A 1 mH coil with Q = 100 is used in a tuned amplifier. The transistor has β = 100 and r_{o} = 100 kΩ and input resistance R_{i} = R_{B} || r_{π} = 10 kΩ.
(a) Determine the value of tuning capacitance (C) such that the amplifier has a maximum gain at f_{o} = 50 kHz. Also calculate the value of this gain for the given L = 1 mH.
(b) Determine the bandwidth of the amplifier.
The small-signal model of the amplifier whose output circuit modifies with the circuit equivalent established.
Given Q_{coil} = 100, f_{O} = 50 kHz and L = 1 mH
(a) f_{O}=\frac{1}{2\pi \sqrt{LC} }
Therefore, C =\frac{1}{(2\pi f_{O} )^{2} L}= \frac{1}{(2\pi \times 50\times 10^{3} )^{2}\times 1\times 10^{-3} }= 10 nF
Q_{coil} =\frac{\omega _{0} L}{r}
Here, r =\frac{\omega _{0} L}{Q_{coil}}= \frac{2\pi \times 50\times 10^{3}\times 1\times 10^{-3}}{100}= 3.14 \Omega
req = Q^{2}_{coil} × r = (100)^{2} × 3.14 = 31.4 k \Omega
Amplifier gain, A_{VO} = –\frac{\beta }{R_{i} }× (r_{o} || r_{eq}); R_{L} =\infty , no load in this case.
R = r_{o} || r_{eq} = 100 k \Omega || 31.4 k \Omega = 23.9 k \Omega
A_{VO} =-\frac{100\times 23.9\times 10^{3}}{10\times 10^{3}}= 239
(b) Q_{circuit} = \frac{R}{\omega _{0} L} = \frac{23.9\times 10^{3}}{2\pi \times 50\times 10^{3}\times 1\times 10^{-3}}= 76.11
BW =\frac{f_{O} }{Q_{circuit}}= \frac{50\times 10^{3} }{76.11}= 657 Hz