Chapter 5
Leakage Inductance
The leading parameter controlling the slope of the pulse rise time passing through a transformer is leakage inductance. The rise time is directly proportional to the leakage inductance magnitude by inspection of the rise time estimator equation:
The leakage inductance factor in the numerator is a magnitude greater than the distributed capacitance factor. The leakage inductance is typically in nanohenries (10-9) and distributed capacitance is in picofarads (10-12). Leakage inductance is a larger magnitude value. 
Leakage inductance is the remaining inductance of the coil after the open circuit inductance OCL is removed by shorting out the secondary winding and measuring the primary winding. This leaves the wire inductance and remaining lines of flux lost to the air by the poor coupling of the windings to each other.
Model:
Calculations
Leakage inductance, Ll, as it appears in relation to other transformer parameters is shown in the previous model. All secondary parameters are reflected back to the primary side by the turns ratio squared. Calculations are performed on the parameters as they appear in the primary side of the model.
Determine wire diameter:
Choose a wire diameter that should fit on the toroid in one layer with the desired number of turns for both windings. Calculate the wire diameter to perform the fit calculations. Formulas are given for different wire insulation thicknesses. Calculations can be run with the thickest insulation first to be assured of a good isolation voltage between the windings of the transformer.
HPN
dW = 0.127602 – 7.507e-3 (x) + 1.546e-4 (x2) –1.107e-6 (x3 )
TPN
dW = 0.1557005 – 9.699e-3 (x) + 2.1299e-4 (x2 )– 1.632e-6 (x3)
QPN
dW = 0.1622482 – 1.007e-2 (x) +2.2073e-4 (x2 )– 1.69e-6 (x3)
where: dw = wire diameter in inches
x = AWG = American Wire Gage size
Calculations here will be run using the HPN wire. Each of the insulation types should be run to check the fit of each on the core ID with the resulting leakage inductances.
EXAMPLE:
For #39 HPN wire, dw = 0.0043 inch
Determine wire inductance per inch of length:
For this size wire, calculate the inductance per inch.
where lW = 1 inch
EXAMPLE:
= = 0.0205 uh/inch
Determine length of wire per turn on a toroid:
L/T = (OD – ID) + 2H
Where:
EXAMPLE:
L/T = (0.115 – 0.067) + 2(0.095) = 0.238 inch
Where: OD = 0.115 in ID = 0.067 in H = 0.095 in
Determine inductance of winding wire:
This is calculated by multiplying the length of one turn around the core by the total number of turns on the winding and the inductance per one inch of wire.
Lw = (L/T) Np (LWIRE)
EXAMPLE:
LW = 0.238 (26) 0.0205 = 0.1267 uh.
Where: N = 26 turns L/T = 0.238 inch
LWIRE = 0.0205 uh/inch
Determine the wire cover on the toroid:
To achieve the lowest leakage inductance on a toroid, the wires of the windings must lie next to each other close to the core with no crossovers. This assures that the wire is very close to the ferrite and there is minimum chance of any lines of flux escaping to the air and not being linked to the core and the secondary winding. Therefore, we will calculate the maximum number of turns that will fit around the inside diameter of the toroid. The ID of the core is key to controlling the leakage inductance. Then, this number of turns is divided by the number of windings on the core to determine the maximum number of turns for each winding.
Insure that NP and NS fit in one layer:
Determine the circumference length that appears in the middle of the wire when it is on the core.
Circ. = (ID – dW)π
EXAMPLE:
Circ. = (0.067 – 0.0043)(3.14159) = 0.1969 in.
Then determine the turns per layer around this circumference.
Turns per layer = Circ. / dW
EXAMPLE:
Turns per layer = 0.198 / 0.0043 = 45.7 turns/layer.
The total winding turns around one layer of the core is:
Total winding turns = NP + NS
EXAMPLE:
Total turns on core = 26 + 26 = 52 total turns
The total turns for primary and secondary windings will exceed one layer.
Because the circuit requires a center-tap on the primary of the transformer, the primary has to be even turns so that each half of the winding is balanced. Thus, the 26 turns is divided by 2 to provide a 13 + 13 turn solution. To maintain the lowest leakage inductance of this coil, it is best to quadfilar wind the windings. Thus, the turns will be 13 turns quadfilar and then the two 13 turn windings will be connected forming a center-tapped 26 turn winding for the primary and also the secondary windings. In most cases, the secondary ceter-tap is just a splice.
Determine wire cover on toroid:
EXAMPLE:
Cover = (26 + 26) / 45.7 = 114% cover
To compensate for wire size and winding variations, it is preferred to achieve a 95% cover.
Determine the Traverse:
Traverse is the angle of cover that the coil wires make in relation to the 360 degrees of the full toroid.
Traverse = Cover (360)
Where the cover is 100%, then the traverse = 1.14 (360) = 410 degrees
This is more than one layer and crossovers are present. Perhaps reducing the wire size would help to provide a single layer of cover. This may improve the parasitic parameter and also produce a tighter standard deviation in production for a better statistical CPK value. Again, the small wire size must be considered for reliability.
Effect of cover on leakage inductance:
Secondary Leakage Inductance
for cover less than 360 degrees)
(for cover greater than 360 degrees)
EXAMPLE:
LP = 0.238 (26) 0.0205 (410/360) = 0.1445 uh
LS = 0.238 (26) 0.0205 (410/360) = 0.1445 uh
Total coil inductance:
LL(total) = LLp + LLsn2 (reflected back to the primary)
EXAMPLE:
LL(total) = 0.1445 + 0.1445 (1) = 0.2889 uh
Determine Service loop additions
To achieve the actual inductance of the wire as well as that in the winding, it is necessary to calculate the inductance in the excess wire coming from the coil to act as service loops.
LL(total) = LL(coil) + 2 (start + finish) LWIRE
EXAMPLE:
LL(total) = 0.2889 + 2 (0.500 + 0.500) 0.0205 = 0.3299 uh
Where: start = 0.500 inch finish = 0.500 inch
Total formula:
For cover less than 360 degrees:
or cover greater than 360 degrees:
