# In a hole with circuits

Discussion in 'Homework Help' started by bob5429, Oct 31, 2007.

1. ### bob5429 Thread Starter New Member

Oct 31, 2007
2
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Hi all,

I am glad i found this forum, unfortunately I have a test due tomorrow that i need help with. The question says to "sketch Vx as a function of Vs for
(-10V<Vs<+10v). Then Show all break point values for Vx at Vs=-10v, Vs=0v Vs=+10v on the sketch"

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2. ### Distort10n Active Member

Dec 25, 2006
429
1
I could just give you the answer, or you could download one of the many flavors of SPICE to simulate the circuit for you.

I assume you know Kirchoff's Voltage Law, and you must also know that the source will be at its given voltage...say 10V. This forward biases D1, if the forward drop is assumed to b 0.7V then what is the voltage at the cathode of D1? This in turn forward biases D2...

3. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
First, remember what diodes do. What happens when Vs is negative? What happens when Vs is positive. What is the voltage drop across a diode when it is forward biased? (You've been given that one.)

4. ### bob5429 Thread Starter New Member

Oct 31, 2007
2
0
my problem is the teacher and our book don't explain how to work problems where a diode is parallel to the resistor.

5. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
When the diode is not foreward biased, current will not flow through it. Ignore it and solve for the resistor.

When a diode is forward biased, voltage across it will be constant - it's resistance changes in proportion to the current flowing through it.

6. ### Distort10n Active Member

Dec 25, 2006
429
1
Ideally the voltage will be constant yes. With more current flowing through the diode you will create a larger forward voltage drop. Not by much, but it does increase.

7. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Yes. And the diode will also conduct just a little before full forward bias is reached. There will also be leakage current.

Most tests at this level deal with "ideal" diodes, though.