First slide

Transistors

Question

Common emitter amplifier circuit, built using an n – p – n transistor, is shown in the figure . Its DC current gain is 250, $R_{c} = 1 k Ω$ and $V_{c c} = 10 V$ . What is the minimum base current for $V_{C E}$ to reach saturation.

Moderate

A

$40 μ A$
B

$10 μ A$
C

$100 μ A$
D

$7 μ A$

Solution

For a common emitter n – p – n transistor, DC Current gain is
$β_{D C} = \frac{I_{c}}{I_{B}}$
At saturation state $V_{C E}$ becomes zero
$\begin{array}{l} V_{C C} - I_{C} R_{C} = 0 \\ I_{C} \approx \frac{V_{C C}}{R_{c}} = \frac{10}{1000} = 10^{- 2} A \end{array}$
Hence , saturation base current
$I_{B} = \frac{I_{c}}{β_{D C}} = \frac{10^{- 2}}{250} = 40 μ A$

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