Introduction to Biasing

In this lab we had to determine which circuit to use in order to light up two LEDs. We must establish the correct voltage across and the current through a component so it operates properly. We are to use a 9V battery, LED1 rated for 5V and 22.75 mA, LED2 rated for 2V and 20 mA, and resistors R1 and R2. We determined the Resistance of the LEDs to be RLED1 = 219.8 ohms and RLED2 = 100 ohms. Using KCby the resistors PR1 = 0.091 W and PR2 = 0.14 W. Resistors are only available in certain discrete values based on these values we chose R1 = 220 and R2 = 470 as they are closest to our values and will ensure that our LEDs are not destroyed. After constructing our experiment we determined the following values.

We determined the Resistance of the LEDs to be RLED1 = 219.8 ohms and RLED2 = 100 ohms. Using KCL, KVL, and Ohm's Law we determined IR1 = 22.75 mA IR2 = 20 mA VR1 = 4V VR2 = 7V R1 = 175.8 ohms R2 = 350 ohms. To ensure we do not destroy our equipment we determined the power consumed by the resistors PR1 = 0.091 W and PR2 = 0.14 W. Resistors are only available in certain discrete values based on these values we chose R1 = 220 and R2 = 470 as they are closest to our values and will ensure that our LEDs are not destroyed. After constructing our experiment we determined the following values.

a. I= ILED1 + ILED2 = 22.75 mA + 20 mA = 42.75 mA

.2 A -hr / .04275 A = 4.678 hr

b. (40.5 - 42.75) / 42.75 * 100 % = -5.3% error

The error in our experiment is attributed to the resistors not being exact to our theoretical resistors.

c. PLED1 = 9.1 mA * 4.73 V = 43.043 mW

PLED2 = 31.4 mA * 2.24 V = 70.336 mW

Pout = PLED1 + PLED2 = 113.379 mW

Pin = 9 V * 39.2 mA = 352.8 mW

113.379/352.8 * 100% = 32.1% efficiency