Current Across a 1st Approx. Diode

Discussion in 'Homework Help' started by Necrocell, Jan 15, 2013.

1. Necrocell Thread Starter New Member

Jan 14, 2013
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Regarding part (b)(i), due to Vd being 0, I am assuming Id is also zero using ohms law, but something about that seems too simple for 5 marks in an exam so I am guessing that I am wrong.

Can anyone care to elaborate?

Much appreciated.

2. t06afre AAC Fanatic!

May 11, 2009
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If Vd is zero. It will mean it act as short. And hence the current in the circuit Vs/R1.

3. Necrocell Thread Starter New Member

Jan 14, 2013
4
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Thanks for clearing that up.

4. mikeleeson New Member

Aug 22, 2012
26
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You have taken Ohm's Law, V= I * R, and assumed that if V=0 then I must be zero. In fact, it is R that is zero (in this first approximation) and so the current can be any value.
Try replacing the two diodes with a short circuit (first approximation) and see what you get when you work out the current.
Then ask yourself, "what can I replace the diodes with to give a voltage drop of 0.7V?". And then see what you get for the current.

5. Necrocell Thread Starter New Member

Jan 14, 2013
4
0
Thanks for the help.

Just to quickly clarify if possible, for part (ii) when its using the 2nd aproxx. Is Id 21.5mA?

My solution came from:

Vd = 0.7V
2x Vd = 1.4V

So the voltage drop across R2 will be the same since it is parallel.

Then the voltage drop across R1 must be 12V - 1.7V = 10.3V

Using KCL, I1 = I2 + Id

Then doing some quick Ohms Law I arrived at Id = 21.5mA

6. Necrocell Thread Starter New Member

Jan 14, 2013
4
0
Apparently 0.7 + 0.7 ≠ 1.7

Thanks for the help again. Ill be using this website alot from now on. Keep up the good work.

May 11, 2009
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