What is the CB Hybrid Model?

We use h-parameter model to analyze the Small signal response. Before moving on to the CB (common base) hybrid model we first understand about the hybrid model. We then first derive the general expression of various transistor parameters in terms of hybrid parameters. Later in this article we will derive expressions of the CB (common base) hybrid model.

The hybrid two port model is shown below.

V_i = h₁₁I_i + h₁₂V_o

I_o = h₂₁I_i + h₂₂V_o

h₁₁ = h_i = V_i/I_o for V_o =0  [Input resistance]

Reverse voltage gain

h₁₂ = h_r = V_i/V_o for I_i =0 

Then Forward current gain

h₂₁ = h_f = I_o/I_i for V_o =0  

Hence,

h₂₂ = h_o = I_o/V_o for I_i =0   [output admittance]

The hybrid equivalent model is shown below

The transistor model has three terminals with two ports.

Input resistance = h_i

Output conductance= h_o

The reverse transfer voltage

h_r= V_i/V_o

h_f= I_o/I_i = forward transfer current ratio

The simplified model is shown below

Simplified Hybrid Model parameters

Finding current gain A_i

Applying KCL at the output of above circuit

I_o =I + h_f I_b = V_o h_o + h_f I_b

V_o = -I_o R_L

Then

I_o = -I_o R_L h_o + h_f I_b

I_o (1 + R_L h_o) = h_f I_b

A_i = I_o/ I_b = h_f/(1 + R_L h_o)

Finding Voltage Gain A_V

Applying KVL at input of the above h-model

V_i = h_i I_b + h_r V_o

But I_b = (1 + h_o R_L)/h_f

I_o = – V_o/R_L

Substituting in above equation and solving for V_o/V_i­­ we get

A_v = – h_f RL/h_i + (h_i h_o– h_f h_r) R_L

Finding Input Impedance Z_i

V_i = h_i I_b + h_r V_o

Then

I_o = – V_o/R_L

A_i = I_o/ I_b

Since

V_i = h_i I_b – h_rR_LA_iI_b

Hence the input impedance is

Z_i = V_i/I_b = h_i – h_rR_LA_i

Finding Output impedance Z_o

It is ratio of output voltage to output current with V_s =0. Then the value of output current becomes

I_o = V_oh_o + h_fI_b

I_i = -h_rV_o/R_s + h_i

Z_o = V_o/I_o = 1/[h_o – (h_f h _r /h_i+ R_s)]

CB hybrid Model Parameters

The common base configuration hybrid model is also shown below. It is the hybrid equivalent of a CB transistor. As we also know input in common base transistors is applied between base emitter terminals. Then the output is collected from the collector emitter junction. Then value of input voltage V_be and output current I_C are as 

V_be =h_ib i_b +h_rb V_C 

i_e = h_fbi_b + h_obV_C 

The hybrid expression can also be obtained from the general hybrid formula discussed above by adding a second subscript letter b which commonly stands for base with the h-parameters.

Current Gain

As derived earlier the value of current gain for CB configuration will be

A_i = -(h_fb / (1+ h_ob r_L)

r_L→ AC load resistance. 

But r_L = R_c|| R_L

Since h_fb is a positive number therefore A_i of a CE amplifier is negative.

Input Resistance

Then value of resistance at the input terminal in the figure is 

R_i = h_ib+ h_rb A_i r_L= h_ib – ((h_rbh_fb )/ (h_ob + (1/r_L)))

The input resistance of the amplifier stage also depends upon the biasing arrangement. Then a fixed biased the stage input resistance is,

R_is= R_i 

Voltage Gain

A_v = A_i  r₁ /R_i

We also know that common base amplifiers have positive current gain. Hence, the voltage gain is also positive. Therefore, the output and input are in phase. Then h-parameter is

A_v = h_fb  r_L /(h_ib + Δh r_L)

Δh = h_ib  h_ob – h_rb  h_fb

Output Resistance

R_o = R_s  + h_ib  /(R_s h_ob + Δh)

Δh = h_ib  h_ob – h_rb  h_fb

Overall Voltage Gain

A_vs = A_v  R_is /(R_s + R_is)

Overall Current Gain

A_is = A_vi R_s /(R_s + R_is)