Tuesday, March 5, 2024

Power-Saving Relay Driver

By T.A. Babu - The author is managing director of Quantum Dots Technology Pvt Ltd, Chennai, and a regular contributor to EFY

In many circuits, the switching action is performed by a relay, which in turn activates an external load. The power consumed by the relay may be unsuitable for battery-powered applications. Here is a simple solution using some inexpensive components to considerably save power.

Circuit and working of Relay Driver

Fig. 1 shows circuit diagram of the power-saving relay driver where resistor R1 and transistor T1 form a standard relay driver circuit.

Circuit diagram of the power-saving relay driver
Fig. 1: Circuit diagram of the power-saving relay driver

Once the relay is energised, its pole is pulled in to make contact with the N/O side, and it holds in that position with typically 75 per cent of its nominal-rated voltage. Power consumed by a relay coil during this holding time equals V²/R, where R is the resistance of the relay coil and V is the voltage. Here resistors R2 and R3, transistor T2 and capacitor C1 lower the power consumption after actuation by applying less than the normal operating power.

94C_Test_Point-2Initially, when power is applied, capacitor C1 momentarily shorts resistor R2 and allows the full voltage across the relay to pull the pole contact, and then slowly the current through the capacitor drops.

AA1_Parts_List-2In the meantime, resistor R2 takes care of the current, ensuring it is just sufficient to hold the relay. The constant current mechanism formed by transistor T2 and resistors R2 and R3 effectively drive the relay at very less power.

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Following calculations will help us understand how additional circuitry around relay driver transistor T1 saves power. The relay used here is a 12V, 400-ohm sugar-cube type. You can calculate the power saved as shown below:

Nominal current required for the relay (I)= 12V/400 ohm = 30mA
Power consumed by the relay=I2R=0.03A×0.03A×400 ohm =360mW

After the introduction of the circuit:

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The current through the coil (I) =VBE/R2=0.6V/47 ohm=12mA
Power consumed by the overall circuit = V×I = 12×0.012=144mW
Power saved=360mW-144mW=216mW

So, we conclude that considerable power can be saved using the additional circuitry.

This makes it fairly simple for anyone to re-design a relay driver to reduce its power consumption without the use of any expensive components.

Construction and testing

An actual-size, single-side PCB for the power-saving relay driver is shown in Fig. 2 and its component layout in Fig. 3.

Switch S1 is used to test the relay driver circuit. You can connect the output of a control circuit, such as a microcontroller, to CON2 for controlling the relay circuit.

CON3 helps in connecting to the electrical load. You can connect the load between N/O and pole contacts or N/C and pole contacts.

Fig. 2: An actual-size, single-side PCB for the power-saving relay driver
Fig. 2: An actual-size, single-side PCB for the power-saving relay driver
Fig. 3: Component layout for the PCB
Fig. 3: Component layout for the PCB

Download PCB and component layout PDFs: click here

Before connecting the load to CON3, verify the test point voltages given in the table. You may reduce the value of R1 as per your requirement.

The project was first published in May 2014 and has recently been updated.



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