Transistor as a Switch

Electronics & ElectricalDigital ElectronicsMiscellaneous

A transistor can be used as a solid state switch. If the transistor is operated in the saturation region then it acts as closed switch and when it is operated in the cut off region then it behaves as an open switch.

The transistor operates as a Single Pole Single Throw (SPST) solid state switch. When a zero input signal applied to the base of the transistor, it acts as an open switch. If a positive signal applied at the input terminal then it acts like a closed switch.

When the transistor operating as switch, in the cut off region the current through the transistor is zero and voltage across it is maximum, and in the saturation region the transistor current is maximum and voltage across is zero. Therefore, both the on – state and off – state power loss is zero in the transistor switch.

Circuit Diagram of Transistor as a Switch

Cut Off State (Open Switch)

When transistor operates in the cut off region shows the following characteristics −

  • The input is grounded i.e. at zero potential.

  • The VBE is less that cut – in voltage 0.7 V.

  • Both emitter – base junction and collector – base junction are reverse biased.

  • The transistor is fully – off acting as open switch.

  • The collector current IC = 0 A and output voltage Vout = VCC.

Saturation State (Closed Switch)

The transistor operating in the saturation region exhibits following characteristics −

  • The input is connected to VCC.

  • Base – Emitter voltage is greater than cut – in voltage (0.7 V).

  • Both the base – emitter junction and base – collector junction are forward biased.

  • The transistor is fully – ON and operates as closed switch.

  • The collector current is maximum

$$\mathrm{I_{C}=\frac{V_{CC}}{R_{L}}}$$ and Vout = 0 V.

Operating Characteristics of Transistor

  • Cut Off Region − To operate the transistor in this region, both the junctions of BJT are reverse biased and the operating conditions of the transistor are as follows − input base current (IB) is equal zero, hence the zero output collector current (IC). The collector – emitter voltage (VCE) is maximum. This results in a large depletion layer on the junctions of the transistor and no current can flow through the device. Hence, the transistor operates as Open Switch i.e. fully – off.

  • Saturation Region − To operate the transistor in saturation region, both the junctions of the BJT are forward biased, hence the base current can be applied to its maximum value which results in maximum collector current. Due to forward biased junctions the width of depletion layer is as small as possible causing minimum collector – emitter voltage drop. Therefore current flowing through the transistor having maximum value, thus the transistor is operated as Closed Switch i.e. fully – ON.

Numerical Example

Find the minimum value of base current required to turn on the transistor for a load current of 150 mA. For the transistor the β = 190. If the input voltage is raised to 10 V. Also calculate the value of RB.

Solution

The base current of the transistor

$$\mathrm{I_{B}=\frac{I_{C}}{β}=\frac{150mA}{190}=0.789 mA}$$

The base resistance of the transistor

$$\mathrm{R_{B}=\frac{V_{in}-V_{cut\:in}}{I_{B}}=\frac{10-0.7}{0.789×10^−3}=11.78 kΩ}$$

raja
Published on 26-May-2021 15:39:51
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