Hello everyone!

This may seem like a stupid question but I need to make sure.

The problem states: Sketch the transfer characteristics (V(o) vs. V(in)) to scale for the circuit. The diode is assumed to be ideal.

So I am pretty sure the diode is gonna act as a short and the second resistor will only account for V(o)

The circuit schematic is attached.

 

So since the diode is ideal, it will act as a short right?!

 

The current /voltage characteristic of the ideal diode has three segments.

for v<0 it is the negative x axis. thus it can be replaced by an infinite resistor.

for v=0 it i=0

for v>0 it is the positive y axis thus it can be replaced by a zero resistance.

see the thick red line here.

http://www.facstaff.bucknell.edu/mastascu/eLessonsHtml/Diodes/Diode1.html

Despite you voltage source being marked + and - I suspect the examiner requires the effect of all three cases on the potential divider formed by the resistors.

Hope this helps.

 

Since the diode is forward biased and is connected across the resistor it will bypass the resistor and thus it will short the resistor across the diode. V(o) which is measured across the other resistor will be equal to V(in).

 

Florance001 I can't see how v(0) will equal v(in) on the "other" resistor. Surely there is a voltage drop accross the "other" resistor?