Analysis of Transistor Amplifier Configurations using Simplified h-parameter Model

Analysis of Transistor Amplifier Configurations using Simplified h-parameter Model

AU : Dec.-16

• In most practical cases it is appropriate to obtain approximate values of current gain, voltage gain, input and output impedances rather than to carry out more lengthy exact calculations.

• As mentioned earlier, if h_oe R_L < 0.1 then we can proceed for approximate analysis.

• In above equation, R_L is the effective load resistance.

1. Analysis of CE Amplifier using Simplified Hybrid Model

• Let us consider the h-parameter equivalent circuit for the amplifier, as shown in the Fig. 6.6.1.

• In approximate analysis, hoe and hre are neglected.

• Fig. 6.6.2 shows the approximate h-parameter equivalent circuit. Here, hreVc is replaced by short circuit and h oe is replaced by open circuit.

Ex. 6.6.1 Consider a single stage CE amplifier with R_s = 1 K, R₁= 50 K, R₂ = 2 K, R_C = 2 K, R_L = 2 K, h_fe = 50, h_fe = 1.1 K, h_oe = 25 µA/V and h_re = 2.5 × 10-4, as shown in Fig. 6.6.3.

Find A_p R_i A_V A_I =         I_L / I_s = A_Vs = V_o / V_s and R_o

Sol. : Since h_oe  R'_L = 25 x 10^-6 × (2 K || 2 K) = 0.025, which is less than 0.1, we use approximate analysis.

Fig. 6.6.3 (a) shows the simplified hybrid model for the given circuit.

Current gain :

(A_I ) = - h_fe = - 50

Input impedance :

Ex. 6.6.2 For CE amplifier with voltage divider bias and bypassed RE, h_fe = 50, h_ie = 1100 Ω,h_re = 2.5 × 10-4, h_oe = 24 µA/V, R₁ = 8.2 Ω. R₂ = 1.6 kΩ, R_C = 1.5 kΩ, R_E = 220Ω, R_s = 1 kΩ, R_L = 4.7 kΩ. Assume that all capacitor trends to infinite. Find Av, AVs, AI, AIs, Rt, Ro, Ri , Ro.

Sol. : Since h_oe R_L = h_oe(R_C||R_L)

= 24 × 10^-6(1.5 K||4.7K) = 0.0273, which is less than 0.1, we use approximate analysis.

a) Current gain     (A_I) = -h_fe = - 50

b) Input resistance         (R_i) = h_ie = 1100Ω

2. Analysis of CC Amplifier using Simplified Hybrid Model

• We have seen the simplified CE model, in which input is applied to base and output is taken from collector, and emitter is common between input and output.

• The same simplified model can be modified to get simplified CC model. For simplified CC model, we have to make collector common and take the output from emitter, as shown in the Fig. 6.6.4.

• The h_fe I_b current direction is now exactly opposite that of CE model because the current h_fe I_b always points towards emitter.

Current Gain :

Input Resistance :

Applying KVL to the outer loop of Fig. 6.6.4 we have,

Important Concept

The above equation shows that input impedance of CC is higher than the CE configuration.

Voltage Gain (A_v) :

Substituting values of A_I and R₁ we get,

Applying KVL to the outer loop of Fig. 6.6.4 we have,

The output resistance R_o of the stage, taking the load into account is given as

R’_o = R_o || R_L  ... (6.6.11)

Ex. 6.6.3 A common collector circuit as shown in Fig. 6.6.6 has the following components : R₁ = 27kΩ, R₂ = 27kΩ, R_E = 5.6 kΩ, R_L = 47kΩ, R₅ = 600Ω , The transistor parameters are h_ie = 1 kΩ, h_fe = 85 and H_oe  = 2 µA/V. Calculate A_T, R_p R_o, A_Vs = V_o / V_s and A = I_o / I_o

Sol. :

• Here, h_oe ×  R'_L = 2 × 10^-6 × (5.6 || 47K) = 0.01, which is less than 0.1. Thus we analyse the circuit with approximate method.

• Fig. 6.6.6 (a) shows the simplified hybrid model for the given circuit.

3. Analysis of CB Amplifier using Simplified Hybrid Model

• The approximate CB model can be drawn by giving input to emitter, taking output from collector and making base common. The Fig. 6.6.7 shows the approximate CB model.

Important Concept

The above equation of CB shows that its current gain is always less than one.

Input Resistance (R_i) :

Important Concept

The above equation of CB shows that its input resistance is very low as compare to CE and CC configurations.

Voltage Gain (A_v) :

Output Resistance (R_o) : R_o = V_o / I_c | _{Vs = 0}

When V_s = 0, the current through input loop I_b = 0, hence I_c = 0 and R_o = ∞

The output resistance R_o of the stage, taking the load into account is given as

R_o = R_o || R_L = ∞ || R_L = R_L  …. (6.6.15)

Ex. 6.6.4 A common base amplifier, as shown in Fig. 6.6.8 has the following components : R_s = 600 Ω, R_C = 5.6 K, R_E = 5.6 K, R_L = 39 K. The transistor parameters are h_ie = 1 K, h_ie = 85 and h_oe = 2 µA/V. Calculate R_i , R_o , A_V, A_VS = V_o / V_S

Sol. :

• Since h_oe × (R_C || R_L) = 2 × 10^-6 × (5.6 K || 39 k Ω) = 9.79 × l0^-3, which is less than 0.1, we use approximate analysis method. The Fig. 6.6.8 (a) shows the simplified hybrid model for the given circuit.