Introduction of Biasing

Introduction
 To establish the correct voltage across and current through component so it operates properly, biasing is required. Biasing is technique used for parallel circuit so that we don't burn the LED in the circuit. In this lab, 2 LEDs with different ratings are connected in paralel.

Prediction:
R_LED1 = 219.8Ω           R_LED2 = 100Ω
I_R1 = 22.75 mA              I_R2 = 20 mA
V_R1 = 4 V                      V_R2 = 7 V
R1 = 175.8 Ω                    R2 = 350 Ω
P_R1 = 91 mW                 P_R2 = 140 mW

The following is the set up of the lab

Based on the measurement that has been made,
R1 = 176.3 Ω                    R2 = 355.7 Ω

We then configure the circuit in 3 configuration:
1) Both LEDs in the circuit
2) LED2 is removed
3) LED1 is removed

The data is as followed:

Config	ILED1 (mA)	VLED1 (V)	ILED2 (mA)	VLED2 (V)	ISupply (mA)
1	19.60 ± 0.05	2.26	14.58 ± 0.01	6.77	26.4
2	19.77 ± 0.01	2.283	X	X	73.4
3	X	X	14.60 ± 0.01	6.83	64.1

Questions

1. The time operation of battery = 0.2 A-hr / 0.0264 A = 7.58hr
2. Based on the discussion, LED 2 (yellow LED) is not viable for error calculation, because the LED itself is not the same spec as it is from the site of purchase. The theoretical value for LED 1 is 20 mA while the actual data comes out to be 19.60mA. The percent error of this is calculated to be 2.0%. The cause of this error is probably because the ammeter itself has an internal resistance which will; reduce the value of the current in the actual circuit. We also ignore the resistance of the cable in this experiment which actually reduce the current too.
3. P battery = 26.4mA * 9 V = 237.6 mW. P LED = 19.60 * 2.26 + 14.58 * 6.77 = 143.0 mW. Efficiency = 143/237.6 *100% = 60%
4. If we repeated using 6V battery, the efficiency will go up because by reducing the voltage of the battery, the power supply will decrease while the power used stays the same which means the ratio increase and efficiency goes up. The efficiency will be the greatest when 5V supply is used since the P_out = 5 * 42.75 = 153.75 mW. P_in = 214 mW. Efficiency = 153.75/214 *100% = 71.9%

Conclusion

     According to the data taken, it is quite agree with the theoretical value. Some of the error in the experiment happens because the yellow LED is not the same as the specification of it. We also ignore the internal resistance from the cable which creates the percent error that we have.