Discussion in 'Homework Help' started by dileepchacko, Feb 27, 2011.

1. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
Dear All

Can you please solve this problem, attached in this Thread. I tried using super position theorem, but there is an diode between the circuit, so that I couldnt apply super position theorem, Can you please help me how to approach this problem. Thanks

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2. ### jut Senior Member

Aug 25, 2007
224
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What are you suppose to solve for??

3. ### Heavydoody Active Member

Jul 31, 2009
140
11
Redraw the circuit for two mesh loops and replace the diode with its model (a .65 V independent voltage source with the positive at node one and the negative at node six).

Jul 31, 2009
140
11

5. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
But how to model the diode as 0.65V for this condition, there is no much current flowing through the diode, the diode may not completely forward biased, The current through the diode approximately 27uA, the diode drop may not be 0.65V in this condition. Thanks

6. ### Heavydoody Active Member

Jul 31, 2009
140
11
Which I suspect is the purpose of such an exercise. So, how do we model a diode which is not forward biased? Also, from a design perspective, if we wanted to have that diode forward biased, what could we change in the circuit to accomplish that goal?

Last edited: Mar 1, 2011

Dec 26, 2010
2,147
302
To determine if the diode is really forward biased or not, evaluate the voltage across the diode connections with the diode removed from the circuit. Does the voltage turn out to be a reverse bias?

Watch out for the sign of the 15V supply by the way. Isn't it -15V? Did you assume that when you got a diode current value of 27μA?

8. ### Georacer Moderator

Nov 25, 2009
5,173
1,284
As it concerns the reversed biased diode, I would just replace it with an open circuit. Leakage currents can be safely ignored.

For the forward biased diode, there are several levels of simplification.
The simplest one is to replace it with a short, with no voltage drop.
A bit more complex is to add a 0.7V DC source in series to it, with its polarity impeding the flow if the current towards the Ground.
Finally, if you have the diode model or some measurements on terminal voltage - diode current, you could add the internal resistance of the diode, in series as well.

Dec 26, 2010
2,147
302
In this particular case the ohmic resistance voltage drop should not be terribly big!

10. ### Heavydoody Active Member

Jul 31, 2009
140
11
This is the response I was hoping to elicit from the OP But, yeah, this is how I have always solved these.

11. ### Georacer Moderator

Nov 25, 2009
5,173
1,284
Oops, sorry for the spoiler. I just thought that it is more important to have a direction to work for, rather than stumble with trial and error. Doing the math and solving the equations is just as instructive in learning the solution of the problem.