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Lab 1 Write-Up - Beck

Page history last edited by Benjamin Tee 8 years, 1 month ago

Michael Beck

EE 47 – Press Play

Lab #1 Write-Up

Executed: 7/3

Due: 7/10


2. Controlling the Brightness of LEDs

a. 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.


max = 25mA for Green LED

VS = 5V

VF = 2.2V for Green LED

I = (VS – VF)/R < Imax

R > (VS – VF)/Imax

   = (5 – 2.2)/.025

   = 112Ω 

b. 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.


112Ω is a minimal resistance. 

c. What is the resistance range of the potentiometer?


The range of the potentiometer is 0 – 10KΩ 

3. Basic LED Circuit with Switch

Next, we'll insert a switch into the circuit. The momentary switches in yourkit are "normal open," meaning that the circuit is interrupted in the idle state, when the switch is not pressed. Pressing the switch closes the circuit until you let go again.

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


It does not matter because the components are in series.

4. Battery-Powered LED with Switch on Breadboard

Now, let's make things a little more portable. Disconnect the breadboard from the power supply. Use the battery clip and a 9V battery to power your LED circuit. You'll need to replace your 150 Ohm resistor for the higher voltage. Be sure to do the calculation below so you know what resistance to use. You can choose a conservative resistance so you are not at the edge of failure.

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


max = 25mA for Green LED

VS = 9V

VF = 2.2V for Green LED

I = (VS – VF)/R < Imax

R > (VS – VF)/Imax

   = (9 – 2.2)/.025

   = 272Ω



Comments (1)

Benjamin Tee said

at 12:17 am on Jul 11, 2012

Great job Beck! It is great that you show the steps in how you derive the answer. Keep it up!

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