Lab 2 : Introduction to Biasing.

3/4/2013

          Introduction: Applying "Biasing" technique to connect electronic components using lower voltage than the voltage supplied by the power source without burning the components.

step 1) Calculate the required baising resistors for each LED.
step 2) Build the circuit with meansuring devices connected to the circuit.
step 3) Measure the current and the voltage across different elements.
step 4) Calculate the effiency and errors in the circuit.

STEP 1

The biasing circuit consist of a 9V power supply , 2 biasing resistors, and 2 LED lights. (2V green LED, and 5V yellow LED)

 

STEP 2

The biasing resistors are approximate at 176 ohm (in series with the 5V LED), and 350 ohm (in series with the 2V LED).

The resistors available to us with the closest values are 220 ohm and 350 ohm.

 

STEP 3

Both LEDs are lit to the approximate same brightness after the voltages across them are "fixed" to their optimal voltages (2 and 5 volt).

 

Voltage across the 5V LED. Error occurs in every circuit. In this case, we are reading a higher voltage than it should be across the 5V LED.

Measuring the current in the branch with the 5V LED. Note tha the DMM is connected in opposite polarity.

 

All measured values for the elements in the circuit.

 

STEP  4

1) Assume that a 9V Alkaline battery has a useful life of 0.2A-hr, with both LEDs in the experiment circuit, how long can the circuit operate before the battery dies?

                                    

                                       0.2 A-hr = 33.64mA * t

                                                   t = 5.95 hr                   The circuit can operate for about 6 hrs.

 

2) Find the percent error between the achieved LED current, and the desired current.

 

                      5V LED:     (22.75-13.65)mA / 22.75mA * 100% = 40%

                      2V LED:     (20.5-20.0)mA / 20.0mA * 100% =2.5%

                

               The achieved LED current was so far off because the actual resistance in the resistors were different from the desired resistance values calculated from the desired current. Moreove, the inexpansive resistors used in the experiment were not within its designed tolerance of 1%.

 

3) Determine the circuit efficiency with both LED in the circuit.

 

            Pout / Pbattery = 0.1452/0.303 * 100% = 48%                Low efficiency.

 

4) Determine whether the efficiency would increase or not if a 6V battery is used instead of 9V. Also, determine what battery would be the most efficient in the circuit.

 

The efficiency increases to 59.9% by replacing the 9V battery with a 6V.

 

The best theoretical value of battery voltage for the circuit would be 5V, such that no resistor is need for the 5V LED but the only for the 2V LED. With fewer resistor and less resistance, the circuit wastes less power to the resistor.

 

 

Bouns!!!!!

 

Why must both LED be in this circuit for it to always work properly??

 

 

The circuit is acutally an example of voltage divider, the current supplied by the battery, Isupply , is increased by adding the parallel LED branch such that the potential drop in R3  will remain at 4V. By doing so, the voltage between the 5V LED and the parallel branch will be at 5V. If the parallel branch is removed, the potential drop in R3 will decrease, and the voltage throughthe 5V resistor will exceed its limited voltage.