Charles Wang
29 June 2011
EE47
Lab 1  Mechatronics Lab Orientation & Equipment Training

Basic LED circuit on breadboard

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)
According to Ohm’s law, voltage = resistance * current which means that resistance = voltage / current.
Red LED: The forward voltage drop of the red LED is 1.85V and there is 5V going into the circuit, therefore, the voltage dropped across the resistor must be 5V  1.85V = 3.15V. For the red LED that has a maximum current of 30mA, which is equal to 0.03A, the resistance is 3.15V / 0.03A = 105Ω.
Yellow LED: The forward voltage drop of the yellow LED is 2V and there is 5V going into the circuit, therefore, the voltage dropped across the resistor must be 5V  2V = 3V. For the yellow LED that has a maximum current of 25mA, which is equal to 0.025A, the resistance is 3V / 0.025A = 120Ω.
Green LED: The forward voltage drop of the green LED is 2.2V and there is 5V going into the circuit, therefore, the voltage dropped across the resistor must be 5V – 2.2V = 2.8V. For the green LED that has a maximum current of 25mA, which is equal to 0.025A, the resistance is 2.8V / 0.025A = 112Ω.
Akil: You're one of the few that included the forward voltage drop of the LED, well done!

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.
It is a minimum resistance because 30 mA is the maximum continuous DC forward current rating for the red LED. Any less resistance would cause the red LED to burst.

What is the resistance range of the potentiometer?
There is 10K written on the side so it must be from 0Ω to 10,000Ω. Akil: exactly. And if you check with the multimeter you might get something slightly different because of the manufacturing tolerances.

Basic LED circuit with switch on breadboard

Does it matter what order the components of your circuit are arranged between power and ground? Why or why not?
Yes, it does matter because the resistor must be placed before the LED light on the positive electrode or else the voltage will not be suppressed and burst the LED. However, the switch can be placed anywhere in the circuit. Akil: The order doesn't matter actually. Remember, think of voltage drops in analogy with changes in graviational potential energy. If I drop a book from 10 ft to 5ft or from 5 ft to 0 ft, it will gain the same amount of kinetic energy. So, in the circuit, the same 1.85 V drop will occur over the LED, 3.15 V over the resistor, and the same current through both  if the order was reversed; they're in series. And also remember, it's current that flows, and too much current is what destroys the LED; we just have to set up the voltage drops so that doesn't happen.

Batterypowered LED with switch on breadboard

Using this battery, what is the minimum resistance required for use with your LED?
The forward voltage drop of the red LED is 1.85V and there is 9V going into the circuit, therefore, the voltage dropped across the resistor must be 9V  1.85V = 7.15V. The red LED has a maximum current of 30mA, which is equal to 0.03A. Therefore, the resistance is 7.15V / 0.03A = 238Ω. Akil: Nice!
Comments (1)
Akil Srinivasan said
at 9:52 pm on Jul 10, 2011
Well done! Very clear calculations. You can embed your photos directly in your wiki page, leaving the actual files in your student folder, and only moving the report 'wiki' page into 'Inbox'
You don't have permission to comment on this page.