Hi again,
Yes the diode needs a positive drop (anode positive, cathode negative)
in order to conduct current. There are a few differences with different
diode models however. Here are three types:
['Vanode' is the voltage at the anode referenced to ground, and
'Vcathode' is the voltage at the cathode referenced to ground]
1. Ideal (perfect diode):
This diode conducts perfectly when Vanode>=Vcathode.
In other words, this diode is a short when Vanode>=Vcathode.
and there is no voltage drop.
This diode is a complete open circuit when Vanode<Vcathode.
We might call this diode the 'perfect' diode, but it's often called 'ideal'.
2. First approximation Ideal Diode (perfect diode with voltage drop):
This diode conducts when Vanode>=(Vcathode+vd) where vd is the
voltage drop of the diode when conducting, considered to be a constant
over all time and for all currents.
This diode does not conduct when Vanode<Vcathode.
Once conducting, the diode acts as an opposing voltage source who's
voltage equals vd, the characteristic voltage drop of the diode.
3. Ideal Diode with Exponential Response:
This diode uses the "Ideal diode equation" and that involves either an
exponential or a log to calculate.
This diode is like the first approximation diode, except that it has a voltage
drop that depends on a log and a current that depends on an exponential:
I=IS*(exp^(v/(N*Vt))-1)
V=N*Vt*ln(i/IS+1)
where N and IS are constants, and Vt is the 'thermal voltage' equal to k*T/q
which equals about 26mv at 27 degrees C.
This diode is used in more advanced studies though, and often includes a
series resistance Rs.
Now for these problems they are probably using either the perfect diode
or the perfect diode with voltage drop.
The two different diodes will make a difference when we go to analyze the
response to a unit step because the one that has a voltage drop (like 0.5
or 0.7) will act like a small battery in series opposing the source and so
that subtracts from the unit step to produce a smaller driving voltage step.
The diode that does not have a voltage drop (diode 1) will not subtract
anything from the driving source, so the unit step remains the same as
if no diode were present.
On the other hand, when we go to analyze the response to an impulse it
doesnt matter what diode we choose (1, 2, or 3 above) because the
voltage drop does not matter. Thus, we can short out the diode when we
analyze the response to an impulse, except that we also remove the short
after t=0+. Thus, the diode conducts only around t=0, after which the cap
charges up instantaneously and and when the impulse goes back to zero
the diode goes to an open circuit.