Lab 2: Introduction to Biasing

Introduction:
The purpose of this experiment is help students know how to analysis the DC circuit by finding the appropriate resistances. We have a 9V battery and want to build a circuit in which include 2 LEDs which have different voltage and current. The objective is to find values of each resistance to properly bias the 2 LEDs.

Procedure:
We built the circuit similar to the diagram

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Step 1:
Calculate the resistance for each LED:

R_LED1 = V/I = 5v/22.75mA = 219.78 Ω

R_LED2 = V/I = 2v/20mA = 100 Ω

Step 2: Perform Design Calculations:

IR1 = 22.75 mA	IR2 = 20 mA
VR1 = 4 V	VR2 = 7 V
R1 = 175.8 Ω	R2 = 350 Ω
PR1 = 0.09 W	PR2 = 0.14W

Step 3: figure out the two closest commercial values for R₁  and R₂ and insert to the circuit.

Color code	Nominal value	Measure Value	Wattage
yellow violet black black brown	470 Ω	460 Ω	1/8 W
Brown black black brown	1000 Ω	980 Ω	1/8 W

Step 4: insert the measurement and record the data:

Config	ILED1	VLED1	ILED2	VLED2	Isupply
1	8.6 ± 0.1 mA	4.96± 0.01 V	6.9± 0.1 mA	2.00 ± 0.01 V	15.7 ± 0.1 mA
2	15.7 ± 0.1 mA	4.97± 0.01 V	X	X	15.7 ± 0.1 mA
3	X	X	15.8± 0.1 mA	1.99± 0.01 V	15.8 ± 0.1 mA

Calculation
a)
The useful life of 9V battery is just 0.2A-hr. The useful capacity is 0.6-0.2 = 0.4 A-hr
According to the figure 2, the current goes through the batter is
I =  I₁ + I₂  = 22.75mA + 20 mA = 42.75 mA
A-hr = A*t , t = (A-hr)/A = (0.4A-hr) / (42.75 mA) = 9.357 hr.

b)
 I_experimental = 15.7 mA

 I_theoretical  = 42.75 mA

%error = (actual -theoretical)*100% / (theoretical) = (15.7 - 42.75)/42.75 = 63.2%

The reason of the error was that we could not get the exactly resistances that we needed, the uncertainty of the power supply and the power supply not rating as same as the theoretical.

c)
P_LED1 = V*I = (4.96 V)*(8.6 mA) = 42.66 mW
P_LED2 = V*I = (2 V)*(6.9 mA) = 13.8 mW
P_out =  42.66 + 13.8 = 56.46 mW
P_power = V*I = (9 V)*(15.7 mA)= 141.3 mW

efficiency = (P_out)/(P_power) = 56.46/141.3 = 40%

d)If we repeated the design with a 6V battery, the efficiency would go up. Because when we reduce the voltage of the battery, we need to reduce the biasing resistors R1 and R2 in order to get appropriate voltage across the LEDs, the lost power will be decreased, and the power out will stay the same. Therefore, a 6V battery will have more efficiency.
The theoretical value of battery voltage would provide the best efficiency is 5V battery. The lost power will be from the biasing resistor R2.
P_lost = (5V - 2V)*(20 mA)  = 60 mW
P_out = (5V)*(22.75 mA) + (2V)*(20mA)  = 153.75

efficiency  = (P_out)/(P_out+P_lost) = (153.75)/(153.75 + 60) = 71.9 %

Bonus:
Design the new circuit as the following picture:

The reason that we have to keep both LEDs in the circuit for the circuit work properly because the current and voltage across other LED will not be properly when one LED is off in the circuit. The LEDs are parallel, the sum resistance will be smaller than one of them alone. When one LED is off in the circuit, the total resistance of the circuit will be increase, the current will be decrease because the power from the power supply is always the same. When the current through one LED decrease, the voltage across it will also decrease (Ohm's laws : V= IR).






Doing the experiment:
The R₃ and R₄ were calculated to be 94 ohm and 150 ohm respectively. 
Obtain equipment and build the circuit.

Data:

Config	ILED1 (mA)	VLED1 (v)	ILED2 (mA)	VLED2 (V)	Isupply (mA)
1	12.64 +/- 0.01	4.93 +/- 0.01	12.50 +/- 0.01	2.08 +/- 0.01	26.22 +/- 0.01
2	18.2 +/- 0.1	2.20 +/- 0.01	X	X	18.2  +/- 0.1
3	X	X	18.1 +/- 0.1	2.18 +/- 0.01	12.8  +/- 0.1