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# Lab 1 Report for Alex Valderrama

last edited by 7 years, 5 months ago

2a. What resistance do you need to limit current to 30 mA (if using red LED) or 25 mA (if using yellow or green)? Be sure to state which color LED you are using. This resistance refers to the total resistance in series with the LED.

Hint: Make sure that you account for the forward voltage drop (Vf) of the LED that you're using.

From the Kingbright data sheet, the red LED has a forward voltage drop of 1.85 volts. Thus since the resistor and LED are in series we can calculate the resistance needed to limit the current to 30mA using V=IR => V/I = R. We know V is 5 - 1.85 = 3.15 and I is .030 A therefore R = 3.15 / .030 = 105 Ohms

2b. Is the resistance from question a) a maximum or minimum resistance? That is, in which direction if you change the resistance (higher or lower) would the LED likely fail.

This is the minimum resistance, if you reduce the resistance then the current will increase and cause the LED to fail

2c. What is the resistance range of the potentiometer?

Using the multimeter I measured the range of the potentiometer to be from 1 Ohm to 9.84 kOhm

3a. Does it matter what order the components of your circuit are arranged between power and ground? Why or why not?

No the order does not matter, as the switch interrupts the circuit whether it comes before or after the LED

4a. Using this battery, what is the minimum resistance required for use with your LED?

Using the same calculation from 2a we change V to be 9 and thus 9 - 1.85 = 7.15 and 7.15 / .030 = 238.3 Ohms, this is the minimum resistance required.

5. Video of circuit