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I have impedance data (\$|Z|\$ and \$\phi\$) of a diode, measured at a fixed frequency but depending on the applied bias. The diode can be modeled as a series resistance \$R_s\$ and a parallel circuit a capacitance \$C_j\$ and \$R_j\$ (related to the junction).

schematic

simulate this circuit – Schematic created using CircuitLab

It's a bit long but it's possible to write down the analytical expression for \$|Z|\$ and \$\phi\$ of this circuit, so that only \$R_s\$, \$C_j\$, \$R_j\$ and \$f\$ (frequency) appear in it. It seems to me there are three unknowns but only two parameters (\$|Z|\$ and \$\phi\$). Am I missing something or is it anyway possible to calculate all three elements from the data?

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  • \$\begingroup\$ When applied to a diode, your model has two components that are greatly affected by DC bias: Rj and Cj. Rs might be considered a fixed-value resistor. \$\endgroup\$
    – glen_geek
    Mar 13, 2020 at 15:28

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A complex impedance can in general be written as:

$$ Z = K_0 \frac{(s + z_1) (s + z_2) \cdots (s + z_n)}{(s + p_1) (s + p_2) \cdots (s + p_m)}, $$

where \$n\$ is the number of zeros and \$m\$ is the number of poles. The number of poles is typically the same as the number of reactive elements of the circuit (not always, but it often is), while the number of zeroes may vary between 0 and the number of poles.

So, if you have a circuit with, say, 4 reactances, you may have as much as 4 zeros and 4 poles, plus the DC term \$K_0\$. Therefore, we’re looking at 9 different terms that need to be determined (\$K_0, z_1, z_2, z_3, z_4, p_1, p_2, p_3, p_4\$).

You could make a Bode diagram out of \$Z\$ (which are two plots, one of the magnitude and the other of the phase of \$Z\$). If you know the rules for drawing this diagram, you could reverse-engineer it later and get these 9 different terms out of it (or, in general, even more). This is very easy to do when the diagram is drawn by hand using asymptotes, and even if you have measured data, you could use system identification tools to fit the best coefficients to it.

Perhaps what you’re really asking for is: if I only have two values (as if \$|Z|\$ and \$\phi\$ were two “points”), isn’t drawing a line through them the best I can do? And a line has only two coefficients, so how is it mathematically possible to get three coefficients out of it?

If this is really what you’re asking, think about a quadratic equation, \$y_2(x) = ax^2 + bx + c\$. Don’t you need three coefficients to represent it? You can take any three points \$(x_1,y_1),(x_2,y_2),(x_3,y_3)\$ and get the three coefficients out the curve. And what if it were a cubic instead, \$y_3(x) = ax^3 + bx^2 + cx + d\$? Again, with 4 points you can obtain the 4 original coefficients \$a,b,c,d\$.

The thing is, a curve is not just a point, of which you need two to make a straight line. A curve contains an infinite amount of points, and theoretically could be generated by equations with as many coefficients as desired. Assuming the equation that represents the curve has \$k\$ coefficients, you could always take a subset of \$k\$ out of the infinite number of points in the curve to determine these coefficients.

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