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Unformatted text preview: EE 101 Handout # 9 Professor Abbas El Gamal Feb 4 , 2003 Homework #3 Solutions 1. Using the given model for the bipolar transistor, the base current is equal to the current through the 50kΩ. This gives us the equation: v cc ( v out + v be ) 50kΩ = i b Now the current through the 1kΩ resistor equals the emitter current which is the sum of the collector and base currents. Using the transistor equations to eliminate the collector and the base currents we have: v out 1kΩ = ( β + 1) i b = ( β + 1) i e v be /v t 1 From the second equation, we get: v be ≈ v t ln( v out 1kΩ( β + 1) i ) Substituting for v be in the first equation, and plugging in v cc = 5V, i = 10 14 A, v t = 26mV, and β = 100, we get: 5 = v out (1 + 50 101 ) + (0 . 026) ln( v out (101000)10 14 ) This equation can be solved iteratively, graphically or by trial and error. In any case you should get v out ≈ 2 . 965V. 2. There are two regions of operation that we discussed for the BJT: off and on. The transistor is not off since i c = i s 6 = 0 does. So we assume that the transistor is on. We can figure out the base current i b : i b = i s /β = 4 . / 50 = 0 . 08 m A Using the exponential diode model you can solve for...
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 Fall '08
 PingLi
 Transistor, Vout, linear region, Vin, Vgs

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