Design of single chip microcomputer driven relay circuit

As mentioned before, the MCU has limited source and sink current, that is, limited output drive capability. How to drive high-power devices such as relays? The answer is simple: use transistors. How to determine the device parameters?

I have a HFD23 5V relay on hand, let's take a look at its parameters.

As can be seen:

1. The resistance of the coil is 125Ω;

2. The power of the coil is 200mW;

3. The rated voltage of the relay is 5V;

From this, the pick-up current of the relay can be calculated in two ways:

1. I = 0.2mW/5V = 40mA;

2. I = 5V/125Ω = 40mA;

Let's look at the parameters of the transistor:

The parameters are explained as follows:

1. PCM is the maximum permissible collector dissipation power;

2. ICM is the maximum allowable collector current;

3. BV (CEO) is the collector-emitter reverse breakdown voltage when the transistor base is open;

4. fT is the characteristic frequency;

5. hFE is the magnification factor;

In order to ensure the stability of the circuit, it is required that:

1. The PCM power of the transistor is at least twice the rated power of the relay, PCM ≥ 0.4W;

2. The ICM current of the transistor is at least twice the pull-in current of the relay, ICM ≥ 80mA;

3. The BV withstand voltage of the transistor is at least twice the rated voltage of the relay, BV ≥ 10V;

It can be seen that these four transistors can meet the needs. For stability considerations, we choose NPN S8050. The control circuit diagram is shown below:

Think about it: In actual application, will there be any problems with the above picture?

Since the coil of the relay is an inductive device, the coil will generate self-induced electromotive force when the changing current passes through the coil. According to Faraday's law, the magnitude of the self-induced electromotive force is proportional to the rate of change of the current passing through the coil (the rate of change of the magnetic flux in the coil). Therefore, when the power supply is disconnected, the current change rate is very large, and the coil will generate a self-induced electromotive force several times higher than the power supply voltage, and superimposed with the power supply voltage, this voltage may cause the transistor pole to be broken down, thereby causing the circuit to collapse.

solution

In order to eliminate the harmful effects of this induced electromotive force, a suppression diode is connected in reverse parallel at both ends of the relay coil to absorb the electromotive force. The sum of the self-inductance voltage and the power supply voltage is a forward bias voltage for the diode, which makes the diode conduct and forms a loop current. The induced high voltage will be released through the loop, ensuring the safety of the transistor. This diode is also called a freewheeling diode. The correct circuit diagram is shown below: