Question 13.2: In the BJT tuned amplifier circuit shown in Fig. 13.32(a), t...

The small-signal model of the amplifier is shown in Fig. 13.32(b).

The dynamic junction resistance, r_{\pi } = \frac{v_{be} }{i_{b} } , which is one of the small signal parameters of the transistor. The coil has a small resistance r, Q_{coil} = 75 and an LC circuit has Q = ∞. The coil and C in parallel can be replaced by its equivalent circuit as in Fig. 13.32(c). Hence, the value r_{eq} is found below:

\frac{\bar{I} _{cap} }{\bar{I} }\left|_{atf= 0}= \frac{r+j\omega L}{(r+j\omega L)+1/j\omega C} \right|_{atf= f_{0} }

\approx \frac{\omega _{O}L }{r};        as r << \omega _{O}L

Q_{circuit} = \omega _{O} r_{eq}C

Q_{coil} = Q_{circuit}

Q^{2}_{coil} = Q_{coil} × Q_{circuit} =\frac{\omega _{O}L }{r}× \omega _{O}r_{eq}C =\frac{r_{eq}}{r}

(a)                                    f_{O} = 250  kHz

BW = 5 kHz

Q_{circuit} = \frac{f_{O} }{BW} = \frac{250\times 10^{3} }{5\times 10^{3} } = 50

f_{O}= \frac{1}{2\pi \sqrt{LC} }

Hence,                             L= \frac{1}{(2\pi f_{O})^{2} C}

= \frac{1}{(2\pi \times 250\times 10^{3} )^{2}\times 470\times 10^{-12} }= 4.06   mH

(b) Using the circuit equivalent established as above, the circuit is redrawn and shown in Fig. 13.32(d).

R = r_{o} || r_{eq}

Q_{circuit} = \frac{R}{\omega ^{'}_{O} L} = 50;           Here \omega _{O} = 2p × 250 × 10^{3}

R = 50 × 2\pi × 250 × 10^{3} × 4.06 × 10^{–3} = 318.71  k \Omega

(c)                              Q_{coil} = \frac{\omega _{O} L}{r}

75= \frac{2\pi \times 250\times 10^{3}\times 4.06\times 10^{-3}}{r}

Therefore,                      r = 85 Ω

Substituting the values

r_{eq} = Q^{2}_{coil} × r = (75)^{2} × 85 = 478.12  k \Omega

Here,                               R = r_{o} || r_{eq}

i.e.                                    318.71=\frac{r_{o}\times 478.12 }{r_{o} +478.12}

Upon solving, we get

r_{o} = 955.91  k \Omega

(d) R_{L} is now connected across V_{o} and so the bandwidth increases by 1 kHz.

BW (new) = (5 + 1) kHz = 6 kHz

Q_{circuit} =\frac{250}{6} = 41.66

R_{eq} = R || R_{L} = Q_{circuit }× \omega _{o} L = 41.66 × 2\pi × 250 × 10^{3} × 4.06 × 10^{–3}

= 265.55

265.55= \frac{318.77\times R_{L} }{318.77+ R_{L}}

i.e.                                     R_{L} = 1592  k \Omega