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## Q. 5.8

Objective: Design the common-base circuit shown in Figure 5.32 such that $I_{E Q} = 0.50 mA$ and $V_{EC Q} = 4.0 V$.

Assume transistor parameters of β = 120 and $V_{E B}(on) = 0.7 V$.

## Verified Solution

Writing Kirchhoff’s voltage law equation around the base–emitter loop (assuming the transistor is biased in the forward-active mode), we have
$V^{+} = I_{E Q} R_{E} + V_{E B}(on) + \left(\frac{I_{E Q}}{1 + β} \right) R_{B}$
or
$5 = (0.5)R_{E} + 0.7 + \left(\frac{0.5}{121} \right) (10)$
which yields
$R_{E} = 8.52 k \Omega$
We can find
$I_{C Q} = \left(\frac{β}{1 + β} \right) I_{E Q} = \left(\frac{120}{121} \right) (0.5) = 0.496 mA$

Now, writing Kirchhoff’s voltage law equation around the emitter–collector loop, we have
$V^{+} = I_{E Q} R_{E} + V_{EC Q} + I_{C Q} R_{C} + V^{−}$
or
$5 = (0.5)(8.52) + 4 + (0.496)R_{C} + (−5)$
which yields
$R_{C} = 3.51 k\Omega$
Comment: The circuit analysis of the common-base circuit proceeds in the same way as all previous circuits.