# Ass8 - Q 1 has =50 and I S = 6.95×10-16 A and V T =25mV...

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1 ECED3201: Introduction to Electronics Fall 2009 Dalhousie University, Department of Electrical and computer Engineering Assignments #8 Problem 1 - [6 points]: For the emitter follower in Fig. 1, the signal source is directly coupled to the transistor base. a) [1.5 pt] If the dc component of v sig is zero, find the dc emitter current. Assume = 100 and r o = . Consider V BE 0.7V for the dc analysis . Now replace the transistor with the T-mode for ac-analysis b) [2.5 pt] Find the input resistance R in . c) [2 pt] Find the voltage gain A v =v o /v sig . Note that the small-signal model of pnp transistor is identical to that of an npn transistor (see lecture 4, page 35) (Hint you will find (a) I E = 1mA, (b) R in = 80 k ) Fig.1 Problem 2 - [14 points]: We wish to analyze the amplifier shown in Fig. 2. Assume the transistor

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Unformatted text preview: Q 1 has =50, and I S = 6.95×10-16 A, and V T =25mV. Also consider Q 1 operates in active region and r o = . a. [4 pt] Calculate the voltage V BE and the current I C . b. [3 pt] Calculate the small signal input resistance R i . c. [3 pt] Calculate the small signal gain A V = v o / v i . 2 d. [4 pt] Propose a modification to the circuit in the dashed box without changing the DC operating point ( I C and V BE ) in order to obtain a gain A V1 = v o / v i =2x A V and verify that the input resistance is not reduced by more than 50%. (Hint you will find (a) V BE = 654.6 mV, (c) A v =-4.53 V/V ) v O Q 1 +5V R E = 500 W R C = 3k W +5V R 1 = 400k W R 2 = 100k W v i +- R i Fig. 2 Due date: November 19, 2009. Late submission will not be accepted....
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Ass8 - Q 1 has =50 and I S = 6.95×10-16 A and V T =25mV...

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