Question 4.6: A double-tuned amplifier has a load as shown in Fig. 4.24(a)...
A double-tuned amplifier has a load as shown in Fig. 4.24(a).
The center frequency of the response is 2 GHz. Effective quality factors of both sides were adjusted to 20 with an appropriate Q-enhancement circuit, which corresponds to a parallel effective resistance of R_{p1} = R_{p2}= 2512 ohm.
(a). Calculate the magnetic coupling coefficient for a maximally flat response and corresponding bandwidth.
(b). Calculate the magnetic coupling coefficient for a 0.5 dB ripple Chebyshev-type
response and corresponding bandwidth.
(c). Plot the frequency responses with PSpice simulation.
(a). The value of the coupling coefficient can be calculated from (4.57):
\left|\sigma\right|=\frac{\omega_0}{2Q}=\omega_0 \frac{k}{2}\Rightarrow k=\frac{1}{Q} (4.57)
k=\frac{1}{Q}=\frac{1}{20}=0.05According to (4.58) the bandwidth is
2\Delta \omega =2\times \left(\sqrt{2}\left|\sigma\right| \right)=\sqrt{2}\frac{\omega_0}{Q}\Rightarrow B=2\Delta f=\sqrt{2}\frac{f_0}{Q} (4.58)
B=\sqrt{2}\frac{f_0}{Q}=\sqrt{2}\frac{2\times 10^9}{20}=141 MHz
(b). From (4.59)
\left(\sigma\right)_C=\frac{\omega_0}{2Q}=0.709\times \omega_0\frac{k}{2}\Rightarrow k=\frac{1}{0.709}\frac{1}{Q}=\frac{\sqrt{2}}{Q} (4.59)
k=\frac{\sqrt{2}}{Q}=\frac{\sqrt{2}}{20}=0.0707From (4.60)
2\Delta \omega =2\times \sqrt{2}\times \omega_0\frac{k}{2}=\sqrt{2}\times \omega_0\times \frac{1}{0.709}\frac{1}{Q}\Rightarrow B=2\Delta f=2\times \frac{f_0}{Q} (4.60)
B=2\times \frac{f_0}{Q}=2\times \frac{2\times 10^9}{20}=200 MHz
(c). Results are given in Fig. 4.24(b) that fit the calculated values.