# Help Understand current

Discussion in 'General Electronics Chat' started by mguptamel, Dec 30, 2013.

1. ### mguptamel Thread Starter Member

Dec 14, 2012
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In attached circuit I want to know why there is 6.5mA of current flow in 2nd leg? Shouldn't all current flow through to leg 1 (only diode attached)?

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2. ### #12 Expert

Nov 30, 2010
16,704
7,354
It depends on the voltage per current curve of the diodes. Find out what part number they are or use this.

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3. ### mguptamel Thread Starter Member

Dec 14, 2012
37
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Is it forward characteristics graph? I am a newbie and using circuit simulator from Falstad. Don't know what type of diode it is. Assuming it's a general or common one.

4. ### #12 Expert

Nov 30, 2010
16,704
7,354
The 1N4000 series forward characteristics graph works for this setup and it confirms the results. The problem is that it is very hard to read because it is so small.

5. ### ronv AAC Fanatic!

Nov 12, 2008
3,403
1,477
Yes, the voltage drop across the first diode is higher because of the higher current.

6. ### mguptamel Thread Starter Member

Dec 14, 2012
37
0
How do I work out that how much current should be flowing on right leg? Help appreciated.

7. ### #12 Expert

Nov 30, 2010
16,704
7,354
Vd1 = Vd2 + 10 x I2

8. ### mguptamel Thread Starter Member

Dec 14, 2012
37
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Vd2 and I2 are 2 unknowns. Lets say Vd1 = 0.7V, then how do I get Vd2?

9. ### #12 Expert

Nov 30, 2010
16,704
7,354
Look at the graph to find out how much current is going through D1 at that voltage, then the rest of the current must be in the other diode and resistor.

Simple to say but difficult to do with any accuracy using such a poor graph as the manufacturer provides.

10. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
That is why they provide us with simulators. Here is an alternate simulation using a free downloadable simulator, LTSpice. In this case, the diodes are simplified, ideal diodes. The results match quite closely with Falstad. Not so much if I replace D1 and D2 with 1N4000 series diodes.

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11. ### mguptamel Thread Starter Member

Dec 14, 2012
37
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Thanks guys for helping me out. However, I am still confused as I don't know how to find current from graph because what forward voltage should I pick up from graph. Forward voltage could be anywhere less than 0.7, for example. Please correct me if I am going in wrong direction

12. ### crutschow Expert

Mar 14, 2008
13,505
3,376
The problem is that the diode voltage versus current is a rather complex non-linear equation so a solution cannot be easily directly calculated. Thus you need to do an iterative solution (which is what a simulator does). You try a calculation with assumed currents for each diode and then see if they match the diode voltage/current curve. If not then you keep adjusting the assumed currents until the diode voltage/current for each diode do match the curve. It's a rather tedious process which is why most of us use a simulator for that purpose.

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