Failure to deliver the correct voltage and current will result in a burned out LED, so that's why the circuit is connected in parallel. Our goal then is to make both LEDs light and take measurements of voltage and current.
The requirements for the LEDs are:
LED1: 5V and 22.75mA
LED2: 2V and 20mA
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| Circuit Diagram |
Step 1: Laboratory equipment
- A regular classroom power supply will be set at 9V
- Although it was suggested to use a breadboard to build the circuit, we built the circuit using cables.
- The resistance on the LEDs (equivalent resistances = R=V/I) were:
R LED1 = 219.78ohms
R LED2 = 100ohms - We have to find out the value of R1 and R2
Step 2: Calculations
KCL is applied to both nodes on the picture above to find:
Ir1 - I LED1 = 0, Ir1 = 22.7mA
Ir2 - I LED2 = 0, Ir2 = 20mA
KVL is applied to both loops to get
Vr1 + 5 - 9 = 4V
Vr2 +2 -9 = 7V
We can apply Ohm's Law to find the needed resistance:
R1 = Vr1/Ir1 = 4/0.0227 = 175.82ohms
R2 = Vr2/Ir2 = 7/0.0200 = 350ohms
We also need to find the maximum power that can be consumed in order to avoid compromising lab equipment
Pr1 = Vr1*Ir1 = 0.09098W
Pr2 = Vr2*Ir2 = 0.14W
Next we had to pick R1 and R2 from restricted-value resistors in the classroom. We picked two fixed resistors; R1 = 220ohms and R2 = 470ohms
Step 3: Measurements
| Color Code | Nominal (ohms) | Measured (ohms) | Wattage | |||
| Red | Red | Black | Black | 270 | 217 | 1/8 |
| Yellow | Purple | Black | Black | 470 | 465 | 1/8 |
| Config | I LED1 (mA) | V LED1 (V) | I LED2 (mA) | V LED2 (V) | I Supply (mA) |
| 1 | 13.37+/-.01 | 6.27+/-.01 | 15.05+/-.01 | 2.14+/-.01 | 28.7+/-.1 |
| 2 | 13.40+/-.01 | 2.96+/-.01 | x | x | 13.36+/-.01 |
| 3 | x | x | 15.09+/-.01 | 2.14+/-.01 | 15.09+/-.01 |
Step 4: Calculations
A). Capacity of 9V alkaline battery is ~0.6Ahr; however, voltage drops significantly as we get to the 'end' of the battery life. Assuming 'useful' life of battery is 0.2Ahr. With both LEDs, how long can the circuit operate before battery voltage goes down too low?
when we have both LEDs;
0.2Ahr = (13.37+/-.01 + 15.05+/-.01)mA
hr = (28.42+/-.014mA)/200mA = 0.1421hrs or 8.53 minutes.
B). What is the percent error between achieved LED current and desired value for both LEDs in the same circuit?
LED1 = 41.23%
LED2 = 75.25%
The cause for high percent errors might have been because the ratings were for individual LED situations. Usage of the LEDs might have affected the current readings as well.
C). From data, determine circuit efficiency when both LEDs are in the circuit.
P LED1 = VI = (6.27+/-.01V)(13.37+/-.01mA) = 0.0838+/-.00015W
P LED2 = VI = (2.14+/-.01V)(15.05+/-.01mA) = 0.0322 +/-.00015W
Ptotal out = 0.116 +/-.00021W
Pin = VI = (9.26+/-.02V)(28.7+/-.1mA) = 0.266+/-.0011W
Efficiency, n = Pout/Pin = 43.6+/-.2%
C). If we repeated the design with 6V battery (changing biasing R1 and R2 but not changing LED parameters), would efficiency go up, down or same?
With 6V,
Vr1 = 1V Vr2 = 4V
R1 = 43.96ohm R2 = 200ohm
Pr1 = 0.02275W Pr2 = 0.08W
n = Pout/Pin = (0.02275+0.08)/(0.1722) = 59.67%
If we had a 5V battery, Vr1 = 5volts battery - 5volts LED1 = 0. As we increase the voltage (i.e. 5.1V, 5.5V, the efficiency gets from 43% - 51.3%.
The most efficient battery would be around 5.9V - 6V.
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