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.
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.
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:
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
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.
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.