Experiment: introduction to biasing

In this experiment, we connet two led in parallel and get the data of its voltage and currents. Then calculating the error with the theoretical value and actual value.

step 1:
calculate the theoretical value.
RLED1: V=5volt and 22.75mA, so it has 219.78ohm
RLED2: V=2volt and 20mA, so it has 100ohm
V power= 9.07V

Theoretical value	Current	Voltage	Resistance	power
R1	22.75mA	4V	175.82ohm	0.091w
R2	20mA	7V	350ohm	0.14w

The resistor didn't have the valur of 175.82 or 350ohm, so connecting in series to get a close value.
R1= 14.9+147.6+99=174.9 with the reisitance on the board, we get R1=172.9ohm
R2= 98+98+149.3=345.3 , but we use the multimeter  get R2=344.1 ohm

step 7:the actual data get from the circuit.

config	I LED1	V LED1	I LED2	V LED2	Isupply
1	14.6mA	6.5V	19.9mA	2.15V	34.5mA
2	14.5mA	6.5V	x	x	14.5mA
3	x	x	19.9mA	2.15V	19.8mA

The LED 1 seems not a 5V LED.
and the resist is 448.27 different with the theoretical value 219.78ohm
LED 2 resistance =108.04ohm

step 9:
a) (0.6A-0.2A) hr/ (22.75+20)mA= 9.357 hr
b) LED 1 error%= (14.6-22.75)/22.75*100=35.82%
    LED 2 error%= (19.9-20)/20*100=0.5%
 The LED 1 is not a 5V LED, the resistance is too large than the theoretical value.
 The  LED 2 is very close to the theoretical value.
c) LED  efficiency= [power out of led1(14.6mA*6.5V)+power out of led2(19.9mA*2.15V)]
  divided by power of supply(9.07V*34.5ma)= 44%
d) If we make the led1 to 5v(reduced the curreny pass through the LED1)
   The current flow to LED1 will be 5V/448.27ohm=11.15mA
   The current flow to LED2 will be 2V/100ohm=20mA
The power out = 5V*11.15mA+2V*20mA=0.09575
The total power= 6V*(11.15+20)mA=0.1869
efficiency= 0.09575/0.1869=51.23%
The efficiency increase.
When the power supply be the 2V or 5V, the efficiency will have highest value for not power lost on resistors except the resistance in the cable.