Experiment 2 complete

For experiment 2 I obtained a 1N4007 Si Diode and an LED. I identified the Anode and Cathode and measured both the LED and Diode in forward bias. The diode gave me a reading of .540V whilst the LED gave me a OL sign. I then measured both components in reverse bias and both components gave me a readout of OL.
I was able to identify the Anode and Cathode by a silver band printed onto the Cathode side of the Diode. The LED had a flat spot on the LED and the positive leg is longer.

I then obtained the data sheet for my 1N4007 Si Diode.

This chart has all the specifications of the 1N4007 diode that i will use in the following circuit.
From this chart i can see that the diode dissipates 2.5W at 25 degrees Celsius. But if i wanted to work out if the diode will be destroyed say for example in a hot day in Brisbane recorded at 42' Celsius, i would then need to do derate the device by 20mW for every degree above 25 as stated in the data sheet.

42 - 25 = 17. This is the number which will be derated. 17 x 20mW = 340mW or 0.340W, So the power dissipation at 42'C equals 2.5 - 0.3 = 2.2 mW. As 2.2 mW is less then the maximum amount of voltage being dissipated by the diode, the diode will not be destroyed.

With this knowledge of my 1N4007 diode i then constructed a circuit with a Vs 5v, 1K as R, and 1N4007 as D.

Using Ohms Law i calculated the amount of amps that will be flowing through the diode. 5V/1000 Ohms = 0.005A or 5mA. Then using my Ampmeter, I calculated the actual amount of amps flowing through the diode and recorded 4.7mA. This reading was as I expected as i had calculated the Diode would have a current of 5mA through it.

Then I was to measure the voltage drop across the Diode, but first I used my measurement which I calculated by subtracting Vd from the resistor which would be around 4.3V from the 5V power supply. Leaving a 0.7 Vd across the diode. When I measured the actual VD when connected to the circuit, I recorded .645V.

Using the data sheet above i obtained the maximum amount of current that could flow through the diode by looking at the Rectified current @ 75'C which is 1 Amp.

Now I calculated the maximum amount of Voltage that could flow through the diode before being destroyed by using the Power Law. Pd=VdxId. We already know that the maximum amount of Pd cant exceed 2.5W as listed in the datasheet, And that the resistor is 1K.

If we use 350V-0.7/1000 Ohms it equals 3.493A. 3.493 x 0.7 =2.44W. 350V is the maximum amount of voltage this diode can handle using a 1K resistor.

Replacing the Diode with an LED i calculated the current by using Ohms law 5V/1000 Ohms = 0.005 Amps or 5mA. I then measured the current flowing through the circuit and recorded 3.05mA.

From this I observed that when I replaced the diode with the LED, the current dropped. If the voltage was the same when I replaced the components and the current decreased, then the resistance has been increased. Therefore the LED has more resistance then the 1N4007 Diode.