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Impedance Measurement
Sketch a circuit that can be used to measure the impedance of a passive component
(resistor, inductor or capacitor) as a function of frequency, f. Assume you have a signal
generator and a multimeter that can count frequency and measure AC voltages.
a
Rref
b
?
c
The mystery component is indicated by a question mark. The resistor above it is a
reference that is known (measured with a good multimeter). Points a, b and c are test
points, where you can put your multimeter (an oscilloscope is really better). It is
assumed that the reference resistor behaves as a pure resistor—no inductance or
capacitance. Such resistors are available as long as the frequency is not too high (they
are called non-inductive resistors, as inductance is the main nonideality for a resistor). It
is also assumed—perhaps wrongly!—that the signal generator will not vary its output
with frequency into this load. This will generally be true as long as the load on the signal
generator is not below the signal generator’s output impedance (typically 50 ohms). As
the lowest impedance that can be seen by the generator is the resistor itself, it’s a good
idea to test whether the resistor “loads down” the generator before beginning. Do this by
adding it and the unknown component to the circuit while watching Vac. If the signal
declines when the components are added, the signal generator is struggling to source the
load with enough current. If that happens, you have to choose a reference resistor of
higher value or follow the signal generator with an amplifier that can source more
current. Look for an amplifier with low output impedance (or, equivalently, high
damping factor).
Kirchoff’s laws require that Vac = Vab + Vbc. We can measure all of these at any
frequency supplied by the signal generator. To identify what kind of component the
unknown is, measure Vbc as a function of frequency. If it stays constant, the unknown
must be a resistor. If Vbc rises with frequency, we can assume that the component is an
inductor. If Vbc falls with frequency, the component is a capacitor.
Now, calculate the current in the reference resistor. This is i = Vab/Rref. Clearly, the same
current flows in any part of the circuit, so Zunknown = Vbc/i.
Then, if the component was a capacitor, we can solve for C from Zcapacitor= (2fC)-1.
If the component was an inductor, we can solve for L from Zinductor= 2fL.
Now, the PROBLEM WITH THIS is that you aren’t really dealing with simple numbers
in AC circuits. This, in turn, is because of the phase shifts between current and voltage in
the components that are not pure resistors. Impedance, current and voltage should all be
treated as complex numbers. The real component indicates the in-phase part. A resistor
has only this component, because current through the resistor and voltage across it are
always in phase. In inductors and capacitors, the imaginary component indicates the
phase angle between voltage and current. Current leads voltage by 90o in any capacitor,
but lags it by 90o in inductors.
For a great treatment of this, see:
1. www.allaboutcircuits.com
2. http://www.phys.unsw.edu.au/~jw/AC.html
In Ref. 1, you will see a “forward” analysis of the problem: given the components, what
is voltage, current and phase across each of them? The reverse analysis of the problem—
given just the frequency-dependent voltages, what must the components be?—is perhaps
more difficult. I will try it eventually. Meanwhile, we’re not without additional, simple
weapons to analyze the problem.
Suppose we have identified by frequency dependence of Vbc that the unknown component
is a capacitor. Then there are methods to estimate its value, independent of all the
complex number stuff. As the frequency goes lower and lower, the capacitor assumes
responsibility for balancing the voltage of the signal generator and current through the
resistor decreases. It reaches zero at f = 0 Hz (i.e., DC). You can monitor this current as
voltage across the resistor, Vab. Thus, Vab will decline as frequency is lowered. The
curve should look something like this:
Vab
fo
f/ Hz
The frequency fo at which the signal has declined significantly (we’ll not worry about
how significantly, but let’s say by about half it’s high-f value) gives C from Rref =
(2foC)-1. I think you could also feed the RC series combination with a pulse (or square
wave) and obtain the RC time constant from the exponential decay across Vab.