Amplified Output Of A Transistor

The transistor is used to amplify the weak signal a CE amplifier connection is shown in the below figure.

The output from the CE amplifier is either taken across the resistance R_C or at terminals 1 and 2. In both cases, the magnitude of the output signal is the same. Since

Output = Voltage drop across R_C = I_CR_C

This method is used in amplifiers having single stage amplification only.

When the output is taken from terminals 1 and 2 the

Output = voltage across the terminals = V_CE = V_CC – I_CR_C

Since V_CC is the DC Voltage, it cannot pass across the coupling capacitor Cc. Only varying output appears across the output terminals. Therefore

Output = -I_CR_C

Thus output is same in magnitude, but negative sign indicates the phase reversal. This method is used in multistage amplifier circuits.

As an amplifier, CE connected transistor can be represented as a 2 port network.

Input Resistance

When a small change of base-emitter voltage (∆ VBE) is applied. It results in a change in base current as ∆ IB at constant collector-emitter voltage, then their ratio is called input resistance R_i.

Mathematically,

R_i = ∆V_BE/∆I_B = V_BE/I_B

So when 1 V is applied and if the base current is 2 mA, then the input resistance of the transistor amplifier will be

R_i = 1 / 2 X 10^-3 = 500 ohms

Output Resistance

When the change in collector-emitter voltage (∆VCE) occurs to produce a change in collector current (∆I_C) at a constant base current, then their ratio is called output resistance (R_O).

Mathematically,

R_O = ∆V_CE/∆I_C

This value is very high due to the reverse biasing of the collector-base PN junction.

Effective Collector Load

It is load offered to the A.C collector current. In the case of single amplifiers, it is the parallel combination of R_C and R_O.

Mathematically,

R_AC = (R_C X R_O) / (R_C + R_O)

Since, R_O is the output resistance and it is large, so

R_C + R_O = R_O

Therefore

R_AC = R_C

Hence in a single-stage amplifier, the effective load is equal to the collector load R_C.

In the case of a multistage amplifier, the input of the next stage is connected to the output of the first stage. So the effective output resistance will now be:

(R_O X R_i) / (R_O + R_i) ≈ R_i

Since R_O is very much more significant than R_i. Now the effective load resistance will be:

R_AC = (R_C X R_i) / (R_C + R_i)

Thus the effective load resistance is decreased corresponding to the value of R_i.

Current Gain

Is the ratio of change in collector current ∆I_C to the change in the base current?

Mathematically,

Current gain = β = ∆I_C/∆I_B

Its value ranges between 20 to 500.

Voltage Gain

Is the ratio of change in output voltage ∆V_CE to the change in the input voltage ∆V_BE.

Mathematically,

A_V = ∆V_CE/∆V_BE = (∆I_C X R_AC) / (I_B X Ri)

= (∆I_C / ∆I_B) X (R_AC/R_i) = β X (R_AC/R_i)

For single stage Ra = R_C and for multistage:

R_AC = (R_C X R_i)/(R_C + R_i)

Power Gain

It is the ratio output signal power to the input signal power.

Mathematically.

A_P = β A_V = Current gain X Voltage gain