Introduction to Biasing

PROBLEM
    Suppose you want to use a 9V Alkaline battery to light to LEDs. The LEDS may be part of an instrument and are being used to indicate whether the device is ON or OFF. LED1 is rated for 5 V and 22.75 mA and LED2 is rated for 2V and 20 mA.

METHOD
    A circuit is to be constructed:

Voltage across LED1 is 5 V w/ current of 22.75 mA , across LED2 is 2 V with current of 20 mA

 We determine the theoretical resistances of LED1 and LED2 using Ohm's Law and the rated current and voltage values for each LED respectively.  RLED1 = 5/.02275=219.8 ohms, RLED2 = 2/.02=100 ohms. Next we determine the values for the bias resistors, R1 and R2 using the current values across each LED. R1=4/.02275=176 ohms, R2=7/.02=350 ohms. Finally, we determine the power dissipated by the resistors - PR1 - 91 W, PR2 - 140 W.


Resistor combinations: R1 = 176 ohms = (1)130 & (1)47 ohm resistors - measure - 176.3 ohms - 1/4 W
                                      R2 = 350 ohms = (1)360 ohm resistor - measure 355.7 ohms - 1/4 W

Data was recorded for three configurations: 1 - both LEDs in the circuit, 2- remove LED2, 3 remove LED1

DATA

Configuration	I LED 1	V LED 1	I LED 2	V LED 2	I Supply
1	14.67 mA	6.8 V	19.62 mA	2.26 V	26.4 mA
2	14.63 mA	6.83 V	x	x	73.3 mA
3	x	x	19.79 mA	2.28 V	64.1 mA.

CALCULATIONS

 a) Capacity of 9V Alkaline batter is approx. 0.6A-hr; however, voltage drops significantly as the battery gets to the "end" of it's life. So let's assume the "useful" life of the battery is 0.2A-hr for a 9V. With both LEDs in the circuit, how long can the circuit operate before the battery voltage goes too low?

Soln: 0.2A-hr = 02.A(3600s)=720  As= 720 C. Since, W=qV , we know there is initally 720(9) J= 6480 J of "useful" energy left in the battery. Additionally, P=W/t >> t= W/P   where P is the total power absorbed (Vi) by the LEDs. Pled1=0.0998 W, Pled2=0.04434 >> P=sum(p1+p2)=0.1441 W. Accordingly, t= 6480/0.1441 J/(J/s)=44956 s*(1hr/3600s)=12.48 hrs.

Answer: Approximately 12.5 hours


b)What was percent error between Achieved LED current and desired value with both LEDs in the circuit? What was the cause of the error?

%err=absval(act. - theo)/(theo) * 100% >>
LED1= 14.67-22.75/22.75*100 = 35.6%
LED2= 19.62-20/20*100 = 1.9%

The error is rather unreasonable. If we analyze our dataset using Ohm's law, we can see that the resistance of the system was different than what we expected. R1 is 463 ohms, which is greater than our Rled1 resistance of 219.8 ohms. Additionally, R2 is 115 ohms, which is again greater than our expected Rled2 resistance of 100 ohms. The error is likely due to the resistance of the cables, which reduced the current flow.

c) Determine the efficiency of the circuit with both LEDs in the circuit.
Pout is power consumed by LEDs, Pin is power supplied by battery
%Eff = Pout/Pin*100% >> 0.1441/0.238*100 = 60.5% efficiency

d) If we repeat the design with a 6V battery - changing the biasing resistors R1, R2, but WITHOUT CHANGING LED parameters - would efficiency go: up, down, or stay the same? What theoretical value of battery voltage would provide the "best" efficiency and what is the efficiency?