So in the case where V-in is a positive voltage,
that means that V-in is attempting to push current around the loop
in this direction from its positive terminal to its negative terminal.
However, the direction of the diode does not allow current to flow in
that direction.
Remember current can only flow from the anode to the cathode of the diode when
it's forward biased.
So when V-in is positive, current should flow in this direction.
However, it's prevented by the diode, the voltage of the cathode
is greater than the voltage of the anode, the diode is reversed biased, or off.
And we can model this circuit as I've drawn here,
with the diode replaced by an open circuit.
Now, in this case where V-in is negative, remember if V-in is negative we can think
of this side as the positive side, and this side as the negative side.
So with V-in negative it's attempting to push current around the loop in
this direction.
That direction is allowed by the direction of the diode.
This voltage is more positive than this voltage.
The diode is on and it can be replaced by a short circuit as I've drawn here.
So these two circuits in combination model the behavior of your original circuit.
This one applies when V-in is negative, this one applies when V-in is positive.
On this slide I've redrawn the two circuits.
Now, it may be apparent to you the relationship between the output and
input voltages for each of these two circuits.
But let's go ahead and
write some equations to determine those relationships.
Because thinking about circuits in this way will help you to analyze more
complicated circuits.
Now, in this circuit you can see that the output voltage is the voltage
between this node and this node.
So to get from this node to this node, I go up by a voltage V-in,
then I go down by 1IR drop by ohm's law across the resistor.
So, we can write that V-out is equal to, we go up from V-in, and