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EE47 - Lab1

Page history last edited by xinyi xie 9 years, 9 months ago

Guilherme Fidalgo Velloso Ferreira

EE47

Lab 1: Orientation and LED

2. Controlling the Brightness of LEDs

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.

In order to limit the current to 25mA on a green LED, using a power supply of +5V and considering a forward voltage drop of 2.2V of the LED, the resistance must be determined. Some calculations using Ohm’s law (V = IxR) are needed:

V = +5 - 2.2 = +2.8 V

I = 25 mA = 0.025 A

V = IxR -> R = V/I -> R = 2.8/.025

R = 112 Ω

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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 resistance from question (a) is a minimum resistance. The LED would likely fail if you lower the resistance, since by doing it the current will increase going over the maximum continuous DC forward current rating for the LED.

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What is the resistance range of the potentiometer?

The resistance range of the potentiometer measured with the digital multimeter goes from 0.26 Ω to 10.05 MΩ.

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3. Basic LED Circuit with Switch

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

No, it doesn’t matter the order the components of my circuit are arranged between power and ground since everything is connected in series and in this type of connections any interruptions between power and ground will interrupt the whole circuit, therefore not lighting up the LED.

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4. Battery-Powered LED with Switch on Breadboard

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

Using this battery (+9V) and in order to maintain the maximum forward current rating for the green LED at 25mA some calculations must be done, still considering the forward voltage drop of the LED to determine the value of the new resistor that should be used:

V = +9 - 2.2 = 6.8 V

I = 25 mA = 0.025 A

R = V/I -> R = 6.8/.025

R = 272 Ω

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