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  • In a common base (CB) configuration, a transistor has a collector current <img alt="\left(\mathrm{I}_{\mathrm{C}}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28%5Cmathrm%7B

In a common base (CB) configuration, a transistor has a collector current \left(\mathrm{I}_{\mathrm{C}}\right) of 4 \mathrm{~mA} and a collector-emitter voltage \left(\mathrm{V}_{\mathrm{CE}}\right) of 10 \mathrm{~V}. If the voltage across the emitter resistor \left(R_E\right) is 0.5 \mathrm{~V}, what is the value of R_E ?

Option: 1

R E=0.125 \, \mathrm{k} \Omega


Option: 2

R E=1 \, \mathrm{k} \Omega


Option: 3

R E=2.5 \, \mathrm{k} \Omega


Option: 4

R E=8 \, \mathrm{k} \Omega


Answers (1)

best_answer

Given:

\begin{aligned} & I_C=4 \mathrm{~mA} \\ & V_{C E}=10 \mathrm{~V} \\ & V_{R E}=0.5 \mathrm{~V} \end{aligned}

Using Ohm's Law, the voltage across the emitter resistor \left(R_E\right) can be calculated as:


V_{R E}=I_E \cdot R_E

Since the emitter current \left(I_E\right) can be approximated as the collector current \left(I_C\right) in a common base configuration.

Substituting the given values, we have:

(0.5) V=(4) m A \cdot\left(R_E\right)

To solve for \mathrm{R}_{\mathrm{E} , divide both sides of the equation by  \text { (4) } m A : 

R_E=\frac{0.5 V}{4 m A}=0.125 k(\Omega)

Therefore, the correct option is a) R_E=0.125 \, \mathrm{k} \Omega

Posted by

Shailly goel

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