Introduction to Biasing

Objective: To determine the voltage and current necessary for electrical loads by using resistors to solve the problem of biasing.

Equipment:

• resistors (Ohms) : 100, 150, 220, 330, 470 //one each
• LED: 5v, 2v                                                         //one each
• 9v power supply                                              //one

Procedure:
Calculate the equivalent resistance of each LED using Ohm's Law with its rated voltage and current.

Given that the the current flowing through the LED is 22.75mA and that the rated voltage is 5v, using Ohm's Law gives,

V=IRled1

5=(.00275)Rled1

Rled1 = 219.78 Ohms

 The same calculation holds for led2

Rled2 = 100.00 Ohms

Given that this is a parallel circuit, the voltage across each branch is 9V. Calculate and determine the resistance needed to drop the voltage for each LED.

Look and measure some resistors needed for the circuit

Color Code          Nominal Value          Measured Value          Wattage

2200                      220 Ohm                    213 Ohm                         1/8 W

3600                      360 Ohm                    352 Ohm                         1/8 W

4700                      470 Ohm                    459 Ohm                         1/8 W

Build the circuit and record data for the following configuration

Configuration 1: Both LED in the circuit

Configuration 2: Remove LED2 from the circuit

Configuration 3: Remove LED3 from the circuit

Configuration         I-LED1          V-LED1          I-LED2          V-LED2          I-supply

1                                 10mA             5.9V               14mA            2.11mA          24mA

2                                 5.9mA            5.9V                   X                     X                 10mA

3                                      X                    X                  14mA            2.11mA         14mA

Bonus 

In this case, both LEDs need to be in the circuit otherwise there will be too much current following through a single LED.

Questions

• a.   If this Fig. 2 circuit used a 9V battery supplying 0.2A-hr, it lasts 0.2/(0.02275+0.02) = 4.68 hr 
• b.   The percent error for the LED1 current (0.014-0.02275)/0.02275=38.5%
• b.   The percent error for the LED2 current (0.010-0.020)/0.020*100 = 50.0%
• c.    Determine circuit efficiency 
• Pin=IV=9*0.04275=0.38475 W
• Pout= 5.9*0.010 + 2.11*0.024=0.059 + 0.051 = 0.110 W
• Efficiency = 0.11/0.38475 * 100 = 27.48% 

• d.   Efficiency would increase because less power would be used since V^2/R. Theoretically, a 5V power supply would be best because only one LED will need a resistor.