mid_term_solution_2007 - Solution to Problem 1 Part A 5V...

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Solution to Problem 1 5 V L C 1 + - C 2 C 3 R 1 R 2 Port 1 Port 2 T Part A Inductor L acts as RF choke and is used for preventing RF signal from leakage to the bias network and allowing DC current flowing into transistor. (0.5) Capacitors C1 and C2 are used to isolate DC bias of transistor T and other component between the RF signal and allowing RF signal flowing into transistor. (0.5+0.5) Capacitors C3 acts as RF ground and are used for isolation between the RF signal and the DC supply. (0.5) The resistors R2 is used for bias the base of transistor and stabilize the bias conditions if current gain changes. (0.5) The resistors R1 are used for bias the collector of transistor and limit the maximum current flowing into transistor T. (0.5)
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Current Gain: Solution to Problem 1 5 V L C 1 + - C 2 C 3 R 1 R 2 Port 1 Port 2 T Part A () 1 cc b c CE VI I R V =+ + 2 20 30 . 7 100 m R ce b b be R V 1 20 52 0 3 100 m mR ⎛⎞ + ⎜⎟ ⎝⎠ By KVL, By KVL, cb II β = 1 99.0 R 2 11.5 RK Check: R>>50 Ω , therefore it can be directly connected to the transmission line (0.5) (0.5) (0.5) (0.5)
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Solution to Problem 1 0 10 ind ZZ > 9 2 1 10 10 50 L π ×× × > × 9 79.58 10 L Minimum inductance = 79.58nH 0 1 10 cap < 9 15 0 2 1 10 10 C < × 12 31.83 10 C Minimum capacitance of all capacitors = 31.83pF Suitable value ~ 10nF Part A Inductor as RF choke should provide large enough impedance. Capacitor as RF short circuit should provide near zero impedance. (0.5+0.5+0.5) (0.5)
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Solution to Problem 1 Part A () 1 cc b c CE VI I R V =+ + 11.5 0.7 120 c ce I VK ce b b be R V = + 59 9 120 c cc e I IV ⎛⎞ + ⎜⎟ ⎝⎠ By KVL, By KVL, 5 99 11.5 0.7 120 120 c II IK + + 21.9771 c Im A = 21.9771 11.5 0.7 120 ce mA 2.8061 ce VV = 21.9771 20 100% 9.4% 21.9771 20 2 c I = + 3 2.8061 100% 6.7% 3 2.8061 2 ce V = + (0.5) (0.5) % error of % error of (0.5) (0.5)
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Solution to Problem 1 Part B (1) (1)
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Solution to Problem 1 For minimum noise figure, 0.15 19 S opt Γ =Γ = ∠ ° 50 Ω is required. Matching Network 0.15 19 S Γ= 50 Ω 50 Ω Solution 1 Solution 2 Part B (1)
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Solution to Problem 1 Solution 1 We start from 50 Ω load, Increase in normalized reactance ( Δ x) = 0.58 – 0 = 0.58 Change in normalized reactance comes from adding a series inductor 0 L x Z ω Δ= 9 20 . 9 1 0 0.58 50 L π × ×× × = 0.51 Ln H = Decrease in normalized susceptance ( Δ b) = 0.44 - 0.08 = 0.36 Decrease in normalized reactance comes from adding a shunt capacitor 0 bZC 9 0.36 50 2 0.9 10 C × × × × 1.3 Cp F = Part B (0.5) (0.5) (0.5) (0.5)
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Solution to Problem 1 For maximum transducer gain, Lo u t Γ =Γ 12 21 22 11 1 S out S SS S S Γ Γ= + −Γ 0.0377 58.0 17.138 120.4 0.15 19 0.704 38.4 1 0.6060 74.8 0.15 19 out °× ° ° + −∠ ° × ° 0.6516 45.401 out ° 0.6516 45.401 L ° Part C (1) (0.5) (0.5)
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Solution to Problem 1 50 Ω Matching Network 50 Ω 50 Ω Solution 1 Solution 2 0.6516 45.401 L Γ =∠ ° Part C
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Solution to Problem 1 Part C (1) (1)
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Solution to Problem 1 Solution 1 We start from 50 Ω load, Increase in normalized reactance ( Δ x) = 1.75 – 0 = 1.75 Change in normalized reactance comes from adding a series inductor 0 L x Z ω Δ= 9 20 . 9 1 0 1.75 50 L π × ×× × = 15.5 Ln H = Decrease in normalized susceptance ( Δ b) = 0.42 - 0.39 = 0.03 Decrease in normalized reactance comes from adding a shunt capacitor 0 bZC 9 0.03 50 2 0.9 10 C × × × × 0.1 Cp F = Part C (0.5) (0.5) (0.5) (0.5)
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22 2 21 11 11 SL T So u t L GS S −Γ = Γ− Γ Γ 27.409 T Gd B = Lo u t
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This note was uploaded on 05/01/2011 for the course ELECTRICAL EE5602 taught by Professor Xuequan during the Spring '11 term at City University of Hong Kong.

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mid_term_solution_2007 - Solution to Problem 1 Part A 5V...

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