Question 10.2: We wish to select appropriate values for CC1, CC2, and CE fo...
We wish to select appropriate values for CC1, CC2, and CE for the common-emitter amplifier, which has RB = 100 kΩ, RC = 8 kΩ, RL = 5 kΩ, Rsig = 5 kΩ, RE = 5 kΩ, β = 100, gm = 40 mA/V, and rπ = 2.5 kΩ. It is required to have fL = 100 Hz.
We first determine the resistances seen by the three capacitors CC1, CE, and CC2 as follows:
RC1 = (RB || rπ ) + Rsig
= (100 || 2.5) + 5 = 7.44 kΩ
R_{CE} = R_{E} || \left[r_{e} + \frac{R_{B} || R_{sig}}{β + 1}\right]
= 5 || \left(0.025 + \frac{100 || 5}{101}\right) = 0.071 kΩ
RC2 = RC + RL = 8 + 5 = 13 kΩ
Now, selecting CE so that it contributes 80% of the value of ωL gives
\frac{1}{C_{E} × 71} = 0.8 × 2π × 100
CE = 28 μF
Next, if CC1 is to contribute 10% of fL,
\frac{1}{C_{C1} × 7.44 × 10^{3}} = 0.1 × 2π × 100
CC1 = 2.1 μF
Similarly, if CC2 is to contribute 10% of fL, its value should be selected as follows:
\frac{1}{C_{C2} × 13 × 10^{3}} = 0.1 × 2π × 100
CC2 = 1.2 μF
In practice, we would select the nearest standard values for the three capacitors while ensuring that fL ≤ 100 Hz. Finally, the frequency of the zero introduced by CE can be found,
f_{Z} =\frac{1}{2πC_{E}R_{E}} = \frac{1}{2π × 28 × 10 ^{−6} × 5 × 10^{3}} = 1.1 Hz
which is very far from fL and thus has an insignificant effect.