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I'm trying to determine the source of approximately 300pf worth of parasitic capacitance in a planar coil project I'm working on. I'm trying to find out how my numerical predictions line up with the manufactured design. I've done an array of different sizes and, I see a shift in resonate frequency that corresponds to ~300pf across all of them

The example below has the following parameters
Layers: 2
Trace Width: 0.205mm
Trace Spacing: 0.152mm
Number of Turns: 10
Inner Diameter: 5.08mm
Outer Diameter: 12.548mm
Board Thickness: 1.13mm
Self Inductance: 0.994uH
Total Inductance: 2.99uH
Parallel C: 1nF
Predicted Resonate Frequency (w/1nF C): 2.899MHz
Measured Resonate Frequency: 2.534MHz

I'm fairly confident in the predicted numbers as they match what the TI Coil Designer outputs If its to be trusted. Its the measured resonant frequency that's got me. I measured the frequency with a Rigol DS1054 in two ways

  1. Apply a sine-wave swept across multiple frequencies and measured the input and output voltage.
  2. Apply a 1Khz square wave and measure the output oscillations

I can account for 13pF from the scope input and 13pF from the probe input. Additionally, from paper I can approximate the capacitance between the two layers as the two plates of a donut-shaped capacitor, but at best, that only accounts for an additional ~13pF, for a total of 39pF.

There's no metal under the surface that would affect it, there is no ground place to couple to. I'm missing something, but I can't think of what it is.

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What about interwinding capacitance? Meaning from adjacent traces in the coil? \$\endgroup\$
    – Aaron
    Sep 18, 2020 at 22:38
  • \$\begingroup\$ page 83 of this paper google.com/… shows somthin like that, but I dont know how I could model that. And even then would that make up the remaining 200 ish pF? \$\endgroup\$
    – Lpaulson
    Sep 18, 2020 at 22:45

1 Answer 1

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Using equation 2 from page 3 of this white paper on distributed capacitance It covers the 300pF you describe.

\$Cd=\LARGE\frac{(\frac{1}{2\pi(SFR)})^2}{L}\$

When \$SFR = 2.899MHz\$ Then \$Cd=1.005nF\$

At the predicted SFR, that is your 1nF parallel cap

When \$SFR = 2.534MHz\$ Then \$Cd=1.315nF\$

At the actual SFR freq there is an additional 300pF!

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  • \$\begingroup\$ This to me, looks like its talking about coupled capacitance through transformers. Your right that Cd = 1.315pF at 2.3534Mhz but that's just rearranging the $SRF = 1 / 2\pi \sqrt{LC}$ it dosen't tell when where the 300pf is coming from, unless I missed somthing. \$\endgroup\$
    – Lpaulson
    Sep 18, 2020 at 23:27
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    \$\begingroup\$ @Lpaulson Yeah,looks like it's bundling all the cap effects together. "The voltage difference between turns, between winding layers and between windings to core create these parasitic elements." \$\endgroup\$
    – Aaron
    Sep 19, 2020 at 0:03
  • \$\begingroup\$ Hmm sounds reasonable, but feels anti-climatic. \$\endgroup\$
    – Lpaulson
    Sep 19, 2020 at 0:17
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    \$\begingroup\$ @Lpaulson I still think it is the turn to turn capacitance, similar to the link you posted in your comment. \$\endgroup\$
    – Aaron
    Sep 19, 2020 at 5:13

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