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Light Emitting Diode

A CE transist...

Question

A CE transistor amplifies a weak current signal because collector current is

A

β

times

I_{b}

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B

β

times

I_{c}

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C

α

times

I_{b}

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D

α

times

I_{c}

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Solution

The correct option is A $β$ times $I_{b}$
In CE amplifier, current gain is given by:
$\frac{I_{c}}{I_{b}} = β$
Therefore, $I_{c} = β \times I_{b}$

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α = δ I_{C} / δ I_{E}

β = δ I_{C} / δ I_{B}

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β

.

Input charcteristics: It is the graph drawn between input current and input voltage when output voltage is kept constant.

Output characteristics : It is the graph drawn between output current and output voltage when input current is kept constant.

(a) Input resitance

(R_{i}

= {[\frac{Change in emitter base voltage (△ V_{B E})}{Change in base current (△ I_{B})}]}_{at constant V_{C E}}

R_{i} = {(\frac{△ V_{B E}}{△ I_{B}})}_{V_{C E}}

(b) Current amplification factor,

(β) = {(\frac{△ I_{C}}{△ I_{B}})}_{V_{C E}}

\frac{△ I_{C}}{△ I_{E} - △ I_{C}}

\frac{\frac{△ I_{C}}{△ I_{E}}}{\frac{△ I_{C}}{△ I_{E}} - \frac{△ I_{C}}{△ I_{E}}}

\frac{α}{1 - α}

\Rightarrow β = \frac{α}{1 - α}

Q. Let i_E, i_C and i_B represent the emitter current, the collector current and the base current respectively in a transistor. Then
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(b) i_C is slightly greater than i_E
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(d) i_B is much greater than i_E.

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