Driving a Relay with a Transistor: Incorrect vs. Correct Methods

I once saw a circuit schematic for a 12V DC relay where the relay was placed on the emitter of its driving transistor. In my opinion, this circuit design was not ideal. The schematic looked roughly like the one in Figure 1.

Why That Circuit Design is Not Ideal

1. High Input Voltage Requirement A circuit with the relay placed on the emitter (whether using a single transistor or a Darlington pair) has a significant drawback. To activate a 12V relay, the voltage across its coil typically needs to be about 75% to 150% of its nominal voltage, which is between 9V and 16V.

To turn a transistor (or a Darlington pair) ON, the base-emitter voltage ($V_{BE}$) must be greater than 0.7V (or about 2 x 0.7V = 1.4V for a Darlington pair). Therefore, for the relay to remain active, the input voltage ($V_{in}$) must be greater than the sum of the relay's operating voltage and the $V_{BE}$ voltage, i.e., $9V+1.4V=10.4V$.

This means the circuit cannot be driven directly by a microcontroller, which typically operates from 1.2V to 5V DC, or by standard logic components like the 74-series ICs. It can only be used with CMOS logic operating at a $V_{CC}$ above 10V.

2. Redundant Component Use While not a serious flaw, using a Darlington pair is a waste of components and PCB space. The operating current of a DC relay coil, whether 5V or 100V, is not very high. On average, the current is around 100mA for a 5V relay, 40mA for a 12V relay, and about 5mA for a 100V relay.

For these current requirements, a small transistor like the 2N3904 (with a maximum collector current of 200mA) is more than sufficient. For a collector current ($I_C$) of 100mA or less, transistors like the 2N3904 generally have a current gain ($h_{FE}$) of over 50. This means a base current ($I_B$) of only about 2mA is needed for an $I_C$ of 100mA, and just about 0.8mA for an $I_C$ of 40mA. A Darlington pair would only be appropriate if the driver ($V_{in}$) is unable to provide such a small amount of current.

The Correct Method: Relay on the Collector

The proper and efficient way to design this circuit is to place the relay on the transistor's collector, as shown in Figure 2.

In this circuit, to turn the transistor fully ON, the input voltage at the base only needs to be slightly above the $V_{BE}$ (around 0.7V). This means with a drive voltage of just 1V, the relay will be activated. A microcontroller or a 74xx family logic IC can easily control this relay.

Crucial Note: The Freewheeling Diode

One important detail that must always be addressed is: every relay in an electronic circuit must have an anti-parallel diode across its coil. The purpose of this diode is to safely dissipate the current generated by the back-EMF when the coil is de-energized (when the relay is turned OFF). Without this anti-parallel diode (also known as a flyback or freewheeling diode), a very high voltage spike can occur across the relay coil, which could damage other components located near the relay.

I hope these simple notes are useful, especially for beginners. Keep building and stay enthusiastic!

TABIK !

(This post is parallel to the status on the FaceBookGroup The Art of Electronics with the same Topic)