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Transistor As A Device - Switch And Amplifier - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Transistor as a device is considered one of the most asked concept.
37 Questions around this concept.

Solve by difficulty

In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be :

45⁰

90⁰

135⁰

180⁰

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In a common base amplifier, the phase difference between the input signal voltage and output voltage is :

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\pi$

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Concepts Covered - 0

Transistor as a device

BJT operation modes-

1.Cutoff mode- In the cutoff mode, both the junctions of the transistor are reverse biased.

As in reverse bias condition, no current flows through the device. Hence, no current flows through the transistor. Therefore, the transistor is in off state and acts as an open switch. The cutoff mode of the transistor is used in switching operation for switch OFF application.

2. Saturation mode-In the saturation mode, both the junctions of the transistor are forward biased.

As in forward bias condition, current flows through the device. Hence, electric current flows through the transistor.

And in this mode Maximum collector current flows and the transistor acts as a closed switch.

The saturation mode of the transistor is used in switching operation for switch ON application.

So we can say that by operating the transistor in saturation and cutoff region, we can use the transistor as an ON/OFF switch.

3.Active mode- In the active mode, one junction (emitter to base) is forward biased and another junction (collector to base) is reverse biased. The active mode of operation is used for the amplification of current.

The transistor can be used as a device application like a switch, amplifier, etc depending on parameters like the configuration used (namely CB, CC, and CE), the biasing of the E-B and B-C junction and the operation region namely cutoff, active region, and saturation.

Transistor as a switch-

When a junction transistor is used in the cutoff or saturation state, it acts as a switch.

For the base-biased transistor in CE configuration shown in the below figure lets try to understand the operation of a transistor as a switch.

Applying Kirchhoff’s voltage rule to the input and output sides of this circuit, we get

$\begin{array}{c}{V_i=\mathrm{V}_{\mathrm{BB}}=\mathrm{I}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}+\mathrm{V}_{\mathrm{BE}} \ldots(1) \ \ \mathrm{and}} \\ {V_0=\mathrm{V}_{\mathrm{CE}}=\mathrm{V}_{\mathrm{CC}}-\mathrm{I}_{\mathrm{C}} \mathrm{R}_{\mathrm{C}} \ldots(2)}\end{array}$

where $V_i \ and \ V_0$ are dc input and output voltage respectively.

The plot of V_o vs. V_i (as shown in the below figure) is called the transfer characteristics of a base-biased transistor.

From the above graph, we can conclude that

If V_i is low and unable to forward-bias the transistor, then V_o is high (=V_CC).
If V_i is high enough to move the transistor into saturation, then V_o is very low (~0).

And when a transistor is not conducting, it is switched off. On the other hand, when it is in the saturation state, it is switched on.

If some low and high states are defined below and above certain voltage levels (i.e cutoff and saturation levels of the transistor).

then we can say for a low input switches the transistor off and a high input switches it on.

And thus transistors can operate as a switch.