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A nonlinear resistor has a bent/distorted \$I\$-\$V\$-curve. We can approximate its \$I\$-\$V\$-curve with a polynomial model such as:

$$I(V)=I_0 + \frac{1}{R_1}V + \frac{1}{R_2}V^2 + \frac{1}{R_3}V^3 + \cdots$$

If all the terms other than \$\frac{V}{R_1}\$ are zero, we end up with a linear resistor of value \$R_1\$.

Assume we measure this resistor in such an arrangement:

schematic

simulate this circuit – Schematic created using CircuitLab

The current through the nonlinear resistor is given by \$\frac{V_C}{R_C}\$ and the voltage across the nonlinear resistor is \$V_{NL}\$. As the resistor is nonlinear, both voltmeters will record harmonic distortion, even if the source is a pure sine and if \$R_C\$ is a perfectly linear resistor:

$$I_\text{res}(V)=I_0 + I_1 \sin(\omega_1 t+\phi_1) + I_2 \sin(\omega_2 t+\phi_2) + \cdots$$ $$V_\text{res}(V)=V_0 + V_1 \sin(\omega_1 t+\phi_1) + V_2 \sin(\omega_2 t+\phi_2) + \cdots$$

What is the relation between the complex Fourier components in the last two equations (e.g. \$I_1\angle\phi_1\$, \$I_2\angle\phi_2\$, ...) and the polynomial approximation of the \$I\$-\$V\$-curve using the coefficients \$R_1\$, \$R_2\$, ... ?

Clearly, there must be a relation as they describe the same phenomenon, namely the nonlinearity of the tested resistor.

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1 Answer 1

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Since \$V_C + V_{NL} = V_2\$, and \$V_2\$ is a sinusoid, hence

\$I_n R_C + V_n = 0\$ for \$n\ge 2 \$ and \$n = 0 \$

To find the relationship between the fourier coefficients and the coefficients of the taylor expansion of the non-linearity, if you consider a slightly simpler problem, that where the voltage across your non-linear resistor is a sinusoid and so the current is given by your Taylor series, then you use the Chebychev polynomials to express each power of the voltage in terms of harmonics:

\$cos^2x = 2 cos x - 1\$

\$cos^3x = 3/4 cos x - 1/4 cos3x\$

etc..

hence you can determine the contribution of each taylor term to each fourier term.

You should be able to extend this to your configuration.

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  • \$\begingroup\$ Did you leave the first two lines in your answer by accident after the edit ? If yes, could you explain them, because I can't make any sense of them. Also pleas note, that the polynomial approximation is generally not a Taylor expansion, i.e. the coefficients would be different from that of a Taylor expansion. Does this affect the validity of your statements ? \$\endgroup\$
    – tobalt
    Jul 11, 2023 at 6:37
  • \$\begingroup\$ The first two lines give the relationship between the \$V_n\$ and \$I_n\$, which is what I initially thought you were after. Also, if your expansion converges then I think it must be the taylor expansion as the taylor expansion is unique, if it doesn't converge then it's not much use, \$\endgroup\$
    – Tesla23
    Jul 11, 2023 at 7:36

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