Ideal Bode plot for 16 28 Given C 50 pF A v 200000 Solution C in C A

# Ideal bode plot for 16 28 given c 50 pf a v 200000

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Ideal Bode plot for Prob. 16-27. 16-28. Given: C = 50 pF A v = 200,000 Solution: C in = C ( A v + 1) (Eq. 16-26) C in = 50 pF(200,000 + 1) C in = 10 μ F Answer: The Miller input capacitance is 10 μ F.
1-81 16-29. Given: C = 100 pF A v = 150,000 R L = 10 k R in = 1 k Solution: C in = C ( A v + 1) (Eq. 16-26) C in = 100 pF(150,000 + 1) C in = 15 μF f 2 = 1/(2 π RC ) f 2 = 1/[2 π (1 k )(15 μF)] f 2 = 11 Hz A v(dB) = 20 log A v (Eq. 16-9) A v(dB) = 20 log(150,000) A v(dB) = 104 dB Answer: See figure below. Ideal Bode plot for Prob. 16-29. 16-30. Given: T R = 10 μs Solution: f 2 = 0.35/ T R (Eq. 16-29) f 2 = 0.35/10 μs f 2 = 35 kHz Answer: The upper cutoff frequency is 35 kHz. 16-31. Given: T R = 0.25 μs Solution: f 2 = 0.35/ T R (Eq. 16-29) f 2 = 0.35/0.25 μs f 2 = 1.4 MHz Answer: The bandwidth is 1.4 MHz. 16-32. Given: f 2 = 100 kHz Solution: f 2 = 0.35/ T R (Eq. 16-29) T R = 0.35/ f 2 T R = 0.35/100 kHz T R = 3.5 μs Answer: The risetime is 3.5 μs. 16-33. Given: C = 1 μF R = 50 Solution: R in = Z in(Stage) in 1 2 || || β e R R R r = R in = 717 f C 1 = 1/(2 π R in C ) f C 1 = 1/[2 π (717 )(1 μF)] f C 1 = 222 Hz Answer: The lower cutoff frequency for the base coupling circuit is 222 Hz. 16-34. Given: C = 4.7 μF R C = 3.6 k R L = 10 k Solution: R = R C + R L R = 36 k + 10 k R = 13.6 k f C 1 = 1/(2 π RC ) f C 1 = 1/[2 π (13.6 k )(4.7 μF)] f C 1 = 2.49 Hz Answer: The lower cutoff frequency for the collector coupling circuit is 2.49 Hz. 16-35. Given: C = 25 μF R C = 3.6 k R L = 10 k Solution: f C 1 = 1/(2 π Z out C ) f C 1 = 1/[2 π (22.4 )(25 μF)] f C 1 = 284 Hz Answer: The lower cutoff frequency for the emitter bypass circuit is 284 Hz. 16-36. Given: 2 pF c C ′ = 10 pF e C ′ = 5 pF Stray C = R 1 = 10 k R 2 = 2.2 k R C = 3.6 k R L = 10 k R G = 50 k β = 200 Solution: r g = R G || R 1 || R 2 r g = 50 ||10 k ||2.2 k r g = 48 C in(m) = 236 pF in(m) 246 pF e C C C = + = Base f 2 = 1/(2 π r g C ) f 2 = 1/[2 π (48 )(246 pF)] f 2 = 13.5 MHz Collector Stray 7 pF e C C C = + = C out(m) = 2 pF R = R C + R L R = 3.6 k + 10 k R = 13.6 k f 2 = 1/(2 π RC ) f 2 = 1/[2 π (13.6 )(1.3 pF)] f 2 = 8.59 MHz Answer: The high cutoff frequency for the base is 13.5 MHz and the collector is 8.59 MHz.
1-82 16-37. Given: g m = 16.5 mS C iSS = 30 pF C oSS = 20 pF C rSS = 5 pF Solution: C gd = C rSS = 5 pF C gS = C iSS C rSS C gS = 30 pF – 5 pF C gS = 25 pF C dS = C oSS C rSS C dS = 20 pF – 5 pF C dS = 15 pF Answer: C gS = 25 pF, C gd = C rSS = 5 pF, C dS = 15 pF. 16-38. Given: R 1 = 2 M R 2 = 1 M R D = 1 k R L = 10 k R G = 50 C in = 0.01 μF C out = 1 μF Solution: R in = Z in(Stage) R in = R 1 || R 2 R in = 667 k f C 1 = 1/(2 π R in C ) f C 1 = 1/[2 π (667 k )(0.01 μF)] f C 1 = 23.9 Hz or 14.5 Hz Answer: The dominant low cutoff frequency is 23.9 Hz. 16-39. Given: R 1 = 2 M R 2 = 1 M R D = 1 k R L = 10 k R G = 50 k C gd = 5 pF (from Prob. 16-37.) C gS = 25 pF (from Prob. 16-37.) C dS = 015 pF (from Prob. 16-37.) Solution: A v = g m r d A v = (16.5 mS)(1 k ||10 k ) A v = 15 C in(m) = C gd (A v + 1) (Eq. 16-40) C in(m) = 5 pF(15 + 1) C in(m) = 80 pF C = C gS + C in(m) C = 25 pF + 80 pF C = 105 pF R = R G || R 1 || R 2 R = 50 ||2 M ||1 M R = 50 Gate f 2 = 1/(2 π RC ) f 2 = 1/[2 π (50 )(105 pF)] f 2 = 30.3 MHz Collector C out(M) = C gd [( A v + 1)/ A v ] (Eq. 16-41) C out(M) = 5 pF[(15 + 1)/15] C out(M) = 5.3 pF C = C dS + C out(m) C = 15 pF + 5.3 pF Drain f 2 = 1/(2 π RC ) f 2 = 1/[2 π (909 )(20.3 pF)] f 2 = 8.61 MHz Answer: The high frequency cutoff for the gate is 30.3 MHz and the drain is 8.61 MHz.

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