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**Unformatted text preview: **Physics 3330 Prelab 7 solutions 1. We want to find voltages and currents for the quiescent (DC) case of the circuit shown in Fig. 7.4 of the lab manual. The first step is finding the base voltage which comes from the voltage divider created by the two resistors R 1 and R 2 so V B = V CC R 2 R 1 + R 2 = 15 V 10 k Ω 47 k Ω + 10 k Ω = 2 . 6 V Knowing V B allows us to calculate V E since the emitter voltage is between 0.6 V and 0.7 V below the base voltage (subject to the constraint that V E can never go below ground). We will use 0.6 V so V E = V B- . 6 V = 2 . 0 V. Knowing V E allows us to calculate I E since there is just one resistor between V E and ground so by Ohm’s law, I E = V E /R E = 2 . 0 V / 1 k Ω = 2 mA. By charge conservation, I E = I C + I B but I B is a factor of h F E smaller than the other currents and since h F E ≈ 200, we can safely ignore I B so I C ≈ I E = 2 mA. With the current I C we can finally find the collector voltage as V C = V CC- I C R C = 15 V- (2 mA)(2 . 74 k Ω) = 9 . 5 V The power dissipated in the transistor in the quiescent state is P = ( V C- V E ) I E = (9 . 5 V- 2 . 0 V)(2 mA) = 15 mW Looking up the datasheet at http://www.colorado.edu/physics/phys3330/phys3330 fa09/pdfdocs/2N3904.pdf we find the max- imum power dissipation is 625 mW. We are well below that limit so this circuit is safe. The same datasheet also indicates that the minimum h F E at I C = 10 mA is 100....

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