negative capacitance ?

Discussion in 'General Electronics Chat' started by bast, Mar 30, 2004.

  1. bast

    Thread Starter New Member

    Mar 30, 2004
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    Hi
    I have measured the capacitance of a capacitor (1my F) and a resistor (1k ohm) in parallel on a HP impedance analyzer. I found that the capacitance went from pos. to neg. at 150000 Hz when performing a sweep!

    If someone has tried this and has an explanation of why this occurs, I would be very grateful.

    Søren
     
  2. Battousai

    Senior Member

    Nov 14, 2003
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    Are you talking about the impedance or the capacitance?

    In reality a capacitor has a lot of parasitics and is modeled as a resistor in series with a capacitor in parallel with a resistor and there are some inductors in there as well, etc. So I would imagine that at very high frequencies you may get some strange effects, but 150kHz is not very high at all.

    Anyway the impedance of a capacitor in parallel with a resistor is: 1/(1+jwRC) or a low-pass filter. At high frequencies the impedance should tends towards zero.

    Also do you know what this capacitor is made out of? Sometimes on-chip people use big transistors as capacitors - in that case it's possible for the capacitance to change (see CV curves), but I don't believe it's possible to become negative.
     
  3. mozikluv

    AAC Fanatic!

    Jan 22, 2004
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    :) hi,

    let me add a little, the parallel circuit will show only the resistance at DC and Z will fall as the freq. is increased. as what battousai said, at high freq. the Z will approach 0 ohms. but at certain intermediate freq. which is determined by fo = 1/2piRC the reactance of the cap will be equal to R and the Z will be1/2 of the resistor value. however this is not the case, the Z will only be about 70% of the R. this is where phase shift comes in and you can calculate that by using vector mathematics (in short trigonometry).

    the equation would be Z = the sq. rt. of (1/Rsqrd. + 1/Xcsqrd), however this equation will work only if capacitive reactance is equal to resistance. now in electrical engineering parlance a leading power factor here is involve because the current is leading the voltage by 90 degrees.

    the above equation i gave is applied to steady state signal and in fact this does not affect sound quality, phase shift in any normal audio system is not audible.

    hope this could enlighten a little :rolleyes:
     
  4. bast

    Thread Starter New Member

    Mar 30, 2004
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    0
    Hi
    Thanks for your replies….
    I know it is difficult to understand my question since my problem properly is related to the understanding of what the apparatus is doing to the data!
    I used a HP 4284A Precision LCR meter and the data I got was the complex capacitance, consisting of a real part and an imaginary part. The real part express the capacitance of the circuit and the imaginary part the resistance.
    I have attached the output file if someone should be interested :blink:

    Thanks...
     
  5. Mike M.

    Active Member

    Oct 9, 2007
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    I was completely baffled by the concept of negative capacitance so I went looking and found this thread. I just was testing the parameters of a Microwave Oven Transformer (MOT) and found this:

    PrimaryWindings:
    R= 0.35 Ohms
    L= 44.4 milli Henries
    C= -0.412 micro Farads

    SecondaryWindings:
    R=88.0 Ohms
    L=19.3 Henries
    C= 4.5 pico Farads

    What is up with the primary windings having a negative capacitance, especially with no source of electricity present (AC or DC)? What is negative capacitance anyways? That sounds as if the induced field is 180 degrees out of phase with the applied field, and larger than the applied field at that! What does that mean if such a thing actually exists? There may be some major implications to develop certain specialty circuits that are focused on that property!!!
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Give some details about how you made your measurements. What kind of instrument did you use to make the measurements? At what frequency were the measurements made? Was the secondary open-circuited when making the measurement on the primary, and similarly for the secondary measurement?
     
  7. Mike M.

    Active Member

    Oct 9, 2007
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    Primary and secondary both had nothing applied and were open circuits. There was no frequency because there was nothing to give a frequency. It was just a transformer sitting on my wood dining room table. As far as what exact instrument I used, I will have to get back to you on that because I am currently at work. I can tell you it was an LCR tester........the most accurate one I could find at that. I just put the alligator clips on the contact electrodes. I tried wiggling, squeezing, and swapping the leads but still got the negative result for capacitance on the primary.
     
  8. beenthere

    Retired Moderator

    Apr 20, 2004
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    A capacitor has a decreasing impedance with an increase in frequency. An inductor has an increasing impedance with increasing frequency, and is pretty much the definition of negative capacitance.

    The transformer you are looking at is a massive animal, designed for 50/60 Hz operation. By the time your sweep is up to 150,000 Hz, I'm not surprised to see anomalous results. At those frequencies, the presense of your body will casue stray capacitive results. Think what Xl is at that frequency.

    If you do the measurements at the designed frequency for the transformer, the results are going to make more sense.
     
  9. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    There would have to be a frequency, because LCR meters work by applying an AC voltage to the inductor (or capacitor or resistor) and measuring the current. If you can, use an oscilloscope to look at the voltage across the LCR meter clips and see what frequency it is. It is quite common for LCR meters to use 1 kHz to make the measurement. If that is what yours uses, then you could well get anomalous results when measuring a 60 Hz transformer. Microwave oven transformers are usually Constant Voltage transformers, and have a resonating capacitor as part of the assembly. This may also be giving unusual readings.
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The capacitor and its series inductance are at resonance at 150kHz.

