Problems with very simple KVL loop

Discussion in 'Homework Help' started by ihaveaquestion, Mar 24, 2010.

May 1, 2009
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2. jlcstrat Active Member

Jun 19, 2009
58
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I might be missing the question, but it looks like if you rearrange your terms to match the book equation it would come out. The book solves for Vd and you solved for Vdo.

3. ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Sorry, that's precisely my question... why doesn't it come out?

If you rearrange my terms you get Vd = Vdo - IR (not what the book gets)

4. jlcstrat Active Member

Jun 19, 2009
58
3
I think it's just the direction of your current.

5. jlcstrat Active Member

Jun 19, 2009
58
3
They're basing the direction off of the source. Which would make it a positive Vdo in the model.

6. ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Thanks jlcstrat...

Anyone else have some input?

7. Jony130 AAC Fanatic!

Feb 17, 2009
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This blue rectangle represents the model of the real diode.
So voltage on the "real" diode is equal:
Vd=Vdo+I*rd
And KVL
Vdd = I*R +Vdo + I*rd

8. rdj New Member

Mar 25, 2010
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If you will make the first equation (Id=4.26) from your point of view, is it same with the book or your will be "Id= -4.26"? if it is negative then your equation will the same with the book.

from my point of view:

assigning polarity to R (rd actually) made your equation different from the book.

notice the direction of the assigned current flow (Id), this serve as my reference:

when the direction of the assigned current flow is along with your loop direction use (+IR).

when the direction of the assigned current flow is against with your loop direction use (-IR).

I use the term assigned current flow because in some cases your have to really assigned the direction of the current flow.

If you get a value of current (I) equal to negative value, this means that the direction of the current is in reverse.

I hope this can help....