Transistor As A Device - Switch And Amplifier

Transistor As A Device - Switch And Amplifier

Edited By Vishal kumar | Updated on Sep 25, 2024 02:57 PM IST

Imagine a dimmer switch in your home that can turn a light on and off and also adjust its brightness. A transistor functions similarly in electronic circuits, serving both as a switch and an amplifier. As a switch, the transistor can turn the flow of electrical current on or off in a circuit, akin to flipping a light switch. This capability is essential for digital electronics, where transistors control the binary signals that make computers and other devices work.

This Story also Contains
  1. Transistor as a Switch
  2. Transistor as a device
  3. Solved Example Based On Transistor As A Device - Switch And Amplifier
  4. Summary
Transistor As A Device - Switch And Amplifier
Transistor As A Device - Switch And Amplifier

When used as an amplifier, a transistor takes a small input current and boosts it to a larger output current, similar to how a dimmer switch adjusts the brightness of a light. This amplification is critical in devices like radios and audio equipment, where weak signals need to be strengthened. By performing these dual roles, transistors are indispensable in a wide range of electronic applications, from simple household devices to complex communication systems. In this article, we are going to cover different types of modes and transistors as a switch with the help of some of the solved examples.

BJT Operation Modes

A Bipolar Junction Transistor (BJT) can operate in three distinct modes depending on the biasing of its junctions: the emitter-base junction (EBJ) and the collector-base junction (CBJ). These modes are:

  • 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 the off state and acts as an open switch. The cutoff mode of the transistor is used in the switching operation for the switch-off application.

  • Saturation mode- In the saturation mode, both the junctions of the transistor are forward-biased.As in the forward bias condition, current flows through the device. Hence, electric current flows through the transistor.In this mode, the Maximum collector current flows and the transistor acts as a closed switch.The saturation mode of the transistor is used in the switching operation for the 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.

  • 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 to amplification the 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 let's 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

Vi=VBB=IBRB+VBE… (1) and V0=VCE=VCC−ICRC… (2)

where Vi and V0 are DC input and output voltage respectively.

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

From the above graph, we can conclude that

  1. If Vi is low and unable to forward-bias the transistor, then Vo is high (=VCC).
  2. If Vi is high enough to move the transistor into saturation, then Vo 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 that a low input switches the transistor off and a high input switches it on.

Thus transistors can operate as a switch.

Transistor as a device

Junction Transistor as an Amplifier-

A device that increases the amplitude of the input signal is called an amplifier.

The transistor can be used as an amplifier in the following three configurations

(i) CB amplifier (ii) CE amplifier (iii) CC amplifier

1. NPN transistor as CB amplifier


(i) ie=ib+iC;ib=5% of ic and iC=95% of ic
(ii) VEE<VCC
(iii) Net collector voltage VCB=VCC−icRL

(iv) Input and output signals are in the same phase

2. NPN transistor as CE amplifier

(i) ie=ib+ic;ib=5% of ie and ic=95% of ic
(ii) VCC>VBB
(iii) Net collector voltage VCE=VCC−icRL
(iv) Input and output signals are 180∘ out of phase.

  • Different Gains in CE/CB Amplifiers
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(1) Transistor as CB amplifier-

(i) ac current gain αac= Small change in collector current (Δic) Small change in emitter current (Δie)

VCE= constan

(ii) dc current gain αdc (or α ) = Collector current (ic) Emitter current (ic) valve of αdc lies between 0.95 to 0.99
(iii) Voltage gain Av= Change in output voltage (ΔVo) Change in input voltage (ΔVi)⇒Av=αac× Resistance gain (iv) Power gain = Change in output power (ΔPo) Change in input power (ΔPc)⇒ Power gain =αac2× Resistance gain

(2) Transistor as CE amplifier

(i) ac current gain βac=(ΔicΔib)VCE= constant
(ii) dc current gain βdc=icib
(iii) Voltage gain : Av=ΔVoΔVi=βac× Resistance gain
(iv) Power gain =ΔPoΔPi=βac2× Resistance gain
- Relation between α and β
β=α1−α or α=β1+β

Solved Example Based On Transistor As A Device - Switch And Amplifier

Example 1: In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be :

1) 450

2) 900

3) 1350

4) 1800

Solution:

Amplifier -

The transistor is used as an amplifier

wherein

in such a circuit, output is amplified as compared to input.

Input Voltage must be forward-biased output voltage must be reverse-biased

so V input must be out of phase of V Output

In a common emitter amplifier, the output will be 180o out of phase with the input voltage waveform. Thus the common emitter amplifier is called an inverting amplifier circuit.

Hence, the answer is option (4).

Example 2: The ratio (R) of output resistance \dpi{100} r_{0{}} and the input resistance r_{i} in measurements of input and output characteristics of a transistor is typically in the range :

1) R~102−103

2) R~1−10

3) R ~ 0.1 − 0.01

4) R ~ 0.1 − 1.0

Solution:

Amplifier -

The transistor is used as an amplifier

wherein

in such a circuit, output is amplified as compared to input.

It's typically in the range of 1~10

Hence, the answer is option (2).

Example 3: The transfer characteristic curve of a transistor, having input and output resistance 100Ω and 100 KΩ respectively, is shown in the figure. The Voltage and Power gain are respectively:

1) 5×104,2.5×106
2) 5×104,5×105
3) 5×104,5×106
4)2.5×104,2.5×106

Solution:

Vgain =(ΔICΔIB)Rout Rin =(5×10−3100×10−6)×103=120×106=5×104 Pgain (ΔIcΔIb)⋅ Vgain =(5×10−3100×10−6)⋅5×104⇒ Vgain =5×104, Pgain =2.5×106


Example 4: In a common emitter configuration with suitable bias, it is given that RL is the load resistance and RBE is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by ( β is current gain, IB,IC and IE are respectively base, collector and emitter currents)

1) βRLRBE,ΔIcΔIB,β2RLRBE
2) βRLRBE,ΔIEΔIB,β2RLRBE
3) β2RLRBE,ΔICΔIE,β2RLRBE
4) β2RLRBE,ΔICΔIB,β2RLRBE

Solution:

Relation between α and β -
β=α1−α - wherein α=ICIEβ=ICIB (current gain ) Current Gain β=ΔIcΔIB Voltage Gain =v0v1=βRLRBE power gain = (voltage gain) (Current gain)

β2⋅RLRBE

Example 5: An NPN transistor operates as a common emitter amplifier, with a power gain of 60 dB . The input circuit resistance is 100Ω and the output load resistance is 10kΩ The common emitter current gain βis : :
1) 102
2) 60
3) 6×102
4) 104

Solution:

Relation between α and β -
β=α1−α-wherein α=ICIEβ=ICIB (current gain) 60dB=10log10⁡(PPo)(PPo)=106= power gain β= current gain So,

(β)2=106×Rin Rout =106×102104β=100 or 102

Summary

The transistors, which are the bridges between the strength of the electronic circuit as an amplifier and a switch, play a vital role in this digital era. The switch-like transistor is an element that influences the electric current from one point to the other. Supposing in the case sixteen of time it regulates currents of a higher magnitude with a minimal current. As an amplifier, it changes the amplitude of a weak signal such that it is capable of driving a heavy load and transmitting over long distances with less distortion. They apply it in discrete applications like digital logic circuits, power regulation, or switching power supplies.

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