Chapter 7

Distributed Capacitance

Distributed Capacitance is the summation of the turn-to-turn shunt capacitance of the same winding on a core.

Schematic:
7a

Formulas and calculations:


To acquire an estimate of the distributed capacitance, it is necessary to determine the average distance between the wires on the core. This is achieved by calculating the wire distances of the wire on the ID of the core (l1) and the OD of the core (l2). Then use the average distance of these two figures.

7b

Calculate the effective circumferences of the OD and the ID of the coil to the center of the wire.
The cover is considered to correctly calculate the distance between wires. A pad is included
by multiplying the result by 0.95.

C1 = 0.95 (ID
core – dw) (cover/360)
C2 = 0.95 (OD
core + dw) (cover/360)

Knowing the length of each circumference, the wire diameter of all the wires on the winding is subtracted from this to leave the open distance between the wires. To determine the distance between two wires, this distance for both the OD and the ID is divided by the total number of turns minus 1.

l1 = [C1 - (Np) (dw)] / (Np – 1)

l
2 = [C2 - (Np) (dw)] / (Np – 1)


The average of these two results is now calculated.

Average distance = lt = (l1 + l2) / 2

Calculate the surface area of the wires facing each other on the coil.

The wire diameter of HPN wire is shown here:

Wire dia. = dw = 0.127602 – 7.507e-3 x + 1.546e-4 x2 – 1.107e-6 x3

The surface area of one half of the wire surface, that surface facing the other winding wire surface, for the length of the winding wire.

Aw = [(OD – ID) + 2H] dw ( 1.5708 ) (NP)

Calculate Capacitance:

Cd = [(0.224 K Aw ) / lt ]

EXAMPLE:

Specifications:

Core OD = 0.115 in. Core ID = 0.067 in. Core H = 0.095 in.
Wire diameter
= dw = 0.0043 in. (#39 HPN) Np = 26 turns
Coil cover = 410 degrees. K = 3.5

Calculate the circumferences of the OD and ID of the core with wire.

C1 = 0.95 (ID – dw) (π) (cover/360)

= 0.95 * (0.067 – 0.0043) * 3.14159 * (410/360) = 0.2129 in.

C2 = 0.95 (OD + dw) (π) (cover/360)

= 0.95 * (0.115 + 0.0043) * 3.14159 * (410/360) = 0.4052 in.

Calculate the distance between the wires:

l1 = [C1 – (Np * dw)] / (Np – 1)

= [ 0.2129 – ( 26 * 0.0043)] / ( 26 – 1) = 0.00403 in.

l2 = [C2 – (Np * dw)] / (Np – 1)

= [ 0.40502 – ( 26 * 0.0043 ) / ( 26 – 1 ) = 0.01172 in.

Calculate the average distance between wires:

lave = ( d1 + d2 ) / 2 = ( 0.00403 + 0.01172 ) / 2 = 0.00788 in.

Bifilar coil = 2 * d ave

Calculate the wire surface areas:

Aw = [ ( OD – ID ) + 2H ] dw ( π / 2 ) NP

= [( 0.115 – 0.067 ) + 2( 0.095 )] 0.0043(3.14159/2 ) 26 = 0.04189 in2

Calculate the turn-to-turn distributed capacitance:

Cp = [ 0.224 K Aw ] / l ave

= [ 0.224 (3.5 ) 0.04189 ( 25 )] / 0.00788 = 4.1681 pf.

The total distributed capacitance for the Bifilar coil:

Cd = 2 * Cp = 2 ( 4.1681 ) = 8.3362 pf.


Measuring Distributed Capacitance:

Because distributed capacity is difficult and very time consuming to accurately measure with conventional test instruments, it is best to estimate it by the following methods. Some waveform analyzers can produce a fairly accurate estimate of this parameter.

7c


The analyzer using the Impedance adaptor test fixture is properly zeroed at the point of connection to the coil by the standard instrument manufacturer’s instructions. Then the coil is connected to the fixture and the instrument is set to the proper capacitance test mode setting.

Conventional test instrument method:

7d

Test waveforms:


7e

Because the schematic consists of the distributed capacitance in parallel with the open circuit inductance and series leakage inductance, then the resonance of this circuit occurs when the impedance of the tank is at the minimum. The resonance is determined by the peaking of the current at the minimum impedance frequency. From the current waveform, there are two peaks present. The first is that of resonance with the OCL. This is broad and of lower amplitude, The second peak is at a higher frequency and is sharp and of higher amplitude. This is the one that the distributed capacitance is calculated from.
Knowing the value of the leakage inductance along with the frequency of the resonance, the distributed capacitance is then calculated using the following formula. The OCL, open circuit inductance, is a low frequency component and is not considered for the high frequency resonance created by the parasitics of leakage inductance and distributed capacitance. Thus we use the following formula.

7f


EMI_EDN