Introduction: Reducing Relay Power Consumption - Holding Versus Pickup Current

Most relays require more current to actuate initially than is require to hold the relay on once the contacts have closed. The current required to hold the relay on (Holding current) can be substantially less than the initial current required to actuate it (Pickup current). This implies that there can be a considerable power saving if we can design a simple circuit to reduce the current supplied to a relay once it has been switched on.

In this instructable we experiment (successfully) with a simple circuit to accomplish this task for one model of 5VDC relay. Obviously depending on the relay type some component values might have to be modified, but the method described should work for most DC relays.

Step 1: Characterize the Relay

To start, I measured the current consumed by the relay at a number of different voltages and also figured out at what voltage the relay would drop out at as the voltage was lowered. From this we can also figure out the relay coil impedance at different voltages using R = V/I. It remains fairly constant in approximately the 137 ohm to 123 ohm range. You can see my results for this relay in the picture.

Because the relay drops out at about 0.9 volts or with about 6 to 7 ma of current flowing, we'll aim to have about 1.2 volts across the coil or about 9 to 10 ma of current flowing in the holding state. This will give a little bit of a margin above the drop out point.

Step 2: The Circuit Diagram

A picture of the schematic is attached. The way the circuit works is that when 5V is applied, C1 is momentarily a short circuit and current flows freely through C1 and R3 into the base of Q1. Q1 is turned on and momentarily puts a short circuit across R1. So essentially we have 5V applied to the K1 coil as pin 1 of the relay will be at almost ground potential due to Q1 being momentarily fully turned on.

At this point the relay actuates. Next C1 discharges through R2 and will be about 63% discharged after 0.1 seconds because 100uF x 1000 ohms give a 0.1 second tau or RC time constant. (You could also use a smaller capacitor and bigger resistor value to get the same result e.g 10uF x 10K ohms). At some point around 0.1 seconds after the circuit has been powered on, Q1 will turn off and now current will flow through the relay coil and through R1 to ground.

From our characterization exercise we know that we want the holding current through the coil to be around 9 to 10 ma and the voltage across the coil to be about 1.2V. From this we can determine the value of R1. With 1.2V across the coil its impedance is about 128 ohms as also determined during characterization. So:

Rcoil = 128 ohms
Rtotal = 5V/9.5ma = 526 ohms

Rtotal = R1 + Rcoil
R1 = Rtotal - Rcoil

R1 = 526 - 128 = 398 ohms
We need to use the nearest standard value of 390 ohms.

Step 3: Breadboard Build

The circuit works well with a 0.1 sec time constant for C1 and R2. The relay actuates and disengages immediately as 5V is applied and removed and latches on when 5V is applied. With a value of 390 ohms for R1 the holding current through the relay is about 9.5 ma as opposed to the measured pick up current of 36.6 ma with the full 5V applied to the relay. Power savings is approximately 75% when using the holding current to keep the relay on.