Question 19.9: Determine the low-frequency effects of the coupling and deco......
Determine the low-frequency effects of the coupling and decoupling capacitors in the circuit of Example 19.7.
The coupling capacitor C
As in Example 19.6, the coupling capacitor introduces a low-frequency cut-off given by
f_{c} = \frac{1}{2\pi CR}where R is the input resistance of the amplifier. In this case, the input resistance is dependent on the current gain of the transistor. If, as in Example 19.8, we know that the input resistance is in the range 2.0–5.2 kV, then we can determine the range of the low-frequency cut-off.
If R = 2.0 kΩ
f_{c} = \frac{1}{2\pi CR} = \frac{1}{2 × \pi × 2.2 × 10^{-6} × 2.0 × 10^{3}} = 36 HzIf R = 5.2 kΩ
f_{c} = \frac{1}{2\pi CR} = \frac{1}{2 × \pi × 2.2 × 10^{-6} × 5.2 × 10^{3}} = 14 HzWe would therefore expect the low-frequency cut-off produced by the coupling capacitor to be in the range 14–36 Hz.
The decoupling capacitor C_{E}
From Equation 19.23, we have
f_{c} \approx \frac{1}{2\pi C_{E}r_{e}} = \frac{1}{2 × \pi × 10 × 10^{-6} × 24} = 663 HzNote that although C_{E} is larger than C it produces a much higher low-frequency cut-off because of the low value of r_{e} . Circuits that are required to work at low frequencies require very large, and expensive, decoupling capacitors.