Explanation Given: The given circuit is shown in Figure 1 Calculation: The expression to determine the value of the Thevenin resistance of the circuit is given by, R T H = ( 38.3 k Ω ) ( 13.5 k Ω ) ( 38.3 k Ω ) + ( 13.5 k Ω ) = 9.98 k Ω The Thevenin voltage of the circuit is calculated as, V T H = ( 13.5 k Ω 13.5 k Ω + 38.3 k Ω ) 10 V = 2.606 V The expression for the collector current of the first transistor is given by, I C 1 = h f e ( V T H − V B E ( on ) ) R T H + ( 1 + h F E ) ( 1 k Ω ) Substitute 9.98 k Ω for R T H , 120 for h F E , 0.7 V for V B E ( on ) , and 2.606 V for V T H in the above equation. I C 1 = 120 ( 2.606 V − 0.7 V ) 9.98 k Ω + ( 1 + 120 ) ( 1 k Ω ) = 1.75 mA The expression to determine the value of the collector voltage of the first transistor is given by, V C 1 = 10 V − I C 1 ( 3 k Ω ) Substitute 1.75 mA for I C 1 in the above equation. V C 1 = 10 V − 1.75 mA ( 3 k Ω ) = 4.75 V The expression to determine the value of the collector current I C 2 is given by, I C 2 = V C 1 − V B E ( on ) 8.1 k Ω Substitute 4.75 V for V C 1 and 0.7 V for V B E ( on ) in the above equation. I C 2 = 4.75 V − 0.7 V 8.1 k Ω = 0.50 mA The expression for the current I E 2 is equal to I C 2 and is given by, I E 2 = 0.50 mA The expression for the transconductance of the first transistor is given by, g m 1 = I C 1 0.026 V Substitute 1.75 mA for I C 1 in the above equation. g m 1 = 1.75 mA 0.026 V = 67.31 mA / V The expression for the small signal resistance is given by, r π 1 = h F E g m 1 Substitute 120 for h F E and 67.31 mA / V for g m 1 in the above equation. r π 1 = 120 67.31 mA / V = 1.78 k Ω The expression for the transconductance of the second transistor is given by, g m 2 = I C 2 0.026 V Substitute 0.50 mA for I C 2 in the above equation. g m 2 = 0.50 mA 0.026 V = 19.23 mA / V The expression for the small signal resistance is given by, r π 2 = h F E g m 2 Substitute 120 for h F E and 19.23 mA / V for g m 2 in the above equation. r π 2 = 120 19.23 mA / V = 6.24 k Ω The small signal model for the given circuit is shown below. Apply KCL at the voltage source V π 1 . v S − V π 1 0.6 k Ω = V π 1 9

Microelectronics: Circuit Analysis and Design

4th Edition

ISBN: 9780073380643

Author: Donald A. Neamen

Publisher: McGraw-Hill Companies, The

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Chapter 12, Problem 12.52P

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The value of the input resistance Rif .

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Chapter 12 Solutions

Microelectronics: Circuit Analysis and Design

Ch. 12 - (a) The open-loop gain of an amplifier is A=5104...Ch. 12 - (a) Consider a general feedback system with...Ch. 12 - (a) A feedback amplifier has an open-loop...Ch. 12 - (a) Consider the circuit shown in Figure...Ch. 12 - (a) The closed-loop gain of a feedback amplifier...Ch. 12 - The gain factors in a feedback system are A=5105...Ch. 12 - Prob. 12.3TYU Ch. 12 - An ideal series-shunt feedback amplifier is shown...Ch. 12 - Consider the ideal shunt-series feedback amplifier...Ch. 12 - An ideal series-series feedback amplifier is shown...

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The value of the input resistance R i f .

Solution Summary: The author explains the value of the input resistance, which is 6.32Omega , and the expression to determine the Thevenin voltage.

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The value of the input resistance Rif .

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Chapter 12 Solutions