I)

At the start the

diode is forward biased, this allows the capacitor to charge to the input peak.

However, when the input starts to decrease below its peak, the capacitor holds its

charge and the diode becomes reverse biased. At this point, the capacitor

starts discharging through the resistor at a set time constant. How much the capacitor

will discharge will depend on the time constant. The capacitor will discharge

more when the time constant is smaller, vice versa the capacitor will discharge

less if the time consent is larger. Then the diode becomes forward biased so

the capacitor charges back to peak of input. Here are a graphical illustration

to justify what I said:

The expected output voltage-time waveform is like a

sine wave. For the input voltage the wave goes up and peaks to a certain point

and then drops down with the same amplitude with a negative value below the x

axis. On the other hand, for the output voltage the wave will always be

positive. Therefor, it would never go below the x axis. The wave would have the

same time period and will bounce back up to the same value when it reaches to the

x axis. Here is a graph I produced by simulating the circuit on task 3 to show

this:

II)

To find the

forward voltage for the diode 1N4001 I figured out the best way would be to

find a data sheet for the following diode. I did a quick search on google which

provided me with a data sheet with diode characteristics graph. Since I know

that we are using 1.5 amp, I used diode characteristics graph to find out the

forward voltage for this diode by making a horizontal line on the graph at 1.5

amps and it give me around 0.95 volts. Therefore, the forward voltage for the

IN4001 diode is 0.95 volts.

Modifying my results using a constant-voltage diode

model I can justify my finding for the forward voltage of 0.95 volts by

simulation on the software Multisim.

I used the circuit that is provided on task 3 and

simulated it. I put 2 cursers on the peak value of the red and green graph. For

the peak value of the red graph I got 14.9608 and for the peak value of the

green graph I got 14.0120. I can the do 14.9608 – 14.0120 which gives me 0.9488

or 0.95 when rounded up. Therefor, the graph justifies my finding of 0.95 volts

for forward voltage.hhh