Of course, you can do that, it depends on the voltage and current of relay, and you can calculate the limiting resistor, you can measure the Vce or just using the voltage about 0.9V~1.1v.

so if its a 12V relay, and coil resistance is 150 ohm. Ordinarily it would draw 12/150=80mA.
If I'd like to decrease it , maybe I'd add 1kohm resistor. So the current would be 12/1150=10mA. Would the relay still switch? There doesnt seem to be anything about it on the datasheet.

 

so if its a 12V relay, and coil resistance is 150 ohm. Ordinarily it would draw 12/150=80mA.
If I'd like to decrease it , maybe I'd add 1kohm resistor. So the current would be 12/1150=10mA. Would the relay still switch? There doesnt seem to be anything about it on the datasheet.

You can't reducing the current of relay itself, only can reducing the offering current, if you cut the current too much then the relay may not active.

Relay not like the LED that you can limiting the current and reducing the brightness.

 

Actually, you can reduce the hold current, as the activation current requirement is generally higher than the hold current. This can save a considerable amount of power, while reducing heating of the relay coil.

A simplistic way to accomplish this is to charge a moderately large capacitor that acts as the coil's current source, via a resistor that is several times the resistance of the coil.

When the low side of the coil is initially grounded via the ULN IC's connection, the capacitor momentarily supplies full voltage from the power supply, causing the relay to actuate. The capacitor discharges to the point to where the charging resistor can keep up.

One possible problem with this technique is trying to cycle the relay too quickly, as it will take time for the capacitor to fully recharge; how long depends on the values of the resistor and the capacitor. You'll have to determine appropriate values via experimentation, as you haven't supplied a link to your relay's datasheet.

 

As a general rule of thumb for small relays, they usually will operate with 75% or 80% of rated current, and will hold at 50%.

ak

 

Actually, you can reduce the hold current, as the activation current requirement is generally higher than the hold current. This can save a considerable amount of power, while reducing heating of the relay coil.

A simplistic way to accomplish this is to charge a moderately large capacitor that acts as the coil's current source, via a resistor that is several times the resistance of the coil.

When the low side of the coil is initially grounded via the ULN IC's connection, the capacitor momentarily supplies full voltage from the power supply, causing the relay to actuate. The capacitor discharges to the point to where the charging resistor can keep up.

One possible problem with this technique is trying to cycle the relay too quickly, as it will take time for the capacitor to fully recharge; how long depends on the values of the resistor and the capacitor. You'll have to determine appropriate values via experimentation, as you haven't supplied a link to your relay's datasheet.

The way you mentioned just like preheating conception, although I try to reduce the activation current, but never try it, I think I need to find some plenty time to try it.

 

Ok, thanks
I'll better not mess with it.