Explanation Given: The given circuit is shown in Figure 1 Calculation: The expression to determine the value of the current I C 1 , 2 is calculated as, I C 1 , 2 = 1 mA 2 = 0.5 mA The expression for the small signal input resistance is given by, r π 1 , 2 = h F E V T I C Q 1 Substitute 100 for h F E , 0.026 for V T and 0.5 mA for I C 1 , 2 in the above equation. r π 1 , 2 = ( 100 ) ( 0.026 V ) 0.5 mA = 5.2 k Ω The expression for the trans-conductance of the first transistor is given by, g m 1 , 2 = I C Q 1 V T Substitute 0.5 mA for I C 1 , 2 and 0.026 V for V T in the above equation. g m 1 , 2 = 0.5 mA 0.026 V = 19.23 mA / V The expression for the small signal input resistance for the third transistor is given by, r π 3 = h F E V T I C 3 Substitute 100 for h F E , 0.026 for V T and 2 mA for I C 3 in the above equation. r π 3 = ( 100 ) ( 0.026 V ) 2 mA = 1.3 k Ω The expression for the trans-conductance of the first transistor is given by, g m 3 = I C 3 V T Substitute 2 mA for I C 3 and 0.026 V for V T in the above equation. g m 3 = 2 mA 0.026 V = 76.923 mA / V The small signal model of the circuit is shown in Figure 2 Apply KCL at node V x . V π 1 r π 1 + g m 1 V π 1 + g m 2 V π 2 + V π 2 r π 2 = 0 From the above circuit the equation of the voltage is given by, V π 1 = V s − V x V x = V s − V π 1 From the above circuit the equation of the voltage is given by, V π 2 = V e 3 − V x Substitute V s − V π 1 for V x in the above equation. V π 2 = V e 3 − ( V s − V π 1 ) V s = V π 1 − V π 2 + V e 3 As the transconductance parameter and the small signal resistance for the transistors one and two are equal, the above equation is written as, V e 3 = V S + 2 V π 2 ....................(2) Apply KCL at node V y . g m 2 V π 2 + V π 3 r π 3 + V y R C 2 = 0 g m 2 V π 2 + V π 3 r π 3 + V π 3 + V e 3 R C 2 = 0 Substitute 19

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

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ISBN: 9781266368622

Author: NEAMEN

Publisher: MCG

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Chapter 12, Problem 12.59P

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The value of the transconductance transfer function.

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Chapter 12 Solutions

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

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 transconductance transfer function.

Solution Summary: The author explains the value of the transconductance transfer function. The expression for the small signal input resistance is given by r_pi 1,2

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Chapter 12 Solutions