# Does (stray) capacitance depend on voltage?

While thinking about how to reduce stray capacitance on my pcb, I got triggered by the question:

Should the stray capacitance depend on the voltage difference?

Acccording to the equation of capacitance(parallel plates:)

$$C = \dfrac{\varepsilon \cdot A}{d}$$

Where $$\\varepsilon\$$ is the dielectric constant , A the effective area and d the distance.

So based on this equation the (stray) capacitance is entirely independent of the voltage.

To me this seems counterintuitive.

If I have a two traces on top of the ground plane (the two traces are far at a inifinite distance of each other, so lets discard trace to trace capacitance), where trace 1 is at 1000V and trace 2 is at 10V.

My intuitions tells me that the parasitic/stray capacitance will be higher between trace 1 and ground compared to trace 2 and ground, basically because the electric field strength is higher.

Is my intuition right?

Should the stray capacitance be dependent on the voltage difference?

No, it shouldn't; it's a physical property related to physical dimensions and the electric permittivity of the materials.

The electric field being higher is immaterial.

capacitance is entirely independent of the voltage

Correct with one caveat - certain parasitic capacitances associated with reverse biased PN junctions will vary their capacitance with applied voltage but, this is because the depletion layer grows and, in effect, the gap between the "plates" increases thus lowering capacitance.

• okay, so there are no other parasitic effect due to the higher voltage in one trace? Commented Mar 26, 2019 at 12:29
• Correct, no effect because neither the dimensions change nor the material electric permittivity. Commented Mar 26, 2019 at 12:51
• Depending on the exact technology, ceramic capacitors may lose as much as 80% of their (differential) capacity at their rated maximum voltage. But this nonlinearity pretty much universally runs opposite to the OP's "intuition" in that the capacitance decreases rather than increases with higher voltage. Commented Jun 10, 2023 at 14:30
• @user107063 the question is about stray capacitance. Commented Jun 10, 2023 at 14:34
• @Andyaka I don't consider "capacitances associated with reverse biased PN junctions" to be "stray capacitance" either: "stray" to me refers to capacitances due to fields outside of the device proper. Commented Jun 10, 2023 at 14:41

Normally the capacitance is independent of voltage. It's a function of geometry and material properties - as can be seen in the equation for a parallel plate capacitor that you cited yourself:

$$C = \varepsilon_{r}\varepsilon_{0} \frac{A}{d}$$

The capacitance can only depend on voltage, if the relative permittivity depends on electric field strength. Such materials do not only exist, but are quite common. In most cases the effect is still negligibly small.

One example of a material for which the effect is highly relevant is liquid crystals. These have two important properties: 1.) They are liquid. 2.) They are dielectrically anisotropous. This means that you can control the effective capacitance by changing the orientation of the molecules by applying an electric field.

Remark: Sometimes it's the geometry that changes with voltage (example: loudspeakers).

• "Effect is small" DC bias at rated voltage of class 2 ceramic capacitors will drop the capacitance by 50% or more , this is a critical issue for DC supply filtering capacitors . Not related to stray capacitance, but illustrates this can be significant when the permittivity of dielectric depends on field Commented Mar 26, 2019 at 15:01

No, the capacitance will remain constant for a fixed geometry.

What will change is the charge. Since $$\ Q = CV \$$ the total charge stored will vary in proportion to the voltage.