# Leakage inductance

(Redirected from Leakage flux)

Leakage inductance derives from the electrical property of an imperfectly-coupled transformer whereby each winding behaves as a self-inductance constant in series with the winding's respective ohmic resistance constant, these four winding constants also interacting with the transformer's mutual inductance constant. The winding self-inductance constant and associated leakage inductance is due to leakage flux not linking with all turns of each imperfectly-coupled winding.

The leakage flux alternately stores and discharges magnetic energy with each electrical cycle acting as an inductor in series with each of the primary and secondary circuits.

Leakage inductance depends on the geometry of the core and the windings. Voltage drop across the leakage reactance results in often undesirable supply regulation with varying transformer load. But it can also be useful for harmonic isolation (attenuating higher frequencies) of some loads.[1]

Although discussed exclusively in relation to transformers in this article, leakage inductance applies to any imperfectly-coupled magnetic circuit device including motors.[2]

The terms inductive coupling factor and inductive leakage factor are in this article as defined in International Electrotechnical Commission Electropedia's IEV-131-12-41, Inductive coupling factor and IEV-131-12-42, Inductive leakage factor.[3]

## Leakage inductance and inductive coupling factor

LPσand LSσ are primary and secondary leakage inductances

The magnetic circuit's flux that does not interlink both windings is the leakage flux corresponding to primary leakage inductance LPσ and secondary leakage inductance LSσ. These leakage inductances are defined in terms of transformer winding open-circuit inductances as well as the transformer's coupling coefficient or coupling factor k,[4][5] the primary open-circuit self-inductance being given by

${\displaystyle L_{oc}^{pri}=L_{P}=L_{P}^{\sigma }+L_{M}}$ ------ (Eq. 1.1)

where

${\displaystyle L_{P}^{\sigma }=L_{P}\cdot {(1-k)}}$
${\displaystyle L_{M}=L_{P}\cdot {k}}$

and

${\displaystyle L_{oc}^{pri}}$ = Primary inductance

${\displaystyle L_{P}}$ = Primary self-inductance
${\displaystyle L_{P}^{\sigma }}$ = Primary leakage inductance
${\displaystyle L_{M}}$ = Magnetizing inductance referred to the primary

Basic transformer inductance measurements

Transformer self-inductances ${\displaystyle L_{P}}$ & ${\displaystyle L_{S}}$ and mutual inductance M are, in additive and subtractive series connection of the two windings, given by,

in additive connection, ${\displaystyle L_{ser}^{+}=L_{P}+L_{S}+2M}$, and,
in subtractive connection, ${\displaystyle L_{ser}^{-}=L_{P}+L_{S}-2M}$
such that these transformer inductances can be determined from the following three equations:[6][7]
${\displaystyle L_{ser}^{+}-L_{ser}^{-}=4M}$
${\displaystyle L_{ser}^{+}+L_{ser}^{-}=2*(L_{P}+L_{S})}$
${\displaystyle L_{P}=a^{2}.L_{S}}$.

The Campbell bridge circuit can also be used to determine transformer self-inductances and mutual inductance using a variable standard mutual inductor pair for one of the bridge sides.[8][9]

Approximate leakage inductances can be obtained by creating a low-impedance short-circuit across each of the respective windings.[10]

It therefore follows that the transformer secondary open-circuit self, magnetizing and leakage inductances are given by

${\displaystyle L_{oc}^{sec}=L_{S}=L_{S}^{\sigma }+L_{M2}}$ ------ (Eq. 1.2)

Inductive coupling factor k is given as

${\displaystyle k=\left|M\right|/{\sqrt {L_{P}L_{S}}}}$, with 0 < k < 1 ------ (Eq. 1.3)

where

${\displaystyle L_{S}^{\sigma }=L_{S}\cdot {(1-k)}}$
${\displaystyle L_{M2}=L_{S}\cdot {k}}$

and

${\displaystyle L_{oc}^{sec}}$ = Secondary inductance
${\displaystyle L_{sc}^{sec}}$ = Secondary short-circuited inductance
${\displaystyle L_{S}}$ = Secondary self-inductance
${\displaystyle L_{S}^{\sigma }}$ = Secondary leakage inductance
${\displaystyle L_{M2}={\frac {L_{M}}{a^{2}}}}$ = Magnetizing inductance referred to the secondary

The electric validity of the above transformer diagram depends strictly on open circuit conditions for the respective winding inductances considered, more generalized circuit conditions being as developed in the next two sections.

