# Hybrid Equivalent of Transistor

Figure(a) shows a transistor amplifier circuit. Such circuit can be connected in any one of the three configuration called common emitter, common base or common collector to a voltage source (V_{s}) and load resistance (r_{i}). The voltages source has an internal resistance (R_{s}) as shown in figure. The load resistance (r_{L})) is the effective or A.C load resistance seen by the transistor at its output. Figure (a) shows the hybrid formulas for input resistance, output resistance, current gain and voltage gain of a transistor.

The expression for transistor amplifier input and output resistances, voltage and current gains may be obtained from the relations:

V_{i} = h_{1}.i_{1} + h_{r}.V_{2} ---------------(1)

i_{2} = h_{f}.i_{1} + h_{o}.V_{2} ----------------(2)

The voltage drop across load resistance (r_{L}) is equal to the voltage across the output terminals of atransistor.

V_{2} = i_{L}.r_{L} = -i_{2}.r_{L} (Since i_{2} = -i_{L})

## Current Gain

It is the ratio of output current (i_{L}) to input currents (i_{1}). Mathematically, the current gain,

A_{i} = i_{1}/i_{1} = i_{2}/i_{1}

Putting the value of V_{2} (equal to –i_{2}.r_{L}) in equation (2)

i_{2} = h_{f}.i_{1} + h_{o}(-i_{2}.r_{L}) = h_{f}.i_{1} –h_{o}.i_{2}.r_{L}

i_{2} + h_{o}.i_{2}.r_{L} = h_{f}.i_{1}

i_{2}(1 + h_{o}.r_{L}) = h_{f}.i_{1}

i_{2}/i_{1} = h_{f}/(1 + h_{o}.r_{L})

A_{i} = i_{2}/i_{1} = -(h_{f}/(1 + h_{o}.r_{L})) -------------------(3)

## Input Resistance

Input resistance value is given by the ratio of input voltage (V_{1}) to the input current (i_{1}).

Mathematically,

R_{i} = V_{1}/i_{1}

Again putting the value of V_{2} (equal to –i_{2}.r_{L}) in equation (1)

V_{1} = h_{i}.i_{1} + h_{r}(-i_{2}.r_{L})

= h_{i}.i_{1} – h_{r}.i_{2}.r_{L}

Dividing the equation (1) by i_{1} on both sides,

V_{1}/i_{1} = (h_{i}.i_{1} – h_{r}.i_{2}.r_{L})/i_{1} =h_{i} – h_{r}(i_{2}/i_{1})r_{L}

Replacing v1/i_{1} with R_{i} and i_{2}/i_{1} by –A_{i} in the above equation.

R_{i} = h_{i} – h_{r}(-A_{i}).r_{L} = h_{i} + h_{r}.A_{i}.r_{L} --------------(4)

Putting the value of A_{i} equal to –h_{f}/(1 + h_{o}.r_{L}) from equation (3) in the above expression.

R_{i} = h_{i} + h_{r} (-(h_{f}/(1 + h_{o}.r_{L})r_{L})

= h_{i} – (h_{r}.h_{f}/h_{o} + (1 / r_{L})) ----------------- (5)

## Voltage Gain

It is defined as the ratio of output voltage (V_{2}) to the input voltage (V_{1}).

Mathematically, the voltage gain,

A_{v} = V_{2}/V_{1} = -(i_{2}.r_{L}/V_{1}) (as V_{2} = -i_{2}.r_{L})

Putting the value of i_{2} (equal to –A_{i}.i_{1}) from equation (2) the above expression,

A_{v} = (A_{i}.i_{1}.r_{L}/V_{1}) = A_{i}.r_{L}(i_{1}/V_{1})

Replacing i_{1}/V_{1} by R_{i} in the above expression,

A = A_{i}.r_{L}/R_{i}

Putting the value of A_{i} (equal to –h_{f}/(1 + h_{o}.r_{L}) and R_{i} (equal to h_{i} – h_{r}/h_{f}(h_{o} + 1/r_{L})) in the above expression and re-arranging,

A_{v} = - ( h_{f}.r_{L}/(h_{i} + (h_{i}.h_{o} – h_{r}.h_{f})r_{1}))

If we replace (h_{i}.h_{o} – h_{r}.h_{f} = ∆h) in the above expression then the voltage gain,

A_{v} = -( h_{r}.r_{L}/ (h_{i} + ∆h.r_{L}))

## Output Resistance

It is obtained by setting the source voltage (V_{s}) to zero and the load resistance (r_{L}) to infinity and by driving the output terminals from a source V_{2} as shown in below Figure (b).

Then output resistance is the ratio of voltage (V_{2}) to the current drawn from the voltage source (i_{2}).

Mathematically, the output resistance,

R_{o} = V_{2}/i_{2} = (V_{2}/(h_{r}.i_{1} + h_{o}.V_{2}) ------------(6)

The value of current (i_{1}) can be obtained by applying Kirchhoff’s Law to the input side of transistor amplifier circuit.

R_{s}.i_{1} + h_{i}.i_{1} + h_{r}.V2 = 0

As i_{1} = -(h_{r}.V_{2})/ (R_{s} + h_{i})

Putting this value of i_{1} in equation (6)

Rearranging the above equation

If source resistance (R_{s}) is zero, then output resistance is,

R_{o} = h_{i}/∆h

## Power Gain

It is defined as the product of voltage gain (A_{v}) and the current gain (∆i).

Mathematically, the power gain is,

∆p = A_{v}.A_{i}

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