Figure 1(a) shows the circuit of a common emitter (CE) amplifier using self-bias and load resistor R_{0} capacitively coupled to the collector. Figure 1(b) gives the a.c. equivalent circuit. Here we have eliminated the biasing circuit consisting of R_{1}, R_{2}, R_{e} and C_{z}.

The R_{1} – R_{2} combination is equivalent to resistance R_{b} (= R_{1} || R_{2}) between base and ground. It is assumed that R_{b} is large in comparison with the input resistance of the amplifier between base and ground and hence R_{b} is neglected in the equivalent circuit. Similarly, the reactance of capacitor C_{z} is so small at the lowest operating frequency that C_{z} effectively bypasses all a.c. components. Hence R_{e}-C_{z} combination is also excluded from the a.c. equivalent circuit. In most of the amplifiers, the a.c. output voltage developed across R_{c} is capacitively coupled to the next stage through the capacitor C_{b} and R_{0} is the effective impedance in the output circuit. Value of capacitor C_{b} is chosen so large that its reactance at the operating frequency is small and may be neglected. Then for a.c. operation, R_{0} comes in parallel with R_{c} and effective load resistance R_{L} = R_{C} || R_{O}. This results in the simple a.c. equivalent circuit of Figure 1(b).

Analysis: For the purpose of analysis, we replace the transistor by its small signal two generator h-parameter model. This results in the equivalent circuit of Figure 2. We assume sinusoidal input. Hence in the equivalent circuit of Figure 2, we have used rms value of voltages and currents namely I_{b}, V_{b}, I_{c} and V_{c}.

* Current Gain or Current Amplification:* Current gain is defined as the ratio of the load current I

_{1}to the input current I

_{b}. Thus,

Current Gain …..(1)

But from figure 2, …..(2)

Also …..(3)

Combining Equation (2) and (3) we get,

Hence current gain …..(4)

* Input Impedance R_{i}:* This is the impedance between the input terminals B and E looking into the amplifier as shown in Figure 2 and is, therefore, given by,

……(5)

From figure 2 …..(6)

But …..(7)

Substituting the value of V_{c} from Equation (7) into Equation (6) we get,

…..(8)

Hence input impedance …..(9)

…..(10)

Where

From Equation (10) we find that the input impedance R_{i} is also a function of load resistance R_{L}.

* Voltage Gain or Voltage Amplification:* It is the ratio of the output voltage V

_{c}to the input voltage V

_{b}. Thus,

Voltage Gain …..(11)

* Output Admittance Y_{0}:* It is the ratio of the output current I

_{c}to the output voltage V

_{c}with V

_{s}= 0. Hence

with V_{S }= 0 ……(12)

On substituting the value of I_{c} from Equation (2) into Equation (12) we get,

…..(13)

But with V_{s} = 0, Figure 2 gives (R_{s} + h_{ie}) I_{b }+ h_{re} V_{c} = 0

Or ……(14)

Combining Equation (13) and (14) we get, ……(15)

Equation (15) shows that the output admittance Y_{0} is a function of source resistance R_{s}. It source impedance is purely resistive, then the output impedance Y_{0} is real i.e. purely conductive.

Output impedance …..(16)

In the calculation of Y_{0}, R_{L }has been considered external to the amplifier. If we include R_{L} in parallel with R_{0}, we get the output terminal impedance Z_{t} given by,

…..(17)

* Overall Voltage Gain Considering R_{s}:* Source voltage V

_{s}applied at the input of an amplifier results in voltage V

_{b}between bae and emitter terminals (input terminals) of the transistor and voltage V

_{c}at the output. Then the overall voltage gain considering the source resistance is given by

……(18)

Figure 3(a) given the driven voltage source V_{s} with source resistance R_{s} in series. This form of equivalent circuit for the energy source known as Thevenin’s equivalent source. This energy source then drives the amplifier represented by its input resistance R_{i}.

Then ……(19)

Hence overall voltage gain …..(20)

If R_{s} = 0, then A_{VS} = A_{V}. Thus, A_{V} forms a special care of A_{VS} with R_{s} = 0.

* Overall Current Gain Considering R_{s}:* We may replace the voltage source V

_{s}with series source resistance R

_{s}by what is known as the Norton’s equivalent source shown in Figure 3(b), consisting of current source I

_{s}with source resistance R

_{s}in shunt. This current source drives the amplifier resulting in I

_{b}at the input terminals of the amplifier and current I

_{L}through the load impedance. Then the overall current gain A

_{Is}is given by:

…..(21)

From Figure 3(b), …..(22)

Hence overall current gain …..(23)

From Equations (20) and Equation (23) we get, …..(24)

Equation (24) is true provided that the voltage source V_{s} and the current source I_{s} have the same source resistance R_{s}.

* Power Gain A_{p}:* From the circuit of Figure 2, the average power delivered to the load imprudence R

_{L}is given by:

……..(25)

Where is the phase angle between V_{C} and I_{L}. ……(26)

If load impedance is purely resistive, then cos and Equation (26) reduce to:

…..(27)

The input power at input port is, …..(28)

The power gain A_{P} is the ratio of power P_{L} delivered to the load to the input power at the input port. Thus,

…..(29)

Table 1 gives the above derivations in tabular from. The expression for A_{V}, A_{VS} and A_{IS} do not involve h-parameters, while expression for A_{I}, R_{i} and Y_{0} involves the h-parameters. These expressions may be used for CB and CC configurations also, of course, on using corresponding h-parameters.

Table 1: Result of small single analysis of low frequency ce amplifier | |