Find V-I characteristic for this circuit with ideal diode

[zenterix]

Homework Statement

For the circuit below, find the input characteristic, $i$ versus $v$, and the transfer characteristic $i_2$ versus $v$. $I$ is fixed and positive.

Relevant Equations

Express your results in graphs, labeling all slopes, intercepts, and coordinates of any break points.

This circuit has an ideal diode, which is modeled as

That is, when the potential difference across the diode is $\leq 0$ then we replace the diode with an open circuit and no current flows through it; when the potential difference is $\geq 0$ we replace the diode with a short circuit.

Initially, I drew the following picture

Suppose $v>e_2$. Then we have

KCL on the two top nodes gives us the equations

$$i=i_1+i_3=\frac{v}{R_1}+i_3$$

$$i_3+I=i_2=\frac{v}{R_2}$$

and putting them together we get

$$i=\frac{v}{R_1}+\frac{v}{R_2}-I=v\cdot R_1 || R_2-I$$

In addition,

$$i_2=\frac{v}{R_2}$$

Graphically,

Next consider the case in which $v<e_2$. We have

and so

$$i = i_1=\frac{v}{R_1}$$

$$i_2=I=\frac{e_2}{R_2}$$

$$i_2+i_3=I\implies i_3=0$$

Graphically,

I am not sure if I solved this problem correctly and there is no solution available in the book I am reading.

 

[zenterix]

The problem asks for THE input characteristic and THE transfer characteristic. In the analysis in the OP, there are two cases and thus a pair of each of these characteristics. Is this correct?

I'd like to add just a few more details of my calculations.

Consider the first case, in which $v>e_2$.

I've added the currents $i_4$ and $i_5$.

First of all, note what $i_3$ is graphically

It is just $i_2$ shifted down by $I$.

To compute $i_4$ and $i_5$ we apply KCL again to the bottom node.

$$i_1+i_5=i_4$$

$$i_2+i_4=I$$

$$\implies i_5=I-i_1-i_2=-i\ \ \ \text{and}\ \ \ i_4=I-i_2=-i_3$$

Graphically, putting all the currents together we have

 

[Gavran]

Suppose $v>e_2$. Then we have

View attachment 350198

KCL on the two top nodes gives us the equations

When the diode is replaced with a short circuit there is only one node on the top. See https://en.wikipedia.org/wiki/Node_(circuits).

$$i=\frac{v}{R_1}+\frac{v}{R_2}-I=v\cdot R_1 || R_2-I$$

The expression $v\cdot R_1||R_2-I$ does not follow Ohm’s law.