How can I get absolute value of impedance?

In summary, the conversation discusses the measurement of real and imaginary parts of capacitance, as well as the impedance and admittance of a capacitor. The goal is to calculate the frequency dependent impedance per length for cables in order to perform more complex measurements. The concept of loss tangents and their effect on capacitors is also briefly mentioned. There is also clarification on using different equations for characterizing the characteristic impedance of coaxial cables.
  • #1
andreass
16
0
Here's the thing. I have 4,7 nF capacitor and AC: 1 kHz < f < 10 MHz.
I measure real and imaginary part of capacitance C*=C'+iC" (C'=C'(f) and C"=C"(f)).
I did 2 measurements - in first capacitor is directly attached to measuring device, in second I use 1 m coaxial cables to attach it.
I need to get impedance per length for cables.
1. How can I get absolute value of impedance??
1.) |C| = sqrt(C'^2+C"^2) and |Z| = 1/|C|/(2*pi*f) or
2.) Z = 1/(2*pi*f*C*) = (C"-iC')/(2*pi*f*(C'+C")) and |Z| = sqrt(C'^2+C"^2)/(2*pi*f*(C'+C"))

And then, I assume, I can subtract absolute value of |Z| for first and second measurement and I should get impedance per meter for my cables !??
 
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  • #2


andreass said:
Here's the thing. I have 4,7 nF capacitor and AC: 1 kHz < f < 10 MHz.
I measure real and imaginary part of capacitance C*=C'+iC" (C'=C'(f) and C"=C"(f)).
I did 2 measurements - in first capacitor is directly attached to measuring device, in second I use 1 m coaxial cables to attach it.
I need to get impedance per length for cables.
1. How can I get absolute value of impedance??
1.) |C| = sqrt(C'^2+C"^2) and |Z| = 1/|C|/(2*pi*f) or
2.) Z = 1/(2*pi*f*C*) = (C"-iC')/(2*pi*f*(C'+C")) and |Z| = sqrt(C'^2+C"^2)/(2*pi*f*(C'+C"))

And then, I assume, I can subtract absolute value of |Z| for first and second measurement and I should get impedance per meter for my cables !??

There's no such thing as a real and imaginary part of capacitance. An impedance can have real and imaginary parts, however.

What are you trying to do? Characterize the complex impedance over frequency for a capacitor? Or something about a coax cable? Can you be a bit more clear in your overall goal?
 
  • #3


berkeman said:
There's no such thing as a real and imaginary part of capacitance. An impedance can have real and imaginary parts, however.

What are you trying to do? Characterize the complex impedance over frequency for a capacitor? Or something about a coax cable? Can you be a bit more clear in your overall goal?
Ok, device measures C' and G depending of frequency.
Y = G + iB, where Y - admittance, G - conductance and B - susceptance
Y = i*2*pi*f*C* -> C* = C' - iC" = C' - iG/2/pi/f
And Z = 1/Y

So in some kind a way there is imaginary part. And both parts can be used for Cole-Cole diagram, but that is not important here.

Here my goal is to calculate frequency depending impedance per length for these cables, so I know it when I do more complex measurements.
 
  • #4


andreass said:
Ok, device measures C' and G depending of frequency.
Y = G + iB, where Y - admittance, G - conductance and B - susceptance
Y = i*2*pi*f*C* -> C* = C' - iC" = C' - iG/2/pi/f
And Z = 1/Y

So in some kind a way there is imaginary part. And both parts can be used for Cole-Cole diagram, but that is not important here.

Here my goal is to calculate frequency depending impedance per length for these cables, so I know it when I do more complex measurements.

I'm still not tracking what you are trying to express about the complex impedance (or admittance) of a capacitor, but whatever.

If your goal is to understand and characterize the Characteristic Impedance Zo of coax cable, you should be using a different set of equations:

http://en.wikipedia.org/wiki/Characteristic_impedance

You will be characterizing the inductance and capacitance per unit length, as well as the resistance and conductance per unit length.
 
  • #5


Here is a Spice model of a 4700 pF capacitor on the end of a 5 ns, 50 ohm cable. (1 meter coax with beta = 0.67 is 5 ns delay). The frequency sweep is from 1 MHz to 1 GHz. See thumbnail. At 1 MHz, the phase angle is 90 degrees (capacitive), and at 10 MHz it is only about 70 degrees (due to 5 ns coax cable). At 60 MHz iimpedance phase goes thru 0 degrees (real impedance).
Dielectrics can have loss tangents, due to imaginary part of permittivity. See
http://en.wikipedia.org/wiki/Loss_tangent
So capacitors can also be lossy.
 

