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

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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