Question 5.8: Objective: Design the common-base circuit shown in Figure 5....
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.
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.