# Relay Minimum Switching Current

I have a relay model RM40-3021-85-1005 (datasheet), and according to datasheet it has a minimum switching current = 10mA

The coil resistance is 125 Ω, and when I connect it directly to 5V it works as expected. The current is $I = \frac{V}{R} = \frac{5}{125} = 40mA$.

Considering I have a limited power source, and as my understanding the relay only needs 10mA and I'm providing 40mA, I tried to include a 100 Ω in series with the coil. So the current will be $I = \frac{V}{R_{coil} + R_{1}} = \frac{5}{125+100} = 22mA$

With the 100 Ω the relay doesn't operate and I believe it is because the voltage over the coil. According to datasheet, the minimum for the coil operate is 3.75V, and with the extra resistor I get 2.75V. $V = R * I = 125Ω * 22mA = 2.75V$

It is not clear to me. Is my understanding correct? If so, how can I operate the relay with a current lower than 40mA? For example 20mA.

As the other replies say, the 40mA is your nominal coil operate current, and the 10mA is the minimum 'wetting' current that is needed to guarantee the specified closed resistance of the contacts. Different contact materials have different characteristics, generally divided into low voltage and high power. Low voltage ones burn easily, high power ones need a significant current to ensure proper contact.

As you have a limited power budget, here is a trick you can use. You notice the 'must release' specification is 5% of the nominal voltage. It is always the case that once a relay has closed, it needs far less current to hold it closed than it took to close it. Often it will tolerate a drop to 25% of the nominal closing current before it drops out. You would need to experiment to find out what yours will do.

The trick is to drive the relay coil through a parallel RC combination. The capacitor gets the nominal switching voltage to the relay, for long enough to close it. The resistor is sized to continue to conduct just enough current to hold it closed. After operation, the capacitor has to discharge through the resistor, so there is an upper limit to how fast the relay can be switched like this.

Considering I have a limited power source, and as my understanding the relay only needs 10mA and I'm providing 40mA

The relay draws 40 mA at 5V applied directly across it. It is the current required by the relay to operate. Your calculation of 40 mA is correct. Soon you add any resistor in series, relay will not be energised due to insufficient current.

according to datasheet it has a minimum switching current = 10mA

This parameter is more towards reliability specification of the relay. This is the minimum current, the relay expects to flow through the now shorted contacts when the relay is turned on (40 mA). The load can be less than that too but relay might not live enough for number of cycles of operation mentioned in the data sheet.

Minimum operating voltage for your specific relay is 3.75 V. This will reduce the current drawn by the relay too. Let all the available voltage be applied across relay. If you have to reduce the overall current consumption, 3 V relay from the same family is available.

For the relay you have, at 5V, 40 mA has to be sacrificed. No other way..

please see Neil answer below/above for parallel RC placement in series with relay for a power budgeted limit.

Look at the datasheet again. The minimum of 10 mA switching current refers to the contact current. This is the minimum load current. The coil specs indicate the requirements for the coil and is indicated in voltage. Ofcourse from there you can calculate the current at the minimum voltage required. In your case this would be 3V/125 ohm = 24mA.

• Where do your "3V/125 ohm = 22mA" come from? According to the specs, that would not be enough to switch the relay.
– Rev
Jan 7 '17 at 13:31
• My mistake I took the 3V version. Answer corrected Jan 7 '17 at 15:01