Simple Transistor Relay Driver Circuit

Transistors are tiny switches that are used in almost all electronics.

But, they cannot handle a lot of power on their own.

A transistor relay driver is like a special adapter for transistors.

It lets a tiny transistor signal control a powerful relay switch.

This way you can use a small weak signal from your circuit to turn on and off things that use a lot of power, like lights or motors.

This is a very important concept for getting started with electronics.

What is a Transistor Relay Driver Circuit:

A transistor relay driver circuit is a circuit that uses a transistor to control the operation of a relay.

Relays are electromechanical switches that can be used to control high power devices or circuits using a low power input.

The transistor in the relay driver circuit amplifies a low power signal to provide the necessary current to energize the relay coil.

Circuit Description:

Parts List:

Below are working of the transistor relay driver circuit:

When a small voltage typically above 0.6V is applied across the base emitter junction of the transistor, it enters into an active state.

This activation is crucial as it enables the transistor to conduct current between its collector and emitter terminals.

The activated transistor serves as a switch for the relay.

With the transistor conducting current flows through the relay coil.

The magnetic field generated by this current causes the relays internal switch to close allowing a larger load to be connected or disconnected.

Upon deactivation of the relay or when the input voltage is removed a back electromotive force EMF is generated across the relay coil.

The freewheeling diode connected in parallel with the relay coil provides a path for this back EMF to circulate preventing it from damaging the transistor.

Essentially, it ensures a safe discharge path for the stored energy in the relay coil.

The base resistor is introduced to control the current flowing into the base of the transistor.

Proper base current is essential to keep the transistor in its active region without overloading it.

This information is critical for sizing components and ensuring the relay operates within its specified parameters.

Formulas and Calculations:

1. Here we will calculate the base resistor value (Rb) for a current limiter circuit using a BJT transistor.

V_in = (V_s – 0.7) * H_Fe / Collector Current

where,

• V_in: Base voltage of the transistor usually in the range of 0.7V to 1.2V
• V_s: Supply voltage (voltage feeding the circuit)
• H_Fe: DC current gain of the transistor (refer to the transistors datasheet)
• Collector Current: Desired maximum current allowed through the collector (the current you want to limit)

Steps to calculate Rb:

1. Rearrange the formula to solve for Rb:

Rb = (Vs – Vin) / (HFe * Collector Current)

Substitute known values:

Replace the variables with known values from your specific circuit:

• V_s: Supply voltage e.g. 12V
• V_in: Base voltage e.g. 0.7V This is typically chosen to ensure the transistor is in the active region.
• H_Fe: DC current gain refer to the transistors datasheet
• Collector Current: Desired maximum collector current e.g. 10mA

Calculate Rb:

R_b = (V_s – V_in) / (H_Fe * Collector Current)
R_b = (12V – 0.7V) / (100 * 0.01A)
R_b = (11.3V) / (1mA)
R_b = 11300Ω = 11.3 kΩ (rounded to two decimal places)

Note:

The resultant value of 11.3 kΩ is in close agreement with the given value of 10 kΩ.

This indicates that, given the expected HFe value of 100, the selected base voltage of 0.7V and collector current of 10mA are appropriate.

The particular device and operating circumstances might affect the transistors actual HFe.

For more precise information on the HFe value, it is advised to refer to the transistors datasheet.

An approximation of Rb can be obtained using this formula.

It could be necessary to adjust the value in light of useful metrics found in the real circuit.

2) Below mentioned is the formula for calculating the current flowing through the relay coil.

Load current or relay coil current = Vs / Coil Resistance

here,

• Load current or relay coil current: This represents the current flowing through the relay coil, which is the current you are trying to calculate.
• Vs: This is the supply voltage applied to the relay coil.
• Coil Resistance: This is the resistance of the relay coil itself, typically found in the datasheet for the specific relay you are using.

Steps to calculate the current:

1) Gather information:

• Find the supply voltage Vs powering the relay coil e.g. 12V
• Locate the coil resistance value from the relays datasheet e.g. 220Ω

2) Apply the formula:

Substitute the known values into the formula:

Current = Vs / Coil Resistance

3) Calculate the current:

In this example:

Current = 12V / 220Ω
Current = 0.054 A (rounded to three decimal places)

Converted to milliamps (mA):
Current = 0.054 A * 1000 mA/A
Current = 54 mA

Therefore, the current flowing through the relay coil in this example is approximately 54 mA.

Note:

For this formula, a coil that is only resistive is used, in actuality relay coils frequently have some inductance, which might have an impact on the switching current flow.

However, this formula gives a decent estimate for simple computations.

It is crucial to confirm that the computed current is within the relay coils operational range.

The coil may get damaged if the current rating is exceeded.

For precise information and suggestions, please consult the relays datasheet at all times.

Circuit Construction:

The circuit diagram presented depicts a transistor operating in the common emitter mode with a relay coil connected between the positive supply and the transistor collector.

Utilize the transistor in the common emitter mode.

Connect the relay coil across the positive supply and the transistor collector.

Apply a small voltage exceeding 0.6V across the base emitter junction of the transistor.

The activated transistor triggers the relay causing it to switch ON.

Integrate a free wheeling diode across the relay coil.

Ensures that the high voltage back electromotive force EMF generated when the relay coil is switched off is shorted through the diode.

Prevents potential damage to the transistor by ensuring the back EMF does not pass through the emitter and collector terminals.

Conclusion:

Mastering the simple transistor relay driver circuit not only lays the foundation for understanding electronics but also equips enthusiasts with a versatile tool for controlling heavier electrical loads.

Through careful construction and application of the provided formulas enthusiasts can confidently engage in the world of electronics.

References

Transistor Relay Driver Demo

Relay driver transistor