## Inductive leakage factor and inductance

Nonideal transformer circuit diagram
Nonideal transformer equivalent circuit

A real linear two-winding transformer can be represented by two mutual inductance coupled circuit loops linking the transformer's five impedance constants as shown in the diagram at right, where,[4][11][12][13]

• M is mutual inductance
• LP & LS are primary and secondary winding self-inductances
• RP & RS are primary and secondary winding resistances
• Constants M, LP, LS, RP & RS are measurable at the transformer's terminals
• Coupling factor k is given as absolute value of coupling coefficient
${\displaystyle k=\left|M\right|/{\sqrt {L_{P}L_{S}}}}$, with 0 < k < 1 ------ (Eq. 2.1)
• Winding turns ratio a is in practice given as
${\displaystyle a=N_{P}/N_{S}=v_{P}/v_{S}=i_{S}/i_{P}={\sqrt {L_{P}/L_{S}}}}$ ------ (Eq. 2.2).[14]

The nonideal transformer's mesh equations can be expressed by the following voltage and flux linkage equations,[15]

${\displaystyle v_{P}=R_{P}i_{P}+{\frac {d\Psi {_{P}}}{dt}}}$ ------ (Eq. 2.3)
${\displaystyle v_{S}=-R_{S}i_{S}-{\frac {d\Psi {_{S}}}{dt}}}$ ------ (Eq. 2.4)
${\displaystyle \Psi _{P}=L_{P}i_{P}-Mi_{S}}$ ------ (Eq. 2.5)
${\displaystyle \Psi _{S}=L_{S}i_{S}-Mi_{P}}$ ------ (Eq. 2.6),
where
• dψ/dt is derivative of flux linkage with respect to time.

These equations can be developed to show that, neglecting associated winding resistances, the ratio of a winding circuit's inductances and currents with the other winding short circuited and at no-load is as follows,[16]

${\displaystyle \sigma =1-{\frac {M^{2}}{L_{P}L_{S}}}=1-k^{2}={\frac {L_{sc}}{L_{oc}}}={\frac {L_{sc}^{sec}}{L_{P}}}={\frac {L_{sc}^{pri}}{L_{S}}}={\frac {i_{oc}}{i_{sc}}}}$ ------ (Eq. 2.7),

where,

• σ is the inductive leakage factor or Heyland factor [17][18][19]
• ioc & isc are no-load and short circuit currents
• Loc & Lsc are no-load and short circuit inductances.
Nonideal transformer equivalent circuit in terms of coupling coefficient k

The transformer can thus be further defined in terms of the three inductance constants as follows,[20][21]

${\displaystyle L_{M}=a{M}}$ ------ (Eq. 2.8)
${\displaystyle L_{P}^{\sigma }=L_{P}-a{M}}$
${\displaystyle L_{S}^{\sigma }=L_{S}-{M}/a}$,

where,

• LM is magnetizing inductance, corresponding to magnetizing reactance XM
• LPσ & LSσ are primary & secondary leakage inductances, corresponding to primary & secondary leakage reactances XPσ & XSσ.

The transformer can be expressed more conveniently as the first shown equivalent circuit with secondary constants referred (i.e., with prime superscript notation) to the primary,[20][21]

${\displaystyle L_{S}^{\sigma \prime }=a^{2}L_{S}-aM}$
${\displaystyle R_{S}^{\prime }=a^{2}R_{S}}$
${\displaystyle V_{S}^{\prime }=aV_{S}}$
${\displaystyle I_{S}^{\prime }=I_{S}/a}$.
Simplified nonideal transformer equivalent circuit

Since

${\displaystyle k=M/{\sqrt {L_{P}L_{S}}}}$ ------ (Eq. 2.9)

and

${\displaystyle a={\sqrt {L_{P}/L_{S}}}}$ ------ (Eq. 2.10),

we have

${\displaystyle aM={\sqrt {L_{P}/L_{S}}}*k*{\sqrt {L_{P}L_{S}}}=kL_{P}}$ ------ (Eq. 2.11),

which allows expression as second shown equivalent circuit with winding leakage and magnetizing inductance constants as follows,[21]

${\displaystyle L_{P}^{\sigma }=L_{S}^{\sigma \prime }=L_{P}*(1-k)}$ ------ (Eq. 2.12)
${\displaystyle L_{M}=kL_{P}}$ ------ (Eq. 2.13).

