Wave-Shaping circuits (Urgent!!)

Discussion in 'Homework Help' started by winzer, Mar 25, 2008.

1. winzer Thread Starter New Member

Mar 25, 2008
7
0
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.

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2. winzer Thread Starter New Member

Mar 25, 2008
7
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So since the diode is ideal, it will act as a short right?!

3. studiot AAC Fanatic!

Nov 9, 2007
5,003
523
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.

4. florance001 New Member

May 18, 2009
1
0
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).

5. howartthou Active Member

Apr 18, 2009
96
0
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?