Question 61: Semiconductors and Logic Gates

Question

For a given transistor amplifier circuit in $$\mathrm{CE}$$ 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

Accepted Answer

In saturation mode $\mathrm{V}_{\mathrm{CE}}=0$
Now
$$
\begin{aligned}
& \mathrm{V}_{\mathrm{CC}}-\mathrm{I}_{\mathrm{c}} \mathrm{R}_{\mathrm{c}}=0 ~~~~	ext{(from KVL)}\\
& \Rightarrow \mathrm{I}_{\mathrm{b}}=\frac{\mathrm{V}_{\mathrm{CC}}}{\mathrm{R}_{\mathrm{C}}}=\frac{1}{1 	imes 10^3}=10^{-3} \mathrm{~A}
\end{aligned}
$$
Given that
$$
\beta=100=\frac{I_c}{I_b}
$$
$\begin{aligned} & \Rightarrow \mathrm{I}_b=\frac{\mathrm{I}_c}{100}=\frac{10^{-3}}{100}=10^{-5} \mathrm{~A} \\ & \Rightarrow \mathrm{I}_b=10 \mu \mathrm{A}\end{aligned}$

Answer

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

Answer

$$\mathrm{I}_{\mathrm{b}}=0.1 ~\mu \mathrm{A}$$

Answer

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

Answer

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

Semiconductors and Logic Gates

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

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