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LAB 7 (2) - LAB 7 1 A 3.15V across the resistor 3.15V/330=...

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LAB 7 1. A. 3.15V across the resistor. 3.15V/330= 9.5mA across the resistor. For 1k resistor there is 3.27V going across. 3.27/1000=.327mA. The LED brightness increases as the current increases (resistance decreases) and the LED brightness decreases as the current decreases. B. Voltage is 0. Current is 0. Voltage across the diode is -5 volts. KVL= 5V-Vr-Vd=0. Yes, the measured voltages are in agreement. 2 A. The LED is on. 330Ohm Resistor = 3.11V 10K Ohm Resistor = 4.28V Ib=3.11/330= 9.42mA Ic=4.28/1000= 4.28mA Measured Voltages VBE=.660V VCE= 50mV Measured VBE is closet to the expected value (.6V). Measured VCE is off from the expected value (.2V). Hfe=Ic/Ib. 4.28mA/9.42mA=.45mA B. VBE=.667V VCE=3.47mV 10k Ohm Resistor=0V 330 Ohm Resistor= 4.33V Ic=0 Ib=4.33/330= 1.31mA C. The LED current is too small for the LED to be visible. D. When the connection point is at (a) LED 1 lights up and LED 2 doesn't. Vice versa when connection point is at (b).
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SOMEONE ELSE'S LAB 1. a) The voltage across the 330 ohm resistor is 2.96 V The current is 8.9 mA The voltage across the 1K ohm resistor was 3.09 V The current is 3.09 mA As the current is decreased the light bulb gets dimmer.
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