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  • #6


Hi BobS.

Fascinating diagram but I'm not sure what it is measuring.

Is it measuring the voltage at the top of the 50 ohm resistor or at the top of the capacitor?

At 1 MHz I would expect the reactance of the capacitor to still be decreasing with frequency ( and = 1/(2 pi F C) ) and the short length of cable (1/300th of a wavelength) to have no effect except to add about 100 pF to the 4700 pF, which would also be negligible.

So I must be misunderstanding what is going on as your impedance seems to be increasing at 1 MHz.
 
  • #7


vk6kro said:
Hi BobS.

Fascinating diagram but I'm not sure what it is measuring.

Is it measuring the voltage at the top of the 50 ohm resistor or at the top of the capacitor?

At 1 MHz I would expect the reactance of the capacitor to still be decreasing with frequency ( and = 1/(2 pi F C) ) and the short length of cable (1/300th of a wavelength) to have no effect except to add about 100 pF to the 4700 pF, which would also be negligible.

So I must be misunderstanding what is going on as your impedance seems to be increasing at 1 MHz.
Here are two thumbnails. The second is a thumbnail of the two circuits I am measuring, one over the other. Both are terminated in 4700 pF. The top one has a zero length transmission line. The bottom one has a 1 meter (5 ns), 50 ohm (e.g., RG-8) transmission line. The voltage source has zero impedance, so the voltage measurements in the two circuits are completely decoupled. The 50 ohm series resistor is to absorb all reflections. The two measurement points are between the 50 ohm resistors and the transmission lines. In the two curves in the next thumbnail, the black curve represents the zero length transmission line, and the red represents the 5 ns delay line.
The frequency sweep in the plot extends from 1 MHz to 1 GHz. The black line represents the 4700 pF capacitor with the zero length delay line. The voltage across it drops monotonically from about 0.5 volts to zero. The phase remains capacitive. The red line represents a 4700 pF capacitor at the end of a 5 ns, 50 ohm delay line. The first zero amplitude point represents a series resonance with the inductance of the delay line and the capacitor. I have checked this by varying both the transmission line length and the capacitor value. The first max amplitude (open circuit) at 50 MHz is a quarter wave "impedance transformer" of the capacitance. The impedance at this point should be (50 ohms)2*jwC. The next zero at 100 MHz is the half-wave point which "moves" (transforms) the capacitor impedance to the measurement point. It is apparent even at 1 MHz, that there is a difference between the two curves, which could be due to the shunt capacitance of the 5 ns delay line. However, at about 5 MHZ, the red line changes from capacitive to inductive. It is very apparent that any measurement with a 5 ns delay line above several MHz is not trustworthy.
 

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  • #8


Hi Bob,
Thanks for clarifying that. I tried it with metre of 50 ohm coax and a 3700 pF capacitor (actual C...marked 4700 pF) and got similar results measuring total Z with an antenna analyser. I got a minimum at about 4 MHz when the impedance with transmission line started to rise again after falling since DC.

MHz...no TL...TL
1.....46....42
2.....28....23
3.....22....17
4.....17....14
5.....15....16
6.....14....18
7.....12....23
8.....11....26
9.....14....30
10....15...33
 
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Related to How can I get absolute value of impedance?

1. What is impedance?

Impedance is a measure of the opposition to the flow of an alternating current (AC) in a circuit. It is represented by the symbol Z and is measured in ohms (Ω).

2. Why is it important to calculate the absolute value of impedance?

Calculating the absolute value of impedance is important because it allows us to determine the total resistance of a circuit, which is necessary for understanding the behavior of the circuit and making accurate predictions about its performance.

3. How can I calculate the absolute value of impedance?

The absolute value of impedance can be calculated using the formula Z = √(R² + X²), where R is the resistance and X is the reactance (the imaginary part of the impedance). This formula applies to both series and parallel circuits.

4. What factors affect the absolute value of impedance?

The absolute value of impedance is affected by several factors, including the resistance and reactance of the individual components in the circuit, the frequency of the AC signal, and the circuit configuration (series or parallel).

5. How can I use the absolute value of impedance in practical applications?

The absolute value of impedance is used in a variety of practical applications, such as designing and analyzing electrical circuits, troubleshooting circuit problems, and selecting the appropriate components for a circuit to achieve a desired impedance value.

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