Objective/Intro:
To simultaneously have two LEDs of different voltages and currents light up without burning out the bulbs while being supplied 9V from a battery source.
 |
| Biasing Solution |
The figure above is the schematic for the biasing solution to allow for the LEDs to light up.
To allow for the biasing to work properly, the values of R1 and R2 must be calculated.
 |
| Calculation of Voltage and Current for R1 and R2 |
 |
| Values of R1 and R2 & Power consumed by each resistor |
The resistor values obtained for the biasing to properly work:
R1 = 175.8 Ω
R2 = 350 Ω
However, these resistors are unavailable in lab, so the values are rounded up to the closest commercially-available resistor values:
R1 = 220 Ω
R2 = 470 Ω
Model:
 |
| Lab Schematic for Biasing Solution |
 |
| Lab Model of Biasing Circuit |
To model the biasing solution, the circuit is connected to a 9V battery supply and the two LEDs are placed in parallel with one another. There is a resistor that is series with each of the LEDs to properly bias the two LEDs.
There are three configurations done for the lab:
Configuration 1: Both LEDs in the circuit
Configuration 2: Remove LED2 (RLED2) from the circuit
Configuration 3: Remove LED1 (RLED1) from the circuit
Configuration Data:
Questions/Calculations:
d. With the 6V battery, the efficiency of the circuit will go up. The "best" efficiency would be obtained by using a battery voltage of 5V.
The efficiency with the 5V battery:
Pin = 5V * Isupply = 5 * ( 27.82*10^-3) = 0.1391 W
Pout = 0.110927W
Efficiency = Pout/Pin * 100 = (0.110927/0.1391 )*100 = 79.75%
Free Mat Assignment 1
No comments:
Post a Comment