diodesplot
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.
Forward biased normal PN junction diode would act as a voltage regulator.so the voltage drop across the diode may be even 0.7V. when we decrease the load resistance(connected across the diode) the current drawn by the diode decrease and the voltage drop also reduces(when a lighter load is connected across the diode it would behave like a good regulator even when there are changes in the supply voltage this is known as LINE REGULATION;but when there is a change in the load it would not hold good LOAD REGULATION).
In other words the dynamic forward resistance of the forward biased diode would increase, if there is a reduction in the current flowing through the diode(the current flowing through the diode would reduce, when the load resistance is decreased)
To create an X/Y plot in Falstad's Java Circuit Simulator, do the following:
Note that you can only measure voltages on the 'Y' axis of an X/Y plot!
Best Answer
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.