt break!!!!!!!!!!

The hold-in current of a relay is indeed smaller than the pull-in current, but it's usually on the order of 20% of the pull-in current, so you're not likely to get there with a 100K resistor. Something like five times the coil resistance in series with the coil is likely to be the maximum that lets the hold-in current flow. You're in the 1K region with the EA2 example, not the 100K range.

[nero1985]

the battery life isnt a problem at all, i just dont want the relay to burn like mr. R.G. said i heard i needed a resistor for the the relay so that it could just get the current it needs

[cd]

Use this circuit as a driver:

http://img36.exs.cx/img36/2599/relay.gif

Adjust the power supply as necesary.