Plot ##i_A## vs ##v_A## characteristics for this nonlinear network

[zenterix]

Homework Statement

Plot the $i_A$ vs $v_A$ characteristics for the nonlinear network shown in Fig. 4.49. Assume the diode is ideal.

Relevant Equations

$i_D = I_s\left (e^{v_D/v_{TH}}-1\right )$ (Diode)

I am quite stumped here.

This is what I did so far.

1) The first thing I did was to forget about the terminals to the left and the diode, and to write the circuit as a Thevenin equivalent circuit. Then I put in the diode. However, the terminals we are interested in are lost when we do this.

2) I then tried using the node method as follows

Here is the v-i relationship for the diode

$i_D=I_S\left (e^{\frac{e_1-e_2}{v_{TH}}}=1\right )$

and here is KCL applied to the $e_1$ node

$i_A-i_D-\frac{e_1}{10^3}=0$

$i_A-i_D=\frac{v_A}{10^3}$

We also have the v-i relationship for the resistors

$\frac{e_2-1}{10^3}=i_D$

$\frac{-e_1}{10^3}=i_D$

I'm a bit stumped at this point.

 

[collinsmark]

Homework Statement: Plot the $i_A$ vs $v_A$ characteristics for the nonlinear network shown in Fig. 4.49. Assume the diode is ideal.
Relevant Equations: $i_D = I_s\left (e^{v_D/v_{TH}}-1\right )$ (Diode)

View attachment 337104
I am quite stumped here.

This is what I did so far.

1) The first thing I did was to forget about the terminals to the left and the diode, and to write the circuit as a Thevenin equivalent circuit. Then I put in the diode. However, the terminals we are interested in are lost when we do this.

2) I then tried using the node method as follows

View attachment 337105

Here is the v-i relationship for the diode

$i_D=I_S\left (e^{\frac{e_1-e_2}{v_{TH}}}=1\right )$

and here is KCL applied to the $e_1$ node

$i_A-i_D-\frac{e_1}{10^3}=0$

$i_A-i_D=\frac{v_A}{10^3}$

We also have the v-i relationship for the resistors

$\frac{e_2-1}{10^3}=i_D$

$\frac{-e_1}{10^3}=i_D$

I'm a bit stumped at this point.

Do you have any clarification about what is meant by the part of the problem statement, "assume the diode is ideal"?

I'm asking because if I was instructed to assume the diode was ideal, I would assume that e_1 = e_2 (labeled in red in your diagram), so long as the current was flowing in the forward direction through the diode (and the diode is an open circuit otherwise). It makes the math a lot easier.

Modeling the diode using i_D = I_S \left( e^{ \frac{e_1 - e_2}{v_{TH}}} - 1\right) is a somewhat/slightly idealized treatment of a practical diode, but it's not what I would call an "ideal diode."

 

[anuttarasammyak]

I observe that for $v_A < 1V$ you can forget about right hand side part of circuit with diode, resistor and battery, so the circuit is with only 1 k ohm resitor,
v_A= 1000\ \ i_A
Then you may proceed to the case $v_A > 1V$.　When v_A is large
v_A \rightarrow (500 + 0)\ \ i_A

 

[scottdave]

I am with @collinsmark about the "ideal diode" - it conducts when forward biased (like closing a switch)

 

[zenterix]

Do you have any clarification about what is meant by the part of the problem statement, "assume the diode is ideal"?

I'm asking because if I was instructed to assume the diode was ideal, I would assume that e_1 = e_2 (labeled in red in your diagram), so long as the current was flowing in the forward direction through the diode (and the diode is an open circuit otherwise). It makes the math a lot easier.

Modeling the diode using i_D = I_S \left( e^{ \frac{e_1 - e_2}{v_{TH}}} - 1\right) is a somewhat/slightly idealized treatment of a practical diode, but it's not what I would call an "ideal diode."

I think this is actually the reason I did not make progress on this problem. I had the wrong idea about "ideal diode".

Here is what it is according to the book:

The ideal diode model is a piecewise linear model for the diode. It approximates the v-i characteristics of a diode using two straight-line segments given by

Diode ON (short circuit): $v_D=0$ for all positive $i_D$

Diode OFF (open circuit): $i_D=0$ for all negative $v_D$

 

[zenterix]

Here is an update on my next attempt at this problem

I considered two cases

Case 1) $v_A>e_2$

When this happens then we can treat the diode as a short circuit. We can analyze the resulting linear circuit using techniques for linear circuits, such as Thevenin method.

Here are the calculations

The resulting relationship between $i_A$ and $v_A$ is $v_A=0.5+500i_A$.

