For a given transistor amplifier circuit in CE configuration $\mathrm{V}_{\mathrm{CC}}=1 \mathrm{~V}, \mathrm{R}_{\mathrm{c}}=1 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{b}}=100 \mathrm{k} \Omega$ and $\beta=100$. Value of base current $\mathrm{I}_{\mathrm{b}}$ is

(D) $\mathrm{I}_{\mathrm{b}}=10 \mu \mathrm{~A}$

(A) $\mathrm{I}_{\mathrm{b}}=1.0 \mu \mathrm{~A}$

(B) $\mathrm{I}_{\mathrm{b}}=0.10 \mu \mathrm{~A}$

(C) $\mathrm{I}_{\mathrm{b}}=100 \mu \mathrm{~A}$

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Current Electricity

For a given transistor amplifier circuit in CE configuration

V_{CC} = 1 V, R_{c} = 1 k Ω, R_{b} = 100 k Ω

and

β = 100

. Value of base current

I_{b}

Answer: (D)

Assuming the transistor is in saturation mode with

V_{C E} = 0

. Using KVL in the collector circuit:

I_{c} R_{c} = V_{C C}

, so

I_{c} = \frac{V _{C C}}{R _{c}} = \frac{1}{1 \times 1 0 ^{3}} = 1 0^{- 3}

A. Using the current gain relation:

β = \frac{I _{c}}{I _{b}}

, we get

I_{b} = \frac{I _{c}}{β} = \frac{1 0 ^{- 3}}{100} = 1 0^{- 5}

= 10 μ

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option