Question 11.16: A silicon BJT small signal amplifier shown in Fig. 11.28 has...

(a) To determine dc bias values:

= +2.595 V

(b) To determine mid-frequency gain:

The mid-frequency circuit model is shown in Fig. 11.29.

Assume                                     V_{s} = 1 V

r_{\pi } = \frac{25\beta }{I_{C}(mA)} = \frac{25\times 50}{1.23}= 1  K \Omega

V_{be}= \frac{V_{s}\times (R_{B} \parallel r_{\pi })}{R_{s}+(R_{B} \parallel r_{\pi })}

R_{B} \parallel r_{\pi }= \frac{9\times 10^{3} \times 1\times 10^{3} }{9\times 10^{3} +1\times 10^{3} }= 0.9  K \Omega

Given                                          R_{s} = 1  k \Omega

Therefore                                   V_{be}= \frac{1\times 0.9\times 10^{3}}{(1+0.9)\times 10^{3}} = 0.474  V

I_{b}=\frac{V_{be}}{r_{\pi }}= \frac{0.474}{1\times 10^{3} }= 0.474  mA

\beta  I_{b} = 50 × 0.474 × 10^{–3} = 23.7  mA

R_{o} = R_{C} || R_{L} = 5 × 10^{3} || 10 × 10^{3} = \frac{10}{3}   K  \Omega

V_{o} = –\beta I_{b} R_{o} = – 23.7 × 10^{3} ×\frac{10}{3}× 10^{3} = –79  V

A_{Vo}= \frac{V_{o}}{V_{s}}= -79

(c) To determine low-frequency cut-off:

\omega _{11}= \frac{1}{C_{C1}\left[R_{s}+(r_{\pi }\parallel R_{B})\right] }= \frac{1}{20\times 10^{-6}(1+0.9)\times 10^{3} } = 26.3  rad/s

\omega _{12}= \frac{1}{C_{C2}(R_{C}+R_{L}) }= \frac{1}{20\times 10^{-6}(5+10) \times 10^{3} }= 3.3  rad/s

\omega _{1p}= \frac{1}{C_{E}\left[(R_{s}\parallel R_{B})+r_{\pi }\right] }= \frac{1}{(150/51)\times 10^{-6}\times (0.9+1)\times 10^{3} } = 179   rad/s

As  \omega _{1p} >  \omega _{11}  >  \omega _{12},                \omega _{L} = \omega _{1p} = 179   rad/s = 28.5   Hz

(d) To determine high-frequency cut-off:

\omega _{21}= \frac{1}{C_{eq} (R_{s}\parallel R_{B}\parallel r_{\pi })}

\omega _{22}= \frac{1}{C_{o} R_{o} }

where                                         C_{eq} = C_{\pi} + C_{\mu} (1 + g_{m} R_{o})

C_{o}= C^{′}_{o}+C_{\mu }(1+\frac{1}{g_{m} R_{o}} )

where                                          C^{′}_{o}= C_{W} + C_{L} = 5  pF

C_{\pi} = 100  pF

g_{m}= \frac{\beta }{r_{\pi }}= \frac{50}{1} = 50 mS

R_{o}= \frac{10}{3} K\Omega

g_{m}R_{o}= 50\times 10^{-3} \times \frac{10}{3}\times 10^{3}= \frac{500}{3}

Substituting the values,

C_{eq}= 100\times 10^{-12}+5\times 10^{-12}(1+\frac{500}{3} )\approx 938.3          pF

C_{o}= 5\times 10^{-12}+5\times 10^{-12} (1+\frac{3}{500} )\approx 10 pF

R^{′}_{s}= R_{B} || R_{s} = 9 × 10^{3} || 1 × 10^{3} = 0.9     k \Omega

R^{′}_{s}|| r_{\pi} = 0.9 × 10^{3} || 1 × 10^{3} = 0.474   k \Omega

Therefore,                                 \omega _{21}= \frac{1}{C_{eq}(R_{s}\parallel R_{B}\parallel r_{\pi }) }

= \frac{1}{938.3\times 10^{-12} \times 0.474\times 10^{3} } = 2.25\times 10^{6}   rad/s  or  4.77  MHz

 \omega _{22}= \frac{1}{C_{o}R_{o} }= \frac{1}{10\times 10^{-12}(\frac{10}{3} )\times 10^{3} }= 30\times 10^{6}  rad/s  or  4.77  MHz

Thus                                            \omega  H = 0.358  MHz,   since f_{21} < f_{22}.