Midterm Exam 1 Solutions 08

E it will suffer less from the early effect than

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Unformatted text preview: ge in the base width due to a change in the reverse bias on the base-collector junction will result in a smaller relative change in the base width in B than in A. d) Solve the bias point of the following PNP transistor (IC, VEB, VEC). Assume I S = 1017 A, β =100, and VA = ∞ [5 pts] VCC = 3V RE = 1 KΩ Assume 10 A (correction made during exam). 1 1 2.12 mA Page 5 ln e) Draw the small-signal model of the circuit in Part d). Specify all the small signal parameters used (e.g., g m , rπ , etc). [4 pts] 81.5 mS 1.226 kΩ Page 6 Problem 3 [30 points]: BJT Amplifiers a) Consider the BJT amplifier shown below with I BIAS = 1 mA . Assume I S = 1017 A, β =100, and VA = 10V . VCC = 3V I BIAS RS = 1 KΩ vin CC vout RL = 10 KΩ RE = 1 KΩ i) Find the value of VBE . [4 pts] Assume 2 V (given during exam) and 10 A (correction made during exam). 2V 1 mA fixed by the current source 1 1.01 V 0.99 V 1 ln 836 mV 1 ii) Is the BJT in the active mode? Why? [4 pts] Yes. The base-emitter junction is forward biased and the base-collector juncti...
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This note was uploaded on 01/24/2014 for the course EE 105 taught by Professor King-liu during the Spring '07 term at Berkeley.

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