# Diode circuit I/V plot

simulate this circuit – Schematic created using CircuitLab

Hi,

I have started studying diodes and In one of the exercises of my book, I have faced a problem.The exercise wants a i/v characteristic, to be more specific it wants me to plot IR1 with respect to Vx. I plotted it, but the manual is suggesting that after Vx=1.8 or 1.7, IR1 becomes constant and equal to (.8/R1). Why does that happen?

P.S : The diode is ideal.

• You need to add a ground reference to be able to run the simulation in the CircuitLab. – tcrosley Apr 9 '14 at 12:43

An ideal diode is a voltage source when forward biased and an open circuit when reverse biased. The voltage source value is called $$\V_\gamma\$$ and is usually around 0.7~0.8V. If you use such a simplification you come out with this model.

Problem is that it is not quite right since a real diode when forward biased has a little but finite resistance. The next step is using the piecewise linear model (green line):

Now when the diode is forward biased you take in account a resistance. This model works for voltages near $$\V_\gamma\$$, if you try to use it with too large bias you'll have completely wrong results. What happens actually when you forward bias a diode with voltages enough larger than $$\V_\gamma\$$ is that the diode burns, so the model fails but that's not a real problem.

Back to your circuit now: I believe your book is introducing the piecewise model with that exercise: the diode characteristic is the one from the first image, and $$\R_2\$$ represents its resistance when it is forward biased, that is $$\\frac{1}{\text{derivative of the green slope}}\$$.

Let's solve the circuit now, starting with a completely ideal diode, i.e. $$\R_2=0\Omega\$$.

When $$\V_x = V_B + V_\gamma = 1.7V\$$ the voltage across the diode cannot change anymore (look at the first graph). You'll get three ideal voltage sources in a loop, that is a bad thing since you can't solve that kind of circuit but let's assume the diode is the "strongest" source: the voltage across it will be $$\V_\gamma\$$ no matter what happens in $$\V_x\$$ or $$\V_B\$$, so calculating the current flowing through $$\R_1\$$ is trivial: $$\I_{R1} = \frac{V_\gamma}{R_1} = \frac{0.8}{R_1}\$$, here is your book result.

What happens if $$\R_2\neq0\Omega\$$? Well, now the voltage across $$\R_1\$$ is $$\V_x-V_b\$$, so $$\I_{R1}=\frac{V_x-V_b}{R_1}\$$, and the diode does not really play a role in all this.

Finally, I think that the exercise is badly written or incomplete since the result it provides is WRONG in any cases. Maybe $$\V_x\$$ has an internal resistance?

Hope at least I helped a bit understanding diodes.

• This analysis appears to be incorrect because it doesn't account for R2 being non-zero. It would no doubt be totally the correct answer for a different question :-). Also, I don't see why 3 ideal voltage sources in series are unsolvable. – gwideman Apr 9 '14 at 9:56
• The final part of my answer starts with "what happens if R2 is not zero? I believe you missed it. If you don't see why they are unsolvable try to solve this circuit: gnd(-V1+)(-V2+)(-V3+)gnd where V1=V2=V3=whatever you want. By solving I mean find all voltages with respect to ground and the current flowing in the only loop present. – Vladimir Cravero Apr 9 '14 at 10:06
• Ah, I see what you're saying about the voltage sources... they are in a loop, not just in series. Fair enough. That said, I realize that you're thinking that D1 + R2 are intended to model a real diode. However, based on the book hinting that IR1 becomes a constant (and not mentioning R2=0), I think this is an exercise a bit less sophisticated than your idea, and it's just intending to demonstrate that a diode has a knee... and R1 in OP's schematic should be across just the diode. But, who knows!? – gwideman Apr 9 '14 at 10:11
• yes, I did not read your answer and I believe your idea that R1 is misconnected is right. And of course I meant in loop, I'm not very precise sometimes, i'll edit my answer – Vladimir Cravero Apr 9 '14 at 10:13

If the circuit you have drawn is correct and the voltage sources ideal, then:

$$I_{R1} = \dfrac{V_X - V_B}{R_1}$$

Regardless of the diode, because R2 will drop the voltage that the diode doesn't, and you don't care about the current on R2 since the voltage sources are ideal.

• Then I suppose there is a problem with my manual, I came to the same result as yours,and since the book said something else, I was confused. – kasra5004 Apr 9 '14 at 9:36

If the book says that, then it's based on a different location for R1, wired across D1. Ie: parallel to D1. The idea is that once the voltage across D1 exceeds the knee voltage of D1 (approx 0.8V), the voltage across R1 will no longer increase.

• Ie: If the book says that IR1 becomes constant... – gwideman Apr 9 '14 at 9:58