# Hybrid Equivalent For CE 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 emitter terminals. The input voltage (V_{be}) and the output current (i_{c}) are given by the following equations:

V_{be} = h_{ie}.i_{b} + h_{re}.V_{c}

i_{e} = h_{fe}.i_{b} + h_{oe}.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 ‘e’ (which stands for common emitter) with the h-parameters and are as discussed below.

## Current Gain

It is given by the relation,

A_{i} = -(h_{fe}/(1 + h_{oe}.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_{fe} 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_{ie} + h_{re}.A_{i}.r_{L} = h_{ie} – ((h_{re}.h_{fe})/(h_{oe} + (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}//R_{B}

If the circuit has no biasing resistances, then 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 emitter amplifier is negative, therefore the voltage gain (A_{v}) is also negative. It means that there is a phase difference of 180^{o} between the input and output. In other words, the input signal is inverted at the output of a common emitter amplifier. The voltage gain, in terms of h-parameters, is given by the relation.

A_{v} = h_{fe}.r_{1}/(h_{ie} + ∆h.r_{L})

Where

∆h = h_{ie}.h_{oe} – h_{re}.h_{fe}

## Output Resistance

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

R_{o} = (R_{s} + h_{ie})/(R_{s}.h_{oe} + ∆h)

Where

R_{s} = Resistance of the source, and

∆h = h_{ie}.h_{oe} – h_{re}.h_{fe}

The output resistance of the stage,

R_{oe} = R_{o} // r_{L}

Overall Voltage Gain

It is given by the relation,

A_{v} = (A_{v}.R_{is})/(R_{s} + R_{is})

Overall Current Gain

It is given by relation,

A_{ie} = A_{i}.R_{s}/(R_{s} + R_{is})

## Example

The h-parameters of a transistor used in a common emitter circuit are h

_{ie}= 1.0 KΩ, h_{re}= 1.0 x 10^{-4}, h_{fe}= 50 and h_{oe}= 100 µmhos. The load resistor for the transistor is 1KΩ in the collector circuit. The transistor is supplied from a signal source of resistance 1000Ω. Determine the value of input and output impedance, voltage and current gains in the amplifier stage.## Solution

## Given Data:

h_{ie}= 1KΩ = 1000Ω

h_{re}= 1.0 x 10^{-4}

h_{oe}= 100 µmhos = 100 x 10^{-6}mhos

R_{c}= 1KΩ = 1000Ω

R_{s}= 1000Ω

### Input resistance of the amplifier stage

We know that thereis no load connected at the output of the amplifier (i.e. R_{L} = 0), therefore the value of A.C load resistance,

r_{L} = Rc = 1000Ω

we also know that current gain of a transistor,

A_{i} = - (h_{fe}/(1 + h_{oe}.r_{L}) = -(50 / (1 + (100 x 10^{-6}) x 1000) = -45.5

And the input resistance of a transistor,

R_{i} = h_{ie} + h_{re}.A_{i}.r_{L} = 1000 + [(1.0 x 10^{-4}) x ( -45.5) x 1000] = 995 Ω

Input resistance of the amplifier stage is,

R_{is} = R_{i} = 995 Ω **Ans.**

### Output Resistance of Amplifier Stage

We know that:

∆h = h_{ie}.h_{oe} – h_{re}.h_{fe}

= [1000 x ( 100 x 10^{-5})] – (1.0 x 10^{-4}) x 50 = 95 x 10^{-3} = 0.095

Output resistance of the transistor looking directly into collector.

R_{e} = (R_{s} + h_{ie})/(R_{s}.h_{oe} + ∆h)

= (1000 + 1000)/[1000 x (100 x 10^{-6})] + 0.95 = 2000/(0.1 + 0.95)

= 10300 Ω

And output resistance of the amplifier stage,

R_{oe} = R_{o} // r_{L} = 10300 // 1000 = 910 Ω Ans.

### Current Gain of Amplifier Stage

We know that the current gain of amplifier stage,

A_{is} = A_{i}.R_{s}/(R_{s} + R_{ie}) = (-45.5) x ( 1000)/(1000 + 995) = -22.8 **Ans.**

### Voltage Gain Amplifier Stage

We know that voltage gain of a transistor,

A_{v} = A_{i}.r_{L}/R_{i} = (-45.5) x (1000)/995 = -45.7

Voltage gain of amplifier stage,

A_{vs} = A_{v}.R_{ie}/(R_{s} + R_{ie}) = (-45.7) x 995/ (1000 +995) = -22.8 **Ans**