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Define Stability factor. Derive the equation for Stability factor. State which biasing technique is more stable? Justify your answer.

Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1

Marks: 10M

Years: May 15

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Stability Factor (S):

The extent to which the collector current $I_C$ is stabilized with varying $I_CO$ is measured by a stability factor S. It is defined as the rate of change of collector current $I_C$ with respect to the collector base leakage current $I_CO$, keeping both the current $I_B$ and the current gain β constant.

S=$\frac{\partial I_C}{\partial I_CO}=\frac{dI_C}{dI_CO}=\frac{\Delta I_C}{\Delta I_CO}$

The collector current for a CE amplifier is given by

$I_C = βI_B + (1+β) I_CO$

Differentiating the above equation with respect to $I_C$, we get

I =$ \frac{β(dI_B)}{(dI_C )}+ \frac{(1+β)(dI_CO)}{(dI_C )}$

Therefore, $\frac{1 -β(dI_B)}{(dI_C )} =\frac{1+β}{S}$

S = $\frac{((1+β))}{((1 - \frac{β (dI_B)}{(dI_C )}}$

From this equation it is clear that this factor S should be as small as possible to have better thermal stability.

Stability factor S’ and S”:

The Stability factor S’ is defined as the rate of change of $I_C$ with $V_BE$, keeping $I_CO$ and β constant.

S’=$\frac{\partial I_C}{\partial V_BE}=\frac{\Delta I_C}{\Delta V_BE}$

The stability factor S” is defined as the rate of change of $I_C$ with respect to β, keeping $ I_CO$ and $V_BE$ constant.

S’=$\frac{\partial I_C}{\partial\beta}=\frac{\Delta I_C}{\Delta\beta}$

The small value of stability factor indicates good bias stability whereas large value of stability factor indicates poor bias stability. Ideal value of stability factor is zero.

For fixed bias stability factor reduces to

S = 1 + β

Since β is large quantity, this is a very poor bias stable circuit.

For collector to base bias stability factor is

S =$\frac{ ((1+β))}{(1+ β( \frac{Rc}{(Rc+RB)}) }$

As can be seen, this value of the stability factor is smaller than the value obtained by fixed bias circuit.

For self-bias,

S =$\frac{(1+β) (1+\frac{RB}{RE})}{(1+ β +\frac{RB}{RE})}$

As can be seen, the value of S is equal to one if the ratio RB/RE is very small as compared to 1. As this ratio becomes comparable to unity, and beyond towards infinity, the value of the stability factor goes on increasing till S = 1 + β. This improvement is the stability up to a factor equal to 1 is achieved at the cost of power dissipation.

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