Hybrid Equivalent For CB Transistor

The figure shows the transistor connected in common base configuration and the figure also shows the hybrid equivalent circuit of such a transistor.

In common base transistor configuration, the input signal is applied between the base and emitter terminals of the transistor and output appears between the collector and base terminals. The input voltage (V_be) and the output current (i_c) are given by the following equations:

V_be = h_ib.i_b + h_rb.V_c

i_e = h_fb.i_b + h_ob.V_c

Hybrid expression

Expression can be obtained from the general hybrid formulas derived in this article Hybrid Equivalent of Transistor by adding a second subscript letter ‘b’ (which stands for common base) with the h-parameters and are as discussed below.

Current Gain

It is given by the relation,

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

Where r_L is the A.C load resistance. Its value is equal to the parallel combination of resistance R_c and R_L. Since h_fb of a transistor is a positive number, therefore A_i of a common emitter amplifier is negative.

Input Resistance

The resistance looking into the amplifier input terminals (i.e. base of a transistor) is given by the relation,

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

The input resistance of the amplifier stage (called stage input resistance R_is) depends upon the biasing arrangement. For a fixed bias circuit, the stage input resistance is,

R_is = R_i

Voltage Gain

It is given by the relation,

A_v = A_i.r₁/R_i

Since the current gain (A_i) of a common base amplifier is positive, therefore the voltage gain (A_v) is also positive. It means that there is no phase difference between the input and output signals of the common base amplifier. The voltage gain, in terms of h-parameters, is given by the relation.

A_v = h_fb.r_L/(h_ib + ∆h.r_L)

Where

∆h = h_ib.h_ob – h_rb.h_fb

Output Resistance

The resistance looking into the amplifier output terminals is given by the relation,

R_o = (R_s + h_ib)/(R_s.h_ob + ∆h)

Where

R_s = Resistance of the source, and

∆h = h_ib.h_ob – h_rb.h_fb

The output resistance of the stage,

R_os = R_o // r_L

Overall Voltage Gain

It is given by the relation,

A_vs = (A_v.R_is)/(R_s + R_is) 

Overall Current Gain

It is given by relation,

A_is = A_i.R_s/(R_s + R_is)

Example

A transistor used in a common base amplifier has the following vaules of h-parameters:

h_ib = 28 Ω, h_fb = -0.98, h_rb = 5 x 10^-4 and h_ob = 0.34 X 10^-6 S

Calculate the values of input resistance, output resistance, current gain and voltage gain, if the load resistance is 1.2 KΩ. Assume source resistance as zero.

Solution

Given Data:

h_ib = 28 Ω
h_rb = 5 x 10^-4
h_ob = 0.34 x 10^-6 S
r_L = 1.2 KΩ
R_s = 0

Input resistance of the amplifier stage

We know that the input resistance,

Ri = hib + hrb.Ai.rL
= 28 + [(5 X 10^-4) X 0.98 X 1200] = 28 + 0.5 = 28 Ω Ans

Output Resistance of Amplifier Stage

We know that:

∆h = h_ib.h_ob – h_rb.h_fb

= [23 X (0.34 X 10^-6)] - [5 X 10^-4) X (-0.98)]

Output resistance

Ro = (Rs + hib) / (Rs.hib + ∆h) = hib/∆h -------------(as Rs = 0)

= 28/5 X 10^-4 = 5.5 X 10^-4Ω = 56 KΩ Ans.

Current Gain of Amplifier Stage

We know that the current gain of amplifier stage,

A_i = -(h_fb/(1+ h_ob .r_L) = -(-0.98)/(1 + [(0.34 X 10^-6) X 1200] = 0.98 Ans.

Voltage Gain Amplifier Stage

We know that voltage gain of a transistor,

A_v = A_i.r_L/R_i = (0.98 X 1200) / 28.6 = 41 Ans.