Case 2: $v_A<e_2$

We can treat the diode as a open circuit and again use Thevenin method.

The result is $v_A=10^3i_A$.

Note that in the thick of things i used Thevenin but after the fact it is clear that that was unnecessary (as anuttarasammyak had previously shown).

Thus we have two different linear behaviors for this nonlinear "ideal diode model" diode.

Part b) of the problem inserts this non-linear two-terminal device in a circuit, as below

and asks for the load line.

It is a simple matter to obtain the load line.

$$\frac{v_T-3.5}{10^3}+i_T=0$$

and the load line is $\frac{3.5-v_T}{10^3}$.

It is not clear to me, assuming everything above is correct, how to decide which case we are in, ie case 1 or case 2 of the diode.

If we are in case 1 then

$$\frac{v_T-3.5}{10^3}+\frac{v_T-0.5}{500}=0$$

and $v_T=1.5V$V.

and if we are in case 2

$$\frac{v_T-3.5}{10^3}+10^{-3}v_T=0$$

and $v_T=1.75$V. This seems to indicate that this is not a possible solution for this circuit.

It seems to me that actually the threshold for the diode is 1V. For $v_A$ below 1V, the diode becomes an open circuit and $e_2=1$ and the entire NLD behaves like a linear resistor.

Once $v_A$ becomes $>1$, now the diode changes behavior and becomes a linear circuit element with v-i characteristic $v_A=0.5+500i_A$.

In the case of part b), for that circuit specifically, we have that $v_A$ (ie, $v_T$) goes to $1.5$V to balance out the equations.

The only weird thing to me is that I had to initially frame the cases in terms of the relationship of $v_A$ to $e_2$ but ultimately it seems that the cases are $v_A>1$ and $v_A<1$.

 

[anuttarasammyak]

The crossing point of these two lines seems to happen.

 

[collinsmark]

Here is an update on my next attempt at this problem

I considered two cases

Case 1) $v_A>e_2$
[...]
The resulting relationship between $i_A$ and $v_A$ is $v_A=0.5+500i_A$.

Case 2: $v_A<e_2$
[...]
The result is $v_A=10^3i_A$.

'Looks right to me.

However -- and this might be important later in the next step -- the way the problem statement was worded, v_A is the independent variable, and i_A is the dependent variable.

In other words, v_A is what you can choose to freely fiddle with, and i_A is what you measure/calculate.

Still in other words, express i_A in terms of v_A, not the other way around.

Make a plot with i_A on the vertical axis and v_A on the horizontal axis.

You'll need to do this anyway when you calculate the "load line."
https://en.wikipedia.org/wiki/Load_line_(electronics)

It is not clear to me, assuming everything above is correct, how to decide which case we are in, ie case 1 or case 2 of the diode.

If we are in case 1 then

$$\frac{v_T-3.5}{10^3}+\frac{v_T-0.5}{500}=0$$

and $v_T=1.5V$V.

and if we are in case 2

$$\frac{v_T-3.5}{10^3}+10^{-3}v_T=0$$

and $v_T=1.75$V. This seems to indicate that this is not a possible solution for this circuit.

??? (Your equations look OK, but I don't see how you came to your conclusion [in boldface].)

[Edit: Oh, wait. Are you saying that v_T = 1.75 V is not correct for the operating point for this circuit? Then yes, I can agree with that. At first I thought were saying that it has no solution at all, which is not true. It has a solution.]

Looking back at your original circuit, which value of v_T causes the ideal diode to transition from an open circuit to a short circuit?

Now, when doing your load-line circuit, you should be able to verify which mode you are in by solving for v_T and comparing it against that threshold.

Also, it might help to draw you load line on the same plot of your i_A vs. v_A. Then look for the operating point, where the two curves intersect (the diode curve and your load-line "curve").

=============== Edit: ==============

It seems to me that actually the threshold for the diode is 1V. For $v_A$ below 1V, the diode becomes an open circuit and $e_2=1$ and the entire NLD behaves like a linear resistor.

Once $v_A$ becomes $>1$, now the diode changes behavior and becomes a linear circuit element with v-i characteristic $v_A=0.5+500i_A$.

In the case of part b), for that circuit specifically, we have that $v_A$ (ie, $v_T$) goes to $1.5$V to balance out the equations.

The only weird thing to me is that I had to initially frame the cases in terms of the relationship of $v_A$ to $e_2$ but ultimately it seems that the cases are $v_A>1$ and $v_A<1$.

So yes, I think you're on the right track. If it helps, make the plot and look for the intersection between the diode curve and the load line "curve."