    Ls=1/((2*pi*Fr)^2*C),

    where Ls is series inductance and Fr is the resonant frequency.

    Using 1mF and 150kHz, Ls is only 1.13nH, which is low and believable. The plot below is the impedance of the network, in amplitude and phase. Note that at 150kHz, the impedance bottoms out at 1 milliohm (the default series resistance of inductors in LTSpice), and then becomes inductive. This will look like negative capacitance to an impedance analyzer unless it has a choice of network topologies to choose from. I have used an HP LNA which had an impedance analyzer attachment, and it had several network topologies to choose from, including the circuit posted below.
    BTW, if you noticed that I plotted negative V/I, I did so because of the way LTSpice plots the direction (phase) of current. Without the minus sign, the phase would have gone from 90 degrees to 270 degrees, which is somewhat confusing.
     
  11. Mike M.

    Active Member

    Oct 9, 2007
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    Does anyone have a thory about my situation that has no frequency associated with it? I can totally understand the 150 KHz situation because EVERYTHING has Inductance and Capacitance and therefore will resonate at some frequency when a voltage is applied. Why would a capacitance measurement on a DEAD circuit using a "BK Precision 875B" LCR tester have a negative capacitance?
     
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Your situation DOES have a frequency associated with it. Your BK meter applies an AC voltage to the component you're testing, and that AC voltage has a frequency.

    If you connect the clip leads to an inductor (or a capacitor at a frequency above its series resonance, where it appears inductive) when the meter has been set to measure capacitance, the meter will show a negative number. Likewise, if you connect the meter to a capacitance when the meter has been set to measure inductance (try it), the number shown will be negative.

    This is because in an inductor, the voltage leads the current. If you connect the meter to a component in which the voltage leads the current and ask the meter to measure the capacitance, it will give you a negative number, because current leads voltage in a (positive) capacitor, but in a negative capacitor voltage leads current. So when the meter sees that the voltage is leading current, but you've asked it to measure capacitance, it returns a negative number because only in negative capacitors does voltage lead current.

    It's possible to synthesize a negative capacitor with an opamp circuit. Such a synthetic component will measure negative even when the meter is set to measure capacitance.

    But, normally, the only time you will get a negative number is when the meter is set to measure the opposite of what you actually have connected.
     
  13. billbehen

    Active Member

    May 10, 2006
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    I favor a stray inductance theory! Once Xc gets to be small at high frequencies, the phase would swing -180deg as lead and wire inductance kicks in.
     
  14. Mike M.

    Active Member

    Oct 9, 2007
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    That actually makes perfect sense and until now I hadn't had the knowledge of how the capacitance part of measurement worked. It sounds like it uses the same method as inductance but the sensed time-based reflected parameters of current and voltage are reversed for each scenario such as to give a positive reading, an overwhelming majority of the time, for the parameter being measured unless the system is more of the other parameter than the one you are testing for such as, capacitance of an inductor or inductance of a capacitor.

    So does that mean when measuring capacitance of an inductor that you would take its inductance value and subtract the capacitance value, if it is negative, to get the real capacitance value? Or likewise when measuring a capacitor, would you take the capacitance value and subtract the inductance value if it is negative? Do you know of any formula that, given a parameter to measure, whether it be an inductor or a capacitor, you can mathematically figure out what the TRUE L of a C is or the TRUE C of an L is, given that both parameters have been tested and are known?
     
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Now you're getting philosophical. The issue of what is the TRUE inductance of some component is open to interpretation.

    What your BK meter (and any LCR meter) does is apply some AC voltage to the component and measure the current. Then it does some airthmetic with the magnitude and phase angle of the measurement and comes up with a value for the inductance (if that's what you set the meter to measure) and any resistance that may be in series or shunt with the inductor.

    But, since every component has resistance, capacitance and inductance, with one of them usually dominant for whatever application it is intended to be used in, you may be able to find circumstances where the parasitic properties become dominant.

    For example, while the primary of a typical 60 Hz transformer may be predominantly inductive, its parasitic capacitance (which ALL inductors have) will resonate with the inductance at some frequency. If you set your meter to measure inductance, but the measurement frequency is above the resonance, then the component will become capacitive, and will show as a negative inductance, and should really be measured as a capacitance.

    Your BK meter probably only has one measurement frequency, and you can't change it. Here is a picture of an obsolete HP Impedance analyzer which could measure inductance, capacitance and resistance at any frequency from 5 Hz to 13 MHz. The instrument cost $30,000 when it was still available from HP.
    http://cgi.ebay.com/HP-4192A-LF-Imp...ryZ25421QQssPageNameZWDVWQQrdZ1QQcmdZViewItem

    Anyway, the point is that a component which is intended to be an inductor (or capacitor, or resistor) doesn't really have a TRUE value. It only has an apparent value at a given frequency. But it would be reasonable to measure it a frequency where the parasitics are negligible and call that the true value. That's why your BK meter probably makes it measurements at 1 kHz.

    If you measure a 60 Hz transformer at a sequence of increasing frequencies, you will find that the apparent inductance decreases steadily until at some frequency (the series resonance), it will become zero. Above that frequency, if the meter is still set to measure inductance, it will measure as a negative inductance. If the meter setting is changed to measure capacitance, then it will measure a positive capacitance.

    Finally, measuring the inductance and then switching the meter to measure capacitance and perhaps subtracting doesn't get you anything. The negative value you get when you set your meter to measure capacitance while it's hooked up to the microwave over transformer is just the same thing, just displayed in another way.

    What the meter does is to apply an AC voltage, measure the current and divide the voltage by the current to get the reactance (X) of the component. Then if the meter is set to measure inductance, it uses the formula L = X/(2*PI*f) to determine what value of inductance to display, where X is the measured reactance. If the meter is set to measure capacitance, then it uses the formula C = 1/(X*2*PI*f) to get the displayed value. Finally, it displays the value as positive or negative depending on whether the voltage leads or lags the current, taking into account whether it's set to measure inductance or capacitance.

    Also, a perfect inductor or capacitor will have a 90 degrees phase shift between current and voltage. But, if the component has any parasitic resistance (and it will), the phase angle will not be exactly 90 degrees. The meter takes that into account, and (typically) calculates the Q of inductors and dissipation factor of capacitors.

    So, if you put a capacitor and inductor in series, the positive reactance of the inductor (which varies with frequency) and negative reactance of the capacitor tend to cancel each other, and what is left is a reactance that is either positive or negative depending on which is larger at the measurement frequency. The meter then sees that resultant reactance and gives you an effective inductance (or capacitance) reading, depending on which one dominates. And, the one which dominates changes with frequency. If the resultant reactance is negative, and you have the meter set to measure inductance, the display will be negative. This means that you should change the meter to measure capacitance.

    You won't be able to tell how much of each you have by measuring the series circuit with an LCR meter, because the apparent value will vary with frequency, and more than one L and C can give the same reading at any given frequency.

    I've attached a .jpg showing the measured inductance of a small 60 Hz transformer versus frequency. You can see that at 20 Hz, it measures about 13.5 Henries, and the inductance decreases with frequency, until at about 4.5 kHz, it becomes zero. This is the frequency where the parasitic capacitance resonates with the inductance of the winding. At this frequency, the component becomes a pure resistance of about 112 K ohms. As the frequency continues to increase, it looks like a negative inductance. This is a clue that we should switch the meter to measure capacitance.
     
  16. Mike M.

    Active Member

    Oct 9, 2007
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    Dammit.........I should have seen it sooner. I am always trying to make things easier but this is one thing that is never going to be easy due to the variability of frequency affecting both components.

    That apparent pure resistance of 112 K ohms at 4.5 KHz is only because the 2 opposite phase angles have converged to zero so the system appears to be a resistor with zero phase shift when in fact it is still reactance with exactly cancelling opposite phase angles at the resonant frequency. If you know the initial DC resistance and subtract that from the resonant reactance, it should be possible to calculate both the true inductance and true capacitance of the circuit given the now known true reactance at resonant frequency. This would give 2 possible solutions (each containing a set of inductance and capacitance values), one set being more capacitive and the other set more inductive but if you are testing a coil of wire, the answer should be the one with the higher inductance and the reverse if you are testing a capacitor.

    I am going to have to play around with some equations for a bit though and get back to you on this. This pretty much boils down to the fact that a single frequency LCR tester is good for nothing when it comes to precision, or even being in the ball park within certain ranges!!!
     
  17. Mike M.

    Active Member

    Oct 9, 2007
    104
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    @ The Electrician

    Actually, I just realized that unless you get an impedence of the original resistance value or infinity at resonance, there are both a series tank as well as a parallel tank acting together, probably due to the 3 dimensional nature of the insulated wires on the transformer. Reactance at resonance cannot be used because it is supposed to be zero or infinity, depending on whether it is series or parallel.

    This is almost like the wave particle duality because everything contains both inductance and capacitance. They are inseparable and the only way to measure their exact value is to have them separated, otherwise at resonance you will have an infinite number of L&C values for solutions to the equation.

    Although..........your graph clearly shows a bottoming out of inductance at a certain frequency which might be able to be used to compare against the resonant frequency and the resultant delta combined with the area under the curve can be manipulated to find the answer.

    I can see that I will probably be working on this for weeks before I find a solution..............sheesh!!!

    And thinking about the area under the curve......If you take the area under the curve from zero to infinity for the inductance measurement and then compare that to the area under the curve from zero to infinity for the capacitance measurement to obtain a ratio, that may actually produce a coefficient that can be applied to either L or C at resonance by some, as yet to be determined equation, to find exact values for both............or it might need to be the ratio of the derivatives of the integrals from zero to infinity. By saying infinity, I mean the highest frequency that can be calculated. The higher the range of frequency, the more accurate the solution becomes.
     
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