# Plannar Inductor Unacounted Parasitic Capacitance

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.

simulate this circuit – Schematic created using CircuitLab

• What about interwinding capacitance? Meaning from adjacent traces in the coil? Sep 18, 2020 at 22:38
• 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? Sep 18, 2020 at 22:45

## 1 Answer

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!

• 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. Sep 18, 2020 at 23:27
• @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." Sep 19, 2020 at 0:03
• Hmm sounds reasonable, but feels anti-climatic. Sep 19, 2020 at 0:17
• @Lpaulson I still think it is the turn to turn capacitance, similar to the link you posted in your comment. Sep 19, 2020 at 5:13