EE332 HW2S Su2010

# EE332 HW2S Su2010 - EE 332 Devices and Circuits II...

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Unformatted text preview: EE 332 Devices and Circuits II Problem #2 Solution Summer 2010 5.5 (a) For this circuit, VBE = 0 V, VBC = -5 V and I = IC. Substituting these values into the collector current expression in Eq. (5.13): 5 I S 5 IC I S 0 exp exp exp 1 . . 025 R 025 1 1 I IC I S 1015 A 2 fA. 1 1 1 R (b) For this circuit, the constraints are VBC = -5 V and IE = 0. Substituting these values into the emitter current expression in Eq. (5.13): I V V V I E I S BE exp BC S BE 1 0 which gives exp exp VT F VT VT V 1 F V exp BE exp BC Substituting this result into IC : . VT 1 F 1 F VT I I V V ICBO S exp BC S BC 1 1 exp . 1 F VT R VT 1 1 1 1 For VBC = -5V , ICBO I S 1015 A 1.01 fA, and 1 101 1 F R 1 1 VBE VT ln 0.025V ln 0.115 V 0! 1 101 F 5.7 VBC - C IC IB + B + V BE 175 A E IE npn transistor 1 V I For VBC 0, I E I S exp BE 1 | I B E 1 | IC F I B F 1 F VT I E 175 A | I B VBE 175A 100 1.73 A | IC 175A 173 A 101 101 175A I E 100 VT ln F 1 0.025V ln 1 0.630 V 101 2 fA F 1 I S 5.10 Using vBE 0 in Eq. 5.13 and recognizing that i iC : 1 v i iC I S exp BC 1, and the reverse saturation current 1 R VT 1 1 ' of the diode connected transistor is I S I S 5 fA 1 6.67 fA 1 3 R 5.26 At IC 5 mA and VCE = 5 V , I B 60A : F IC 5mA 83.3 I B 60A IC 7 mA 87.5 I B 80A IC 10 mA 100 I B 100A At IC 7 mA and VCE = 7.5 V , I B 80A : F At IC 10 mA and VCE = 14 V , I B 100A : F 5.74 For example : 0 mA, 14 V and 1 V IC F I B FO CE B | We need two Q - points from the output characteristics. 1 I VA 5 mA, 5 V 14 5 10mA = FO .1mA and 5mA = FO .06mA y ields 1 0 1 0 VA VA 14 5 100 = FO and 83.3 = FO Solving these two equations yields 1 1 . VA VA FOO 79 and nd VA 6 V . V F 72. 4.02 a VA 37. 39.89 ...
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