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Unformatted text preview: UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences Homework #4 Solutions Due Tuesday, September 25, 2007 EE 105 Fall 2007 Prof. Liu Problem 1. PNP BJT Biasing a) I will use R C to refer to the 500 Ω resistor at the collector of the BJT. V EB = 2 . 5 − 1 . 7 = 0 . 8 V I C = I S parenleftBig e V EB /V T − 1 parenrightBig = 1 . 15 mA V X = I C R C = 500 I C = 577 mV b) I C = I S parenleftBig e V EB /V T − 1 parenrightBig parenleftbigg 1 + V EC V A parenrightbigg V EC = V CC − V X = V CC − I C R C I C = I S parenleftBig e V EB /V T − 1 parenrightBig bracketleftbigg 1 + 1 V A ( V CC − I C R C ) bracketrightbigg I C = 1 . 49 mA V X = 745 mV Problem 2. Input Resistance These can be solved by inspection if you recall that looking into the base, you see r π +(1+ β ) R E (if R E is a resistor present in the emitter) and looking into the emitter, you see 1 /g m . If you don’t recall these resistances, you can derive them from the small signal model. a) R in = R 1 + ( R 2 bardbl r π ) b) R in = 1 g m bardbl R 1 c) R in = r π 1 + (1 + β 1 ) r π 2 Problem 3. Output Resistance The resistance looking into the collector of Q 2 is 1 g...
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 Fall '07
 KingLiu
 Electrical Engineering, Voltage divider, Resistor, Electrical resistance, R1 R2 R1, VT R1 R1

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