# Help Analyzing a Nonlinear Circuit

Thread Starter

#### Pulsed

Joined Sep 10, 2012
41
Dear all, I am stuck on a problem that requires me to analyze the circuit attached, which involves a nonlinear component.

The question asks:

"In the circuit Lem found this strange element that was connected to a power supply of 5V with a series resistance of Rs=4.7k and it was shunted with a parallel resistance Rp=8.2k"

"What is the current (in Amperes) through the strange object when it is in operation in the circuit?"

And

"What is the voltage (in Volts) across the strange object when it is in operation in the circuit?"

This is my attempt:

The voltage at node e is a simple voltage divider, therefore Ve = (Rp*V)/(Rp+Rs) = 3.17v

Therefore Current through the nonlinear element is equal to current through Rs MINUS Current through Rp.

Current through Rs = (Ve-V)/Rs = -0.0003, Current through Rp = 0.0006.

0.0003-0.0006=-0.0009A Through the Nonlinear element??

How do you analyze this circuit?

Many thanks in advance for the time and effort you take to help me and all those with questions, it's really appreciated!!

The answers are 0.000797045101089A and 0.797045101089V respectively, but I don't understand.

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Thread Starter

#### Pulsed

Joined Sep 10, 2012
41
I think I understand this question better now... it is a question of graphing...

The question was double barreled and had a graph containing the nonlinear V I characteristics of the unknown element.

So I have done the following: Created an equivalent Thévenin circuit, now the open circuit voltage is 3.178v and the shorted current is 1mA.

Plotting a RED line over the V I characteristics of the nonlinear element where the red line passes through 3.17829v on the V axis and 1ma on the mA axis, the coordinate where the two intercept should give me the current through the device and the voltage across the device given it's resistance at the point.

How do I go about putting this info into a graphing calculator?

How do I plot a line which passes through 3.17829 and 1?

I plan on just superimposing the red line on that first segment which is simply x=y as that is the line segment the red line passes through, and then reading off the values at that point.

I have GraphCalc. http://www.graphcalc.com/

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#### WBahn

Joined Mar 31, 2012
30,058
"What is the current (in Amperes) through the strange object when it is in operation in the circuit?"

And

"What is the voltage (in Volts) across the strange object when it is in operation in the circuit?"
If this is all your given, then the answer is basically, "What do you want them to be? Tell me and I will make that happen."

The voltage at node e is a simple voltage divider, therefore Ve = (Rp*V)/(Rp+Rs) = 3.17v
It's not a simple voltage divider. A simple voltage divider has two resistors in series. Rp and Rs are not in series, since there is this "strange object" in parallel with Rp.

Therefore Current through the nonlinear element is equal to current through Rs MINUS Current through Rp.
This much is certainly true. And notice how this emphasizes that Rs and Rp are NOT in series, since if they were they would have the SAME current.

How do you analyze this circuit?
You need to have something that gives you the current-voltage characteristic of the "strange object". You then analyze the circuit's behavior with the object removed and replaced by a linear load that you can vary. You analyze the circuit and plot the current through the load as a function of the voltage across the load. This plot is called a load line. You then overlay that on a plot of the current-vs-voltage characteristic for the non-linear device. Since, when the object is in the circuit instead of the load, the voltage across the object and the current through it has to simultaneously satisfy both load lines, the only points it can be are those where the lines intersect.

The answers are 0.000797045101089A and 0.797045101089V respectively, but I don't understand.
There's no point or value in giving answers with, supposedly, a dozen or so sig figs. It is hard to read and only adds no information (in fact, if anything it removes information). Unless something really argues for something else, it is usually sufficient to give answers to 3 sig figs (of which a leading 1, if you have it, is generally not considered significant).

#### WBahn

Joined Mar 31, 2012
30,058
I think I understand this question better now... it is a question of graphing...
I should have looked to see if there was a followup before I finished up and submitted. I started to replay this morning but had to go away and just got back to it.

How do I go about putting this info into a graphing calculator?
That probably depends on the graphing calculator, but why would you want to? You've already graphed it!

I plan on just superimposing the red line on that first segment which is simply x=y as that is the line segment the red line passes through, and then reading off the values at that point.
Except that y=x is nonsensical. It's like saying that the length of my desk is equal to the weight of my bed. Apples and oranges; y is a voltage and x is a current.

There are a couple of ways to find the intersection. First, you have two points on each load line that connect a straight line segment on each line that passes through the point of interest. Just use those points to determine the equations for the two lines and then solve for the point of intersection. Another is to note that the characteristic of the strange object, within the vicinity that matters, is indistinguishable from a simple resistor with a resistance of 1kΩ. So replace the nonlinear element with a 1kΩ resistor and analyze the resulting linear circuit.

Thread Starter

#### Pulsed

Joined Sep 10, 2012
41
Thank you for the time you have taken!

So I have done the following:

Converted the two point on the graph into an equation: y=-(3.178*x)+3.178

Then, using GraphCalc I superimposed this on the line "y=x" because this is linear like the resistor? (1,1 - 2,2 - 3,3)?

As you can see the point at which the two intersect are approx 0.76V and 0.00076A respectively?

"That probably depends on the graphing calculator, but why would you want to? You've already graphed it!".

The problem is I just used paintbox to draw a line by estimating where 3.178v is in the Y-axis for easy visualization. I did not come up with the answers 0.000797045101089A and 0.797045101089V, they were given to me.
From what I can see this is an awkward method for solving a circuit like this.

----

But let me see if I am correct in my methodology up until this point...

You have the original circuit in my first post which involves the nonlinear element.

(1) - You rip out that component and measure the Thévenin voltage, which is (8200r*5v)/(4700r+8200r)=3.178v .

(2) - You then calculate the Thévenin resistance which is (4700r*8200r)/(4700r+*8200r)=2987r .

(3) - Put these values into a Thévenin equivalent circuit (Vth = 3.178v and Rth = 2987r) .

(4) - Short the circuit at it's output and calculate the current though Rth. I=3.178v/2987r= appx 0.001A

(5) - Plot the V I characteristics on a graph and superimpose the curve on the Nonlinear characteristics of the element http://www.mathsisfun.com/algebra/line-equation-2points.html .

(6) - Read off the values of V and I at the point where the two curves intersect, this gives your Voltage across and current through the nonlinear element.

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#### WBahn

Joined Mar 31, 2012
30,058
Yes, I think you have a good handle on the concept.

But solving for it exactly, for this particular nonlinear element, is real easy because you can just treat it as being a linear element within the region of interest. From the graph, you know that

y = 1V/mA * x

for the nonlinear element. For the rest of the circuit, you have

y = 3.178V - (3.178V/1.06mA)x = 3.178V(1 - x/1.06mA)

So now you have two equations and two unknowns. Set them equal to eliminate y.

y = 1V/mA * x = 3.178V(1 - x/1.06mA)

x(1V/mA + 3.178V/1.06mA) = 3.178V

x = 3.178V / (1V/mA + 3.178V/1.06mA) = 0.759mA

y = 1V/mA * 0.759mA = 0.759V

the

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