Interfacing Relay to 8051

Relays are devices which allow low power circuits to switch a relatively high Current/Voltage ON/OFF. For a relay to operate a suitable pull-in & holding currentshould be passed through its coil. Generally relay coils are designed to operate from a particular voltage often its 5V or 12V.

The function of relay driver circuit is to provide the necessary current (typically 25 to 70ma) to energize the relay coil.

Figure 1 shows the basic relay driver circuit. As you can see an NPN transistor BC547 is being used to control the relay. The transistor is driven into saturation (turned ON) when a LOGIC 1 is written on the PORT PIN thus turning ON the relay. The relay is turned OFF by writing LOGIC 0 on the port pin. A diode (1N4007/1N4148) is connected across the relay coil; this is done so as to protect the transistor from damage due to the BACK EMF generated in the relay's inductive coil when the transistor is turned OFF. When the transistor is switched OFF the energy stored in the inductor is dissipated through the diode & the internal resistance of the relay coil. Normally 1N4148 can be used as it is fast switching diode with a maximum forward current of 300ma. This diode is also called as free-wheeling diode.

The LED is used to indicate that the RELAY has been turned ON. The resistor R1 defines the current flowing through the LED thereby defining the LED’s intensity.

Resistor R2 is used as a Series Base Resistor to set the base current. When working with 8051 controllers I have noted that it’s not compulsory to use this resistor as the controller has internal 10k resistor which acts as a base resistor.
Microcontrollers have internal pull up resistors hence when a port pin is HIGH the output current flows through this internal pull up resistor. 8051 microcontrollers have an internal pull up of 10KΩ. Hence the maximum output current will be  5v/10k = 0.5ma. This current is not sufficient to drive the transistor into saturation and turn ON the relay. Hence an external pull up resistor R3 is used. Let us now calculate the value of R3. Normally a relay requires a pull in current of 70ma to be turned ON. So our BC547 transistor will require enough base current to make sure it remains saturated and provide the necessary collector current i.e. 70ma. The gain (h_fe) of BC547 is 100 so we need to provide at least   70ma/100 = 0.7ma  of base current. In practice you require roughly double the value of this current so we will calculate for 1.4ma of base current.

Base Current(1.4ma) =o/p current of controller (0.5ma) + 5v/R3

From the above equation the value of R3 comes out to be 5.55KΩ. Typically I use 4.7KΩ resistor.

Whenever 8051 microcontroller is turned ON initially the controller is in reset state and all the controller pins are HIGH which would result in TURNING ON the relay every time power is turned ON or if there is a power fluctuation. This may also damage the device connected to relay so as to avoid this problem another transistor Q2 has been added between the controller & the previous transistor. This transistor acts as an inverter.Figure 2 shows the upated Relay Driver Circuit. So now when a High is applied from the controller the TRANSISTOR Q2 turns ON so the base of transistor Q1 gets 0 so the transistor Q1 turns OFF so the relay turns OFF. And when a LOW is applied from the controller the TRANSISTOR Q2 turns OFF so the base of transistor Q1 gets high voltage through the resistor R2 so the transistor Q1 turns ON which turns ON the relay.

So basically Q2 & R2 acts as an inverter.

If you want to connect more relays to microcontroller then you can use ULN 2003 for connecting seven relays or ULN 2803 for connecting eight relays. Figure 3 shows how to connect a relay to microcontroller using ULN 2003/ULN 2803.  These IC’s are high voltage, high current Darlington transistor arrays with open collector outputs and free-wheeling clamping diodes hence there is no need of a diode across the relay. Also there is no need of the series base resistor as the IC has an internal resistor of 2.7KΩ