Question 6.14: Let us assume that a 50 Ω coaxial transmission line cable ne...
Let us assume that a 50 Ω coaxial transmission line cable needs to drive the input of amplifier with an input resistance of 10 kΩ. In order to maximize power transfer from one stage to the next, the output impedance of the driving stage, the transmission line in this case, must match the input impedance of the receiving stage, the CB amplifying stage. The output impedance of the CB stage must match the input impedance of the circuit downstream. Figure 6.39a shows this downstream circuit simply as R_{in}. Refer to Figure 6.39a to view the circuit set up. Design a common base amplifier configuration that presents an input impedance of 50 Ω to the coax signals. The amplifier output impedance must be 10 kΩ to match the input impedance R_{in} of the stage that need to be driven.
From Equations: (6.86) and (6.88) we know that:
R_{out}=R_C (6.86)
R_{in}=(1/g_m)//R_E=\frac{R_E}{1+g_m R_E} (6.88)
Ignoring the Early effect of the BJT, and
R_{in}=(1/g_m)//R_E=\frac{R_E}{1+g_m R_E} (6.94)
We need the CB stage R_{in} to be 50 Ω, and we need an R_{out} = 10 kΩ. Since R_{out} = R_C, if we obtain a BJT with a g_m of 1/53 Ω or 0.0188679 S and R_E = 1 kΩ, it yields
R_{in}=(1/g_m)//R_E=53//1000≈50 Ω
And since we are ignoring the Early effect, r_o → ∞, thus R_{out} simple equals R_C, hence:
R_{out}=10 kΩ
Finally, the CB amplifier needs to be biased such that a g_m of 0.0188679 S is obtained. Since g_m = I_C/V_T and V_T = 0.026 V at room temperature, then we need a collector biasing current of
I_C=g_mV_T=0.0188679\times 0.026=0.49 mA
The selection of resistors R_1 \text{ and } R_2 are left as an exercise to the reader.