# Hybrid Equivalent For CB Transistor

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

In common emitter 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}.r_{1}/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= 28_{ib}Ω,h= -0.98,_{fb}hand_{rb}= 5 x 10^{-4}h= 0.34 X_{ob}10S^{-6}Calculate the values of input resistance, output resistance, current gain and voltage gain, if the load resistance is 1.2 K

Ω. 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^{-4}**Ω** = 56 K**Ω** 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.

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