Introduction to Biasing

Purpose


The purpose of this lab is to design a proper bias circuit to light up two LEDs for preventing them from burning out.


List of materials

• 9 V power supply
• LED 1 (5V 22.75mA)
• LED 2 (2V 20mA)
• Resistors
• Multimeter
• BreadBoard

Process

1. We determine the value of resistor by using given current 22.75 mA and 5 V across LED 1, and 20 mA, 5 V across LED 2. R(LED 1)=5/22.5*1000 = 220 ohm, and R(LED 2)= 2/20*1000= 100 ohm.
2. Next decide each resistor R1$ R2 need to be used in series with each LED. R1= (9-5)/22.75*1000=176 ohm. R2= 7/20*1000=350 ohm.
3. In order to get those values of resistance, we have to use two 150 ohm and 100 ohm to get 175 phi, and two 100 ohm and one 150 ohm to get 350 ohm. So, we pick up three "Brown, Green, Brown Gold" resistors and three "Brown, Black, Brown, Gold" resistors. The actual value resistance are 151.1 ohm and 98.5 ohm, and their wattage are 1/4 W. 
4. Build the circuit as following Figure 1.

Figure 1. Biasing LED circuit

• I(LED1)=13.7mA, V(LED1)=6.55V, I(LED 2)= 19.3mA, V(LED2)=2.18V, I(supply)=3.29V

     5. Next we remove LED 2 from the circuit

Figure 2. Remove LED 2

•    I(LED1)=18.93mA, V(LED1)=6.56V, I(supply)=3.29V

     6.  Remove LED 1

Figure 3. Remove LED 1

• I(LED 2)= 19.25mA, V(LED2)=2.19V, I(supply)=3.29V

Calculation

1. Useful 9V battery life is 0.2 A-hr, so we want to calculate how long can the circuit operate before the battery goes to low. 

• I(total)=13.7+19.3=33mA=>0.2A*Hr=33mA*t  =>  t=6.1 hr

    2.  We figure out the percent error between LED current and the desired value for both LEDs.

• % error LED1 = (13.7-22.75)/22.75*100% = -39.8%
• %error LED 2 = (19.3-20)/20*100% = -3.5%

    3.  the power efficiency is pout/(Pin+pout)*100 % = 0.079/(0.079+0.297)*100% = 21.01%

    4. pin= 6 *(22.75+20)*10^-3= 0.255W, pout = 0.154W. power efficiency=    
        0.154/(0.154+0.225)=40.62 goes up.