Question
Define $\alpha $ and $\beta$. Derive the relation between then.
✓
Answer
The dc common-base current ratio or current gain $(\alpha _{dc})$ is defined as the ratio of the collector current to emitter current.
$\alpha_{ dc }=\frac{I_C}{I_{ E }}$
The dc common-emitter current ratio or current gain $\left(\beta_{ dc }\right)$ is defined as the ratio of the collector current to base current.
$\beta_{ dc }=\frac{I_C}{I_B}$
Since the emitter current $I _E= I _B+ I _C$
$ \frac{I_{ E }}{I_C}=\frac{I_{ B }}{I_{ C }}+1$
$\therefore \frac{1}{\alpha_{ dc }}=\frac{1}{\beta_{ dc }}+1 $
Therefore, the common-base current gain in terms of the common-emitter current gain is $\alpha_{ dc }=\frac{\beta_{ dc }}{1+\beta_{ dc }}$
and the common-emitter current gain in terms of the common-base current gain is
$\beta_{ dc }=\frac{\alpha_{ dc }}{1-\alpha_{ dc }}$
For a transistor, $\alpha_{ dc }$ is close to but always less than $1$ (about $0.92$ to $0.98$ ) and $\beta_{ dc }$ ranges from $20$ to $200$ for most general purpose transistors.
$\alpha_{ dc }=\frac{I_C}{I_{ E }}$
The dc common-emitter current ratio or current gain $\left(\beta_{ dc }\right)$ is defined as the ratio of the collector current to base current.
$\beta_{ dc }=\frac{I_C}{I_B}$
Since the emitter current $I _E= I _B+ I _C$
$ \frac{I_{ E }}{I_C}=\frac{I_{ B }}{I_{ C }}+1$
$\therefore \frac{1}{\alpha_{ dc }}=\frac{1}{\beta_{ dc }}+1 $
Therefore, the common-base current gain in terms of the common-emitter current gain is $\alpha_{ dc }=\frac{\beta_{ dc }}{1+\beta_{ dc }}$
and the common-emitter current gain in terms of the common-base current gain is
$\beta_{ dc }=\frac{\alpha_{ dc }}{1-\alpha_{ dc }}$
For a transistor, $\alpha_{ dc }$ is close to but always less than $1$ (about $0.92$ to $0.98$ ) and $\beta_{ dc }$ ranges from $20$ to $200$ for most general purpose transistors.
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