lab4 tables - differ from the emitter and collector...

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Table 1 Vcb=5V Vbe(volts) Vre I 0 4mv 181uA .2 .16 7.27uA
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.4 64 2.9mA .5 653 29.6ma .55 3412 155ma .6 11414 518.8ma Table 2 1ma 2ma 3ma 4ma 5ma Vcb Vrc Ic Vrc Ic Vrc Ic Vrc Ic Vrc Ic -.55v -.58 580uA -.66 660uA -1.7 1.7mA -2.5 2.5mA -3.35 3.35mA -.5 -.58 580uA -1.22 1.22mA -2.15 2.15mA -3.0 3.0mA -3.65 3.65mA 0 -.62 620uA -1.23 1.23mA -2.17 2.17mA -3.0 3.0mA -3.6 3.6mA 5 -.71 710uA -1.41 1.41mA -2.46 2.46mA -3.4 3.5mA -4.2 4.2mA 10 -4.3 430uA -4.3 4.3mA -4.77 4.77mA -7.0 7.0mA -5.58 5.58mA Questions: 3. Ie=3mA Vrc Ic a -1.7 1.7mA 566mA -2.15 2.15mA 716mA -2.17 2.17mA 723mA -2.46 2.46mA 820mA Alpha is different in different regions of Vrc because the current ratio between the current collector increases along the voltage. 4.the variable Ic and Ie are considered to be in the same linear region because they can be used to find each other. Once you have solved for Ic, you can find Ie and visa versa. Once you find the value for both Ic and Ie, it can be used to solve for alpha, thus can be used to find beta.
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Data:
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Conclusion: This lab was very helpful in learning how the common base characteristics
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Unformatted text preview: differ from the emitter and collector characteristics. For a common base, the base is grounded while the input voltage and current the emitter and collector. When we designed the circuit, the input voltage for the collector could not be obtained if its above 5 volts, so any data below 5volts is obtainable. As shown in table one and two, 15V and 25V cannot be used. For the first circuit, we measured the voltage across the emitter to find the current across the emitter. For the 2 nd graph, we built the circuit and measured across the resistor to find the voltage across the collector to solve for the current across the collector. With all of these information collected, we can use it to create a graph that is similar to the characteristics shown in figure 1 and 2....
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lab4 tables - differ from the emitter and collector...

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