EE2005-Tutorial03-Solution - NATIONAL UNIVERSITY of...

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NATIONAL UNIVERSITY of SINGAPORE Department of Electrical and Computer Engineering EE2005 – Electronics Tutorial 3 Solution Q1. Given β = 150, and I S = 10 -15 A (a) I C = β I B = 150 × 12 μ A = 1.8 mA I C = I S T BE V e V Rearrange, V BE = V T ln( I C /I S ) With V T = kT/q 0.025V, V BE = 0.025V ln(1.8×10 -3 /10 -15 ) = 0.705V I C = 1 + β I E Rearrange, I E = 1 + I C = 1.81 mA (b) For the I C to increase by a factor of 10, 10 ' ' = = T BE T BE V V V V C C e e I I Hence, 10 ' = T BE BE V V V e . Taking natural logarithm, we have Δ V BE =V BE ’- V BE =V T ln10 = (0.025V ) ln 10 = 0.0576 V
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Q2. The DC collector current I C is 1.5 mA. In order for the voltage at the collector V C to be 6 V with respect to ground, there must be a 3 V voltage drop across resistor R C . Hence, I C · R C = 3 V R C = 3V/(1.5 mA) = 2 k Ω The DC emitter current I E is given by I E = [( β +1)/ β ] I C = [76/75] · 1.5 mA = 1.52 mA. The DC base current I B = I C / β = 1.5 mA/75 = 0.020 mA = 20 μ A. For the emitter voltage
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This note was uploaded on 10/24/2011 for the course ELECTRICAL ee2021 taught by Professor Tan during the Spring '11 term at National University of Singapore.

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EE2005-Tutorial03-Solution - NATIONAL UNIVERSITY of...

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