EE302Lecture16 - (0.0005+0.005)(100) = 0.55 volts. 1 The...

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Today we discussed the HW 6 problems and the transistor problem below: In the NPN transistor problem below, we are to select a value of R that will give the collector C a voltage of 7 volts. We are given all resistors except R, and given the transistor Beta = 10, and the supply voltage of 12 volts DC. The problem is solved by first calculating the current through the 1000 ohm resistor on the collector. This current is (12-7)/1000 = 0.005 A. Then we calculate the base current using the current gain which is 10. The base current is 1/10 th the value of the collector current or 0.0005 A. The collector and base currents add together to give the total current out of the emitter. The emitter current goes through the 100 ohm emitter resistor to ground. We calculate the voltage drop across the 100 ohm resistor as
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Unformatted text preview: (0.0005+0.005)(100) = 0.55 volts. 1 The silicon transistor has a 0.7 voltage drop from the base to the emitter. This means that the base is 0.7 volts greater than the emitter. Summing the 0.7 volts and the emitter 0.55 voltages gives the base voltage as 0.7+0.55 = 1.25 volts on the base. Knowing the base terminal has a voltage of 1.25 volts allows us to calculate the current through the 2000 ohm resistor as 1.25/2000 = 0.000625 A. The sum of the current through the 2000 ohm resistor and the transistor base lead is 0.000625+0.0005 = 0.001125 A. The R needed to give this current for a voltage drop of 12 to 1.25 volts is: (12 – 1.25) / 0.001125 = 9555 ohms or R = 10k ohms as a standard value. 2...
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This note was uploaded on 05/01/2008 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas.

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EE302Lecture16 - (0.0005+0.005)(100) = 0.55 volts. 1 The...

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