Lab Report 5

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University of Illinois, Urbana Champaign *

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212

Subject

Electrical Engineering

Date

Feb 20, 2024

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pdf

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3

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Kitchen Sink Names & NetID: Ravyn Edran (redran2), Josh Frye (jcfrye2), Aidan Stahl (ahstahl2), Tuna Tuncer (ttuncer2), & Zhengze Cao (zhengze2) 28 September 2023 Lab Report 5 Introduction The purpose of this lab is to determine if an LED is considered Ohmic. We test the claim that LED bulbs are ohmic across the entire voltage output range of the DAC & can be treated as ideal resistors. This is done by connecting an LED and resistor in series (figure 1) and measuring the voltage at different points across the system. Then we can conclude that LEDs also follow Ohm’s law. This can be found by determining if the LED has a consistent and non-changing resistance. Our team expects the LED to be ohmic since it consumes power like a resistor. Group Dynamic In this lab the first action we took to improve our ability to work as a group was switching around who did what. This time the people who normally wrote most of the lab reports collected data and visa versa. This was done to ensure that everyone was gaining experience in all areas of the lab. The second action we took we allow other group members to create the procedure for the lab. Normally the same people create the procedure for the lab and carry it out, however this time we decided to switch who did what to ensure that everyone has an understanding of what was actually being done in the lab. Methods To complete this lab we set up the resistor and the LED in series with each other. We collected the voltage above the resistor and between the resistor and LED (figure 2). We will determine
the current passing through the resistor using Ohm’s law. Due to the fact that the resistor and LED are in series, they will have the same current. Using this data and the voltage of the DAC we will be able to determine if the LED has a constant resistance and is therefore Ohmic. If the voltage and therefore current stay the same then the LED will be Ohmic. 𝑉 = 𝐼𝑅 𝑉 1 = 𝑉 ?𝐴? 𝑉 1 = 𝑉 𝑅2 + 𝑉 𝑅1 = 𝐼𝑅 2 + 𝐼𝑅 1 In series: 𝐼 1 = 𝐼 2 Results
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