Transistor as Amplifier

One of the key characteristics of a transistor is that it can be used as an amplifier. Transistors can act as amplifiers while they are functioning in the active region or when it is correctly biased. The need for a transistor as an amplifier arises when we want to increase or amplify the input signal. A transistor can take in a very small weak signal through the base junction and release the amplified signal through the collector.

Transistor amplifiers are used frequently in RF (radio frequency), OFC (optic fibre communication), audio amplification, etc. In this lesson, we will discuss how a transistor works as an amplifier.

Common-Emitter Configuration

For a transistor to work as an amplifier, we usually use the common-emitter configuration. The figure below shows how the transistor is set up when it is connected to a circuit as an amplifier.

In the figure given above, the input is connected in forward-biased, and the output is connected in reverse-biased. The input signal is applied on the base-emitter junction, and the output is taken through the load in the emitter-collector junction. There is also an application of DC voltage in the input circuit for amplification. Besides, a small change in signal voltage results in the change of emitter current, which is mainly due to the low resistance in the input circuit.

Also Read: Forward Bias

The output is taken across the load connected to the output side. The load can be in any combination of R, L or C. The load resistance is of high value, which causes a large voltage drop. Overall, the weak signal is thus amplified in the collector circuit.

However, to work as an amplifier, the transistor has to work in the active region of the output voltage versus the input voltage curve, as seen in the figure below.

In the above graph, we have AB as the cut-off region, BC as the active region, and from C, we have a line parallel to X-axis, which is the saturation region.

Gain in Power, Current and Voltage

From the figure that is given above, it is clear that on the output side, V₀ = V_CC– I_CR_C, where V₀ is the output voltage, I_c is the collector current, R_c is the load resistance, and V_cc is the fixed voltage on the output side.

If we consider ∆V₀ and ∆V_i as small changes in output and input voltages, respectively, then ∆V₀ / ∆V_i is called the small-signal voltage gain, A_v of the amplifier.

Therefore, ∆V₀= 0 – R_c ∆I_C

The gain in terms of voltage when the changes in input and output currents are observed is called voltage gain.

Similarly, in the input side,

V_in = I_B R_B + V_BE

Or

∆V_in = ∆I_BR_B + ∆V_BE ~ ∆V_in = ∆I_BR_B (∆V_BE <<∆I_BR_B)

Or,

A_v = ∆V₀ / ∆V_in = – R_c ∆I_C / R_B ∆I_B = -β_ac R_c / R_B

where β_ac = ∆I_C / ∆I_B is the AC current gain.

When there is a gain in terms of current due to the changes in input and output currents, it is called current gain. β value can range between 20 and 500.

The power gain of an amplifier is defined as the product of the current gain and voltage gain of the amplifier. It is also defined as the rate of change in output power to the change in input power.

Mathematically, A_p = β_ac x A_v.

Use of Capacitors

We have a coupling capacitor that is used to couple the AC component and decouples the DC component contained in the input signal of the amplifier. At the output of the amplifier, the DC component will be invariantly present due to the amplification process. This is being removed by the coupling capacitor at the output, and hence we will have a pure AC signal being supplied to the load connected at the output.

Also Read: Capacitor Types and Capacitance

Similarly, we do have an Emitter-bypass Capacitor C_E. When an AC signal is applied to an amplifier circuit, the variable current will flow through resistors connected at the collector and emitter, i.e., R_C and R_E. This current in R_E will develop a variable voltage drop across R_E and provides additional negative feedback to the emitter junction.

This will result in an overall reduction of voltage gain associated with the amplifier. The capacitor C_E connected across R_E provides a short circuit path for the AC signal and reduces the effect of additional negative feedback due to the AC signal and a corresponding reduction in voltage gain.

Role of Resistance

The resistors R₁ and R₂ form the voltage division circuit to supply the DC voltage to the base of the transistor. The resistors R_C and R_E control the collector and emitter currents, respectively. Proper selection of these resistors helps us to control the value of the limiting currents. These resistors provide the required junction voltages between E-B, C-B, C-E and currents I_E, I_B and I_C to work the transistor in the active region of the output characteristics.

The emitter resistor R_E produces the following changes in the performance of the CE amplifier:

• It causes bias stabilization.
• It causes the current gain to remain essentially unaltered.
• Increases the input and output impedances.
• It stabilizes the voltage gain.

Also Read: Transformer

The Need for CE Configuration

We usually employ CE configuration for transistors as amplifiers because it provides large values of current gain, voltage gain and power gain. Moreover, there is a phase shift of 180 degrees between input and output. It implies the output signal will be an inverted amplified version of the signal given in the input.

As we come to the end of the lesson, we should remember that a transistor amplifier, in order to function properly, must have the following things:

• High input impedance
• High gain
• High slew rate
• High bandwidth
• High efficiency
• High stability
• High linearity

Solved Problems

1. Consider a CE- transistor made to work as an amplifier. The audio signal voltage across the collector resistance of 2 KΩ is 2 volts. Suppose the current amplification factor of the transistor is 100, and the base resistance is 1 KΩ. Determine the input signal voltage and base current.

Answer:

Given R_C = 2 KΩ = 2000 Ω; V_C= 2V; β_ac = 100; R_B = 1KΩ = 1000 Ω

Collector current, I_c= V_c/R_c = 2/2000 = 1 mA

I_B = V_B/R_B = V_B/1000 = V_B mA

β_ac = 100 = I_C/I_B = 1/V_B

V_B  = 1/100 = 0.01 V

Therefore, I_B= V_B m A = 0.01 / 1000 = 10 * 10^-6 A = 10 µA.

2. 2 amplifiers are connected in a series (cascaded). The voltage gain of the first amplifier is 10, and the second is 20. The input signal is given as 0.01 V. Calculate the output of the AC signal.

Answer:

Total voltage gain is A_V = A_V1 * A_V2 = ∆V_o / ∆V_i

∆V_o = ∆V_i * A_V1 * A_V2 = 0.01 * 10 * 20 = 2V

Semiconductors