lab 6 2010 bipolar junction transistor - transistor switch

Lab 6 2010 bipolar - LAB 6 BIPOLAR JUNCTION TRANSISTOR TRANSISTOR SWITCH I Purpose and Background This laboratory has three objectives 1 To measure

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L AB 6: B IPOLAR J UNCTION T RANSISTOR – T RANSISTOR S WITCH I Purpose and Background This laboratory has three objectives: 1. To measure the collector characteristic (V CE - I C curve) for a bipolar junction transistor; 2. To build and test three transistor switch circuits from the simplest one-transistor switch to a more complex two-transistor switch exhibiting hysteresis. 3. To diagnose simulated transistor circuits to identify possible faults in the circuits. The npn BJT transistor is operated with the base-emitter pn junction forward-biased and the base-collector junction reverse-biased. In normal operation, the base-emitter voltage is approximately 0.7V. The BJT is a current amplifier device: a small base current is amplified into a larger collector current. The β DC of the transistor is defined as I C /I B ; it is ~ 100. In a transistor switch, the transistor is operated either in cutoff or in saturation. In cutoff, the collector current is 0 and the transistor acts like an open switch (V CE is maximal). In saturation, the collector current is maximal and the transistor acts like a closed switch (V CE is small: V CE(sat) , about 0.2V). The data sheet for the 2N3904 BJT transistor are attached to the handout. II Collector Characteristic Curve Prelab: Consider the circuit shown on the following graph: VBB 5 V RB VCC 100 2N3904 1. Compute the values of R B that give a base current of 50 μ A, 100 μ A, 150 μ A. Enter the values in a table. For each R B , determine the single standard 10% precision resistor or combination of two standard 10% precision resistors that when added yield the value nearest to R B . (A table of resistor values is available at the end of your textbook in appendix.) Ib = (5-.7)/Rb Rb= (4.3)/Ib Ib(micro A) Rb Real Rb 50 =4.3/50x10^-6= 86 k ohms 82k+3.9k
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100 =4.3/100x10^-6= 43 k ohms 2.7k+1.5k 150 =4.3/150x10^-6= 28.666 k ohms27k +1.5k 2. The data sheet of the transistor lists two maximum values for V CE(sat) ; which value would you use for this circuit given the expected base currents. Use 200 mV 3. For each base current, what is the largest V CC that places the transistor in saturation? Vc=.2
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This note was uploaded on 04/04/2010 for the course BME 302L taught by Professor Maarek during the Spring '08 term at USC.

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Lab 6 2010 bipolar - LAB 6 BIPOLAR JUNCTION TRANSISTOR TRANSISTOR SWITCH I Purpose and Background This laboratory has three objectives 1 To measure

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