The nonideal transformer can be simplified as shown in third equivalent circuit, with secondary constants referred to the primary and without ideal transformer isolation, where,

iM = iP - iS' ------ (Eq. 2.14)
iM is magnetizing current excited by flux ΦM that links both primary and secondary windings.
iS' is the secondary current referred to the primary side of the transformer.

Refined inductive leakage factor derivation

a. Per Eq. 2.1 & IEC IEV 131-12-41 inductive coupling factor k is given by

${\displaystyle k=\left|M\right|/{\sqrt {L_{P}L_{S}}}}$ --------------------- (Eq. 2.1)

b. Per Eq. 2.7 & IEC IEV 131-12-42 Inductive leakage factor ${\displaystyle \sigma }$ is given by

${\displaystyle \sigma =1-k^{2}=1-{\frac {M^{2}}{L_{P}L_{S}}}}$ ------ (Eq. 2.7) & (Eq. 3.7a)

c. ${\displaystyle {\frac {M^{2}}{L_{P}L_{S}}}}$ multiplied by ${\displaystyle {\frac {a^{2}}{a^{2}}}}$ gives

${\displaystyle \sigma =1-{\frac {a^{2}M^{2}}{L_{P}a^{2}L_{S}}}}$ ----------------- (Eq. 3.7b)

d. Per Eq. 2-8 & knowing that ${\displaystyle a^{2}L_{S}=L_{S}^{\prime }}$

${\displaystyle \sigma =1-{\frac {L_{M}^{2}}{L_{P}L_{S}^{\prime }}}}$ ---------------------- (Eq. 3.7c)

e. ${\displaystyle {\frac {L_{M}^{2}}{L_{P}L_{S}^{\prime }}}}$ multiplied by ${\displaystyle {\frac {L_{M}.L_{M}}{L_{M}^{2}}}}$ gives

${\displaystyle \sigma =1-{\frac {1}{{\frac {L_{P}}{L_{M}}}.{\frac {L_{S}^{\prime }}{L_{M}}}}}}$ ------------------ (Eq. 3.7d)

f. Per Eq. 2.14. Eq. 3.1, Eq. 3.2, Eq. 3.5 & Eq. 3.6

${\displaystyle \sigma =1-{\frac {1}{(1+\sigma _{P})(1+\sigma _{S})}}}$ --- (Eq.3.7e)

All equations in this article assume steady-state constant-frequency waveform conditions the k & ${\displaystyle \sigma }$ values of which are dimensionless, fixed, finite & positive but less than 1.

## Refined inductive leakage factor

Referring to the flux diagram below at left, primary driven and the secondary open-circuited, the following equation holds:[22]

Magnetizing and leakage flux in a magnetic circuit
σP = ΦPσM = LPσ/LM ------ (Eq. 3.1)

In the same way,

σS = ΦSσ'M = LSσ'/LM ------ (Eq. 3.2)

And therefore,

ΦP = ΦM + ΦPσ = ΦM + σPΦM = (1 + σPM ------ (Eq. 3.3)
ΦS' = ΦM + ΦSσ' = ΦM + σSΦM = (1 + σSM ------ (Eq. 3.4)
LP = LM + LPσ = LM + σPLM = (1 + σP)LM ------ (Eq. 3.5)
LS' = LM + LSσ' = LM + σSLM = (1 + σS)LM ------ (Eq. 3.6),

where

• σP is primary inductive leakage factor
• σS is secondary inductive leakage factor
• ΦM is mutual flux (main flux) / LM is magnetizing inductance
• ΦPσ is primary leakage flux / LPσ is primary leakage inductance
• ΦSσ is secondary leakage flux / LSσ is secondary leakage inductance

The leakage ratio σ can thus be refined in terms of the interrelationship of above winding-specific inductance and Inductive leakage factor equations as follows