Name: Semiconductor Devices vs Vacuum Tubes
Uploaded: 2019-07-19
Duration: 5 min 16 s
Description: Learn in a detailed manner about the uses of semiconductor devices such as P-N junction diode, transistor etc. Also, learn the working principle of vacuum tu...

Question

Deduce the relation between the current amplification factors $α$ and $β$ of a transistor.

Accepted Answer

The relation between the current amplification factors α and β of a transistor.
The current amplification factor or current gain of a transistor is the ratio of output current to the input current. If the transistor is connected on base mode the current gain α= $ \frac{I_{C} }{I_{E}}$ and if the transistor is connected in common emitter mode, the current gain α= $\frac{I_{C} }{I_{B}}$ .
Figure shows a NPN transistor connected in the common base and common emitter configurations. Since, 95% of the injected electrons reach the collector, the collector current is almost equal to the emitter current. Almost all transistors have a, in the range 0.95 to 0.99.
We know that
α= $\frac{I_{C} }{I_{E}}$ =IB+ $\frac{I_{C} }{I_{B} + I_{C}}$ (∵ $I_{E} = I_{g} + I_{C}$ )

$\frac{1}{α} = \frac{I_{B} + I_{C }}{I_{C}} = \frac{ I_{B}}{I_{C}} + 1$
$\frac{1}{α} - 1 = \frac{1}{β} $
$∴ β = \frac{α}{1 - α}$
Usually β lies between 50 and 300. Some transistors have β as high as 1000.

Deduce the relation between the current amplification factors $α$ and $β$ of a transistor.

Important Questions on Semiconductor Electronics

Learn and Prepare for any exam you want !

Deduce the relation between the current amplification factors α and β of a transistor.

Important Questions on Semiconductor Electronics

Learn and Prepare for any exam you want !

Deduce the relation between the current amplification factors $α$ and $β$ of a transistor.