Experiment No.2 Diodes n LED part 2

  Identifying a diode without using the multimeter: the only way is looking @ the sign on the cover. It often reads: 1N4007 and major black color, and there is a bit of a different color, so that's where the cathode is.
  Now a diode is to be wired in a circuit which has 1 resistor, 1 diode or 1 LED, and supply of 12V

                   
Knowing the specification for diodes is very important because when hooking up with Supply, it might destroy the component. In the data sheet for 1N4007 it shows: Average Rectified Current @ Absolute Maximum Rating of 75'C IS 1.0 Amps: this means that the operational Amperage which can safely flow through the diode is 1.0 Amps without overheating n eventually breaking the diode. This also explains why we have to put a resistor in series with the diode in order to regulate current flow in the whole circuit to lower than 1 Amp, for example: a 1K resistor. Using Ohms law, its easy to figure out that we shouldn't put a diode straight with the supply voltage because as R is minimum, I will be REALLY big and will certainly greater than 1Amp.
Next, our job is to figure out the Amperage actually flowing through the circuit:
      Since the actual resistance through a diode is inconsistent to determine, we have to resort on Ohms Law n series circuit law. In a series, current flow equally through every component. So, we need to find the voltage drop across the resistor. We already know 0.66 V of the diode in forward, otherwise we can measure it too. Obtain that, and subtract with the Supply we have the Vd of the resistor. And take that value divide by the resistance it self then we got the total Amperage. Both calculated(theoretical) and measured value are pretty much 0.005A, just as we expected. This is wayy below our limit of 1.0 A, so its safe.

By replacing an LED into the circuit, it makes no different than some light emitted and more Voltage Drop is drawn for the LED. The LED uses more Voltage to emit lights that is actually electromagnetic radiation, so to me it's reasonable that it's power dissipation is higher, hence requires more Vd, not necessarily because of resistance.

Again, the operation of a diode, can be comprehend by basic electron flow and conventional current. Conventional current says electricity flow from positive and negative, but what actually happens is electrons flow from negative to positive. If we apply the case of diode where there are P-N layers; holes that attracts electrons and vice versa, then the statement of conventional current is not entirely incorrect! Holes are actually atoms those are missing electrons, thereby positively charged. So the conventional flow is saying positive charges flow from positive to negative.