Introduction to Biasing

In this lab experiment we powered two different LED's by a 9V source. The LED's were different in their rating one was rated at 2V and 20mA and the other 5V and 22.75mA. To meet the LED's specification we cannot simply put them in series or parallel since in both ways as explained by the manual they will burn out fast. One solution is to add resistors to the circuit to ensure these specifications are met.

From LED's ratings we can obtain their resistance.

R_LED1=219.8 Ω

R_LED2=100 Ω

so in the circuit if we replace the LED's with equivalent resistors we can determine R1 and R2 values easier.

 

since resistors only exist in discrete values we had to choose a combination of resistors in order to get as close as possible to desired values.
 For R1 we chose a 100, three 22 and a 10 Ω resistor, and for R2 a 220 and a 130 Ω resistor.
We measured the voltages across and currents through the LED's in 3 different configurations.
Configuration 1:Both LED's in the circuit
Configuration 2: Remove LED2, 9V supply
Configuration 3:  Remove LED1
Configuration 4:  Both LED's in the circuit, 6V supply
in all 3 configurations the current through the power supply was also measured. 

 In the table above the value of the current through LED2 is missing but it is 20.3 mA.

The reason for the relatively huge uncertainties is that I did the experiment twice and measured the values each time a few times.  

Assuming the useful life of a 9V battery to be 0.2A-hr, with both LED's in the circuit the battery will last 6.45 ± 0.85hr or 387 ± 51min.

Following is the error percentage between the theoretical and actual value:

 

 the huge error percentage for LED1 comes from several different places. One source of error could be due to loss of power to wires and that not every single element's value was close to its theoretical value. But this source of error applies to both LED's. The only thing uncommon between the two LED's is that the assumption for their ratings might be wrong. On different websites like ebay or amazon the forward voltage varies between 2.5-5V and current from 20-30 mA. This gives rise to a huge miscalculation in analysis of the circuit.

efficiency of the circuit is:

P_out=P_LED1+P_LED2=85.34±5 + 42.81±5 mW

P_in=P_Battery=279±5 mW

  

Efficiency of the circuit when the battery is 6V is:

P_out=P_LED1+P_LED2=25±5 + 41.2±5 mW = 66.2 mW

P_in=P_Battery=95.4±5 mW

As the voltage developed by the battery goes down the efficiency goes up since to deliver the needed specifications to the loads we need less resistance and eventually the most efficiency circuit is produced when the output voltage is 5V since in this case there is no need for any resistance in the circuit leg that contains the LED2. As a result all the power dissipated in that leg will only be due to the load.

In that case R1 is 0 and R2 is 150 ohm to meet the specifications and the efficiency becomes:

P_out=P_LED1+P_LED2=113.75 ±5 + 40±5 mW = 513 mW

P_in=P_Battery=213.75±5 mW

efficiency = 72%