In the BJT tuned amplifier circuit shown in Fig. 13.32(a), the bandwidth is 5 kHz and the voltage gain has a maximum value at 250 kHz, when the tuning capacitor is adjusted to 470 pF. Assume the coil Q as 75. Determine the (a) coil inductance, (b) output impedance at resonant frequency,

Question

In the BJT tuned amplifier circuit shown in Fig. 13.32(a), the bandwidth is 5 kHz and the voltage gain has a maximum value at 250 kHz, when the tuning capacitor is adjusted to 470 pF. Assume the coil Q as 75. Determine the (a) coil inductance, (b) output impedance at resonant frequency, (c) internal resistance ro of the BJT, and (d) load resistance if the bandwidth increases by 1 kHz.

Accepted Answer

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

The dynamic junction resistance, [latex] r_{\pi } = \frac{v_{be} }{i_{b} } [/latex], which is one of the small signal parameters of the transistor. The coil has a small resistance r, [latex]Q_{coil}[/latex] = 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 [latex]r_{eq}[/latex] is found below:

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

[latex]\approx \frac{\omega _{O}L }{r}[/latex]; as r << [latex]\omega _{O}L[/latex]

[latex]Q_{circuit} = \omega _{O} r_{eq}C[/latex]

[latex]Q_{coil} = Q_{circuit}[/latex]

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

(a) [latex]f_{O} = 250 kHz[/latex]

BW = 5 kHz

[latex]Q_{circuit} = \frac{f_{O} }{BW} = \frac{250 imes 10^{3} }{5 imes 10^{3} } = 50[/latex]

[latex]f_{O}= \frac{1}{2\pi \sqrt{LC} } [/latex]

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

[latex]= \frac{1}{(2\pi imes 250 imes 10^{3} )^{2} imes 470 imes 10^{-12} }= 4.06 mH[/latex]

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

[latex]R = r_{o} || r_{eq}[/latex]

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

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

(c) [latex]Q_{coil} = \frac{\omega _{O} L}{r}[/latex]

[latex]75= \frac{2\pi imes 250 imes 10^{3} imes 4.06 imes 10^{-3}}{r}[/latex]

Therefore, r = 85 Ω

Substituting the values

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

Here, [latex]R = r_{o} || r_{eq}[/latex]

i.e. [latex]318.71=\frac{r_{o} imes 478.12 }{r_{o} +478.12}[/latex]

Upon solving, we get

[latex]r_{o} = 955.91 k \Omega [/latex]

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

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

[latex]Q_{circuit} =\frac{250}{6} = 41.66[/latex]

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

= 265.55

[latex]265.55= \frac{318.77 imes R_{L} }{318.77+ R_{L}} [/latex]

i.e. [latex]R_{L} = 1592 k \Omega [/latex]

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

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