Component Impedance Using Complex Maths

Here is a practical application of complex maths equations.

This is in fact a very useful technique that you can use to characterize components, or even an antenna, at predetermined frequencies.

If you've been tinkering with electronics you may be familiar with Resistors and Ohm's law.
i.e. R = V / I You may now be surprised to know this is all you need to solve for complex impedance's too! All impedance's are essentially complex, that is, they have a Real and an Imaginary part. In the case of a Resistor the imaginary (or reactance) is 0, correspondingly there is no phase difference between V and I, so we can leave them out.

A quick summary on complex numbers. Complex simply means that the number is made up of two parts, an real and an imaginary. There are two ways to represent complex numbers, for example in the figure above, a point could be defined by the Real and Imaginary values, such as where the yellow and blue lines meet. For example if the blue line were at 4 on the X axis, and 3 on the Y axis, this number would be 4 + 3i, i indicates that this is the imaginary part of this number. Another way to define the same point would be by the length (or amplitude) of the red line as well as what angle it makes with the horizontal. In the example above this would be 5 < 36.87.

Or a line with a length of 5 at an angle of 36.87 degrees.

In the equation above all of the parameters, R ,V and I can be thought of as having an imaginary part, when working with resistors this value is 0.

When working with inductors or capacitors, or when an phase difference can be measured (in degrees) between signals, the equation remains the same but the Imaginary part of the number must be included. Most scientific calculators make working with complex maths very easy, in this tutorial I will work thru an example on a Casio fx-9750GII.

First, a recap on the resistor voltage divider equation.

As per the figure -

The voltage at Y is current i multiplied by R2

i is voltage X divided by the sum of R1 and R2

When R2 is unknown we can measure the other values, X, Y, R1 and re-arrange the equation to solve for R2.

Scientific calculator

Signal generator

Oscilloscope

Let's assume we want to calculate the inductance of the Device Under Test (DUT) at 1MHz.

The signal generator is configured for a sinusoidal output of 5V at 1MHZ.

We are using 2k ohm resistors, and the oscilloscope channels are CH1 and CH2

We get the waveforms as shown in the figure. A phase shift can be seen and measured on the oscilloscope to be leading by 130ns. The amplitude is 3.4V. Note, the signal on CH1 should be 2.5V as it is taken at the output of the voltage divider, here it is shown as 5V for clarity, as this is the value we must also use in our calculations. i.e. 5V is the input voltage to the divider with the unknown component.

At 1MHz the period of the input signal is 1us.

130ns gives a ratio of 0.13. Or 13%. 13% of 360 is 46.6

The 5V signal is given an angle of 0.. as this is our input signal and phase shift is relative to it.

the 3.4V signal is given the angle of +46.6 (the + means it is leading, for a capacitor the angle would be negative).

Now we simply enter our measured values into the calculator.

R is 2k

V is 5 (EDIT - V is 5, later on in the equation is used X! result is exactly the same as i has X as 5 in my calculator)

Y is our measured voltage with the phase angle, this number is entered as a complex number, simply by specifying the angle as shown on the calculator screen

now the equation

(Y * R) / (X - Y)

is typed into the calculator, this is exactly the same equation that we use to solve resistor voltage dividers :)

The calculator yielded the result

18 + 1872i

The 18, is the real part of the impedance and it has an inductance of +1872 at 1MHz.

Which works out to 298uH as per the inductor impedance equation.

18 ohms is higher than the resistance that would be measured with a multimeter, this is because the multimeter measures resistance at DC. At 1MHz there is skin effect, in which the inner part of the conductor is bypassed by the current and it only flows on the outside of the copper, effectively decreasing the cross area of the conductor, and increasing it's resistance.