Day 3- Introduction to Biasing

There is time that we would like to connect two LED's to a device for an back up if one of the LED has failed. But the question is that the two LED's have the different ratings, one is 5V the other one is 2V
if we connect the LED in series, the LED with a lower voltage will burn out really fast and short the circuit. If we connect the LED in parallel, one of the LED will still have too much voltage across and then burn out

The goals of this experiment is to determine suitable power ratings for each resistor and then determine how long the 9V battery can power the circuit. Also, we are given further practice in measuring resistance, voltage, and current.

We have the schematic that illustrate the connection between the parts

 Where the current through the 5V LED is 22.75mA and the current through 2V LED is 20mA

First, we need to determine the equivalent resistance of each LED using Ohm's Law

R_LED1= 5/22.75mA=220Ω
R_LED2=2/20mA=100Ω

Next, we will need to determine the necessary values of resistors R1 and R2

I_R1	22.75mA
I_R2	20mA
V_R1	4V
V_R2	7V
R_1	175.8Ω
R_2	350Ω
P_R1	0.091W
P_R2	0.14W

Since we are limited by the certain discrete values in resistors, we need to determine the equivalent resistance by connecting multiple resistors in parallel, or series.

Possible Resistors		Measured
R_1 (Ω)	2//150+100	172
R_2 (Ω)	150+2*100	347

Next, we will need to set up the circuits by using wires, alligator clips, electronic parts, and breadboard to demonstrate the circuit in class

After turning on the power, we obtain the picture

 Where the yellow LED "should" be rated at 5V and the green LED is rated at 2V

By measuring any currents and voltages across the LED we obtained the following data
Also, we were told to made some modification on our circuits
Configuration 1 : Both LEDs in the circuit
Configuration 2 : Remove LED2(remove R_LED2) from the circuit
Configuration 3: Remove LED1(remove R_LED1) from the circuit

*Configuration 2 and 3 illustrate when one of the LED failed but the circuit still work

Config	I_LED1(mA)	V_LED1(V)	I_LED2(mA)	V_LED2(V)	I_supply(mA)
1	14.8	6.46	19.7	2.15	34.4
2	14.6	6.48	X	X	14.6
3	X	X	19.7	2.18	19.6

1. If the capacity of a 9V Alkaline battery is approximately .6A-hr, assuming the useful life of the 9V battery is just .2A-hr. With both LED's in the circuit on, how long can the circuit operate before the battery voltage goes too low

I_supply=0.0344A
T=A-hr*I = 5.8Hr

2. The % error b/w the achieved LED current and the desired value with both LEDs in the circuit? What was the cause of this error?
I_led1=35%
I_LED2=1.5%

The reason is the yellow LED is actually rated at 10V but not 5V, thus the current value will be off from the theoretical value

3. Determine the circuit efficiency when the both LEDs are in the circuit

Efficiency = 44%


4. If we operated the battery  to 6V, the circuit will not operate due to not sufficient voltage to drive both LEDs


Bonus: the reason why we must need both LEDs be in circuit without on of the LED will cause shortage on one of the branch, then cuz the current goes to high to burn out the LED