ET212_Week 4 Lab_IngramJ

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Grantham University *

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ET212

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Electrical Engineering

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Jan 9, 2024

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Jason Ingram GID: G00151729 Lab 4: Transistor Fundamentals Grantham University Date: 11/5/2023
Introduction: The purpose of this lab is to understand the fundamentals of a transistor and analysis of an analog circuit. When doing the lab, you must analyze and calculate measurements for the following v E , I E ,V RC ,V C , V CE . The lab will be compared to what was calculated and measured followed by the following review questions that will be answered at the end of the lab: Compare the calculated and measured values in the table and analyze the performance of the transistor. Discuss whether the values are the same or different. If they are different, provide the reasoning and explain how to reduce this difference between calculated and measured values. Explain what happens when the transistor changes to a PNP transistor. How does the change in the transistor affect the current and voltage in the circuit? Equipment/Components: NPN transistor: 2N3904 4 Resistors: 2-40Ω, 1kΩ, 470Ω Jumper wires Breadboard myDAQ instrument device Screwdriver Screw Terminal Connector USB Cable Multimeter probes. Procedure: The first part of the lab is to analyze the figure given in the directions of the lab and make calculations for v E , I E ,V RC ,V C , V CE . Once the calculations have been made the next step is to build the circuit on the breadboard using the transistor and resistors. Once the circuit has been built use the jumper wires connect the breadboard to the myDAQ hardware and use the channel +15v to supply the voltage for the circuit. Using the multimeter take measurements for V E ,V RC ,V C ,V CE , currents : I C , I E . When all measurements have been obtained table the values and compare them between calculated and measured.
Calculations: V E = I E 1 10 3 I E = ( 150 + 1 ) .00006179 V RC = I C R C V E = .00933 1000 I E = 151 0.06179 mA V RC = 00933 470 V E = 9.33 V I E = .00933 = 9.33 mA V RC = 4.39 V V C = 15 I C R C V CE = V C V E V C = 15 ( .00933 470 ) V CE = 10.61 9.33 V C = 15 7.0132554 V CE = 1.28 V V C = 10.61 V Circuit design:
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Execution/Results: VE Circuit IE circuit V RC circuit
VC Circuit V CE Circuit Analysis: Calculated Measured V E 9.33v 9.0V I E .00933 or 9.33mA 14.624V V RC 4.39v 14.767V V C 10.61v 233.302mV V CE 1.28v 14.933V Compare the calculated and measured values in the table and analyze the performance of the transistor. Comparing the two measurements that were obtained from this lab there was a big significant difference. I believe the reason for this was the resistors that were used on the breadboard. Discuss whether the values are the same or different. If they are different, provide the reasoning and explain how to reduce this difference between calculated and measured values. Looking between the results, there is a difference between the two.
Explain what happens when the transistor changes to a PNP transistor. How does the change in the transistor affect the current and voltage in the circuit? The PNP transistor has a base terminal that is negative with respect to the emitter. When switching the base, the current flows when it is a negative base voltage. In other words, the low voltage will make the transistor become an open circuit. Conclusion: In this lab we built several circuits to measure v E , I E ,V RC ,V C , V CE . We successfully measured and calculated the values that were obtained in the table above in the lab report. It was concluded that even though everything was built there was a difference between the calculated values and measured values.
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