# Stuck on a simple piecewise linear element problem

See the schematic of interest on a picture. $$\V_S = 5 \,\rm{V}\$$, $$\R_S = 4700 \,\Omega\$$, $$\R_P = 5800 \,\Omega\$$. The cucumber-shaped element is a nonlinear element. The only thing we know about it is that it behaves like a nonlinear resistor (see voltage versus current relationship on the graph). My goal is to find out the element's $$\V_X\$$ (Volts) and $$\I_X\$$ (Amps) in this circuit. Here is why i am stuck:

I derived possible resistances Rx of the element from the graph above and then tried to apply a simple current divisor first with Rx = 1K to see if the Ix is within 1mA range and then with Rx = 2K and again check if the Ix is within 2mA range. This does not work

I don't understand whether or not i am wrong with such a logic.

edit:


We have only two possible values for RX=dV/dI. And the Thevenin of VS, RS, and RP is VTH=VS*RP/(RS+RP) and RTH=RP ∣∣ RS, with RX now in series with RTH. Plug in the two possible values for RX and you will find two boundary currents, both of which are well within one of the segments. You only have one alternative, now. The answer is captured by the boundaries. Then you get two possible ix currents ix = 0.76 when Rx = 1k (which is right) and ix = 0.6 when Rx = 2k. The reason 1k current turned out to be right is because the two boundaries both lie on the same segment. So that segment is the one that applies. It is convenient that this segment also crosses through 0.0

• Your approach seems reasonable so why didn't it work? I don't know because you haven't said and you also haven't said what the answer should be. Apr 7 '20 at 11:56
• In both ways it gives me an i value which is within [-1, 1] mA range. Moreover in both ways it is wrong @RodrigodeAzevedo. Apr 7 '20 at 12:31
• The linear relationship gives a current, $\small I_x<1mA$.
– Chu
Apr 7 '20 at 17:48
• @Chu i guess that the property of my non-linear devise is that it behaves exactly like a graph suggests. So that Ix can be > 1mA but this region corresponds to a different linear dependency Apr 7 '20 at 20:52
• @jonk How can you possibly plug in Rx when the element is affine, rather than linear? You cannot replace the element with a resistor. An extra voltage source is needed. Apr 8 '20 at 18:56

One possible way is to use a Thevenin's theorem and replace this circuit with the equivalent circuit that contains only a single RTH resistor and nonlinear element in series with the RTH resistor and Vth voltage source.

And use the load line to find the operation point.

The equivalent circuit will look like this:

Where:

$$V_{TH} = V_S \times \frac{R_P}{R_S + R_2} \approx 2.76V$$

$$R_{TH} =R_S||R_P \approx 2.6k\Omega$$

All this means that the current that flows through a nonlinear element cannot be greater than this: $$I = \frac{V_{TH}}{R_{TH}} \approx 1.06mA$$

• Would you please give me a clue on how to use a thevenin theorem in my case. I am a bit lost because of the non-linear thing in the middle. Apr 7 '20 at 12:36
• @NEOdinok I edit my answer
– G36
Apr 7 '20 at 12:47
• You are wrong, see the v-i relationship. The graph alone tells us that the current can be up to 2mA Apr 7 '20 at 14:46
• If you say so. You must be right.
– G36
Apr 7 '20 at 14:53
• @NEOdinok G36 tells you straight up, though he doesn't explain enough (technical English may take him time, though.) It's obvious G36's approach is the simplest way to the answer. And the answer then just falls out of the process if you'd apply it. Why don't you see how it is that the Thevenin equivalent of $R_P$ and $R_S$ and voltage source $V_S$ is a simplifying and appropriate method? Are you ignorant about its application in the simplest of circuits?
– jonk
Apr 8 '20 at 6:40

I think that a way of answering this question is to find the Thevenin voltage and resistance for the circuit shown below and then use a load line.

Having found the Thevenin voltage, V_th, and resistance, R_th, you can add a load line for the Thevenin equivalent circuit to the V_x vs I_x graph for the non-linear element.

The load line will have an intercept of V_th on the V_x axis, a gradient of -R_th and an intercept of V_th/R_th on the I_x axis.

From that you can decide where the intercept of the two graphs, $A$ or $B$, is and hence find the values of the voltage and current at that intercept.

Another way:
To find the position of the load line one can find the open circuit voltage across R_p (which is the Thevenin voltage) and the short circuit current, V_S/R_S,. These two values define the position of the load line.

• This is not the simplest method. G36 points that one out. (My opinion, of course.)
– jonk
Apr 8 '20 at 6:41
• @Jonk I take your point and have removed the word "simplest" as it is a matter of judgement. It is however an equivalent method. Apr 8 '20 at 6:53

You can use iteration to get the value. Start with an assumption that RC (R cucumber) is 1500 because 3V/2mA = 1500R The starting point doesn't matter much. Solve for RC and RP in parallel (RC||RP). Calculate VRPRC, the voltage across RP and RC in parallel. Then use that voltage to determine what RC would actually be at that voltage. The formaula, from the chart,

RCactual = IF(VRPRC<1,1000,2*VRPRC/(VRPRC+1)*1000)


The first iteration estimates RCactual at 1005R. So use that as the starting point for iteration 2. Iteration 2 finds that RCactual = 1000R. Use that as a starting point for iteration 3. Iteration 3 finds that RCactual =1000R. The assumption and the result agree. So RC = 1000. You now have the voltage across RC||RP and can find any voltage and current you need.