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cancelling inductance on a long wire pair

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Mattski

Fri Aug 20 2010, 08:23PM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

You can get an estimate of series inductance and shunt capacitance from a calculator like this:

Using it with guessed values I'm seeing ~9pF total capacitance of the wire with ~210nH inductance for 152ohm characteristic impedance.

The transmission line can be represented as a circuit made of pi-sections, and if sufficiently short in terms of wavelengths then one section may be enough:

A wavelength in your transmission line is lambda=c/sqrt(eps_r) where c = speed of light, eps_r is effective relative dielectric constant which is somewhere between air (1) and whatever the wire insulation is (2-4?). Using some guesses your line is about 0.15 wavelengths at 100kHz which is pretty short, so with one pi-section I think you'll be okay. There's a formula to check that at above Mathworks link. Basically what it means is ~4.5pF shunt capacitance at start of the wire with 210nH series inductance. The other 4.5pF is in parallel with your 5.6uF cap, so it is negligible.

So how do you fix this? Well where and how are you measuring voltage a current? Schematics would be dandy, much easier to help when we know how your system is set up, otherwise we're kind of shooting in the dark, like that 56k resistor which is important. A software fix sounds like a fine idea to me. Otherwise you're going to need to implement some time delay/phase shift with analog circuits so that the voltage and current measurements have the same shifts at the point they enter the ADC.

klugesmith

Fri Aug 20 2010, 08:51PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

IamSmooth wrote ...

radiotech wrote ...

and if you feed it through this 30 CM wire, you have a new problem,
As Klugesmitth said before "It will contribute a phase error directly proportional to frequency"

I know there is a problem. What does one do about it? Right now, I simply compensate for it in the software algorithm.

The "phase error proportional to frequency" was based on the constant delay of a terminated transmission line. When you specified the line length and working frequency, we saw that the transmission delay (about 1.7 ns) is negligible. 0.00017 cycles at 100 kHz (Mattski misplaced the decimal point).

It sounds like you are dealing with a RC lowpass filter, as radiotech pointed out, if there is a 56K resistor at the source end of the twisted pair. For that circuit, the phase delay increases from 0 to 90 degrees beween roughly 0.1x and 10x the corner frequency.

I figure your line capacitance is about 15 pF (1.7 ns / 110 ohms). With another 5 pF for receiver input capacitance, the corner frequency is [Rich whips out slide rule again] around 140 kHz, give or take an octave.

Practical remedy: reduce the value of your 56K series resistor by a factor of 10 or more, and/or move it from the near end to the far end of the twisted pair.

IamSmooth

Fri Aug 20 2010, 11:23PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

Klugesmith wrote ...

Practical remedy: reduce the value of your 56K series resistor by a factor of 10 or more, and/or move it from the near end to the far end of the twisted pair.

How does moving it to the far end help?

klugesmith

Sat Aug 21 2010, 01:23AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

IamSmooth wrote ...
How does moving it to the far end help?

If we correctly understand your circuit, the delay is due to capacitive loading on the downstream side of the 56K series resistor. With such a high impedance, it doesn't take many pF to be significant at 100 kHz.

If the driven end of twisted pair is connected directly across your 5.6 uF tank capacitor, the capacitor voltage will "instantly" and continuously appear at the far end of the twisted pair. There would still be a delay due to chip input capacitance downstream of the 56K resistor.

What is the 56K resistor for? If it is the big end of a voltage divider, then our corner frequency analysis has been wrong. Then R*C should be calculated using the Thevenin resistance of the voltage divider.

IamSmooth

Sat Aug 21 2010, 01:59AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

I've been at work all day so I was not able to post a photo until now. The 56k resistor is to make sure the current does not exceed the ratings of diode clamps further down. It also has to be large enough so it can handle the power.

The capacitor can see over 400-500v rms. The capacitive voltage is the source of the signal I am measuring. The diodes are 1n4148 and clamp at 15v and -15v. I forgot their wattage rating, but it is probably 500mW. The phase of the drive signal and the measured capacitor voltage are compared. I have tried different values for the resistor and I can clearly see that the lower the value of R the less of a phase error. Will moving the position of the resistor help?

Anyway, does the image help?

radiotech

Sat Aug 21 2010, 02:54AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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If you have 600 volts at 100 kHz then there is 1900 Amps flowing in the tank capacitor of 5 ufd.

Why dont you use 3 resistors 27K ---2K---27K Connect the twisted pair across the 2 K and the end free ends of the 27K resistors across
the 5ufd tank cap. Mount the three resistors at the tank capacitor end and connect a 2 K at the other end. The output voltage across the far end 2 K will be 11 volts with 600 volts input. No phase shift.
Use 5 watt resistors for the 27 K. (noninductive)

klugesmith

Sat Aug 21 2010, 03:34AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

IamSmooth wrote ...
I have tried different values for the resistor and I can clearly see that the lower the value of R the less of a phase error.

With R values in the 56K range, the phase error when you have a scope probe on the node will be significantly more than the phase error when it's not being probed!
Unless you are using a special low-C probe, or probe downstream of the receiving buffer or comparator.

IamSmooth

Sat Aug 21 2010, 03:42AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

I know how one resistor will introduce a phase shift with the capacitance. How does this voltage divider eliminate it?

Is it that instead of having the 56k resistor in quadrature we just have a 1k equivalent resistor?

radiotech

Sat Aug 21 2010, 05:39AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

The transmission line is balanced and operates at 1000 ohms. Read all the previuos posts to see why this wont cause you any phase shift.

The other thing is, from a strictly symmetrical view point, you had the microprocessor dancing on the end of a dodgy whip effectively earthing a high power floating loop. One oops and it might have perished. Also the 56 K resistor sytem draws about 6 watts off your work circuit. You're caught between a rock and a hard place in trying to ground the work coil because any fault current will light up your ground mesh with 100kHz current. How elegant would be a fibre optic path back to the processor !

IamSmooth

Sun Aug 22 2010, 12:33AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

Radiotech,

when I asked how the resistors got rid of the shift I thought it was somehow getting rid of the small shift imposed by the wire. What I see is what I thought would happen - the current through the resistor is in phase with the voltage, as expected, eliminating the entire phase shift of my 5.6uf capacitor. I need to 90 capacitor phase shift; it is the additional shift caused by the wire and resistors that is undesirable. It is hard for the software algorithm to know if a voltage value is to the right or left of 0 degrees. When it was 90 degrees, I knew to go up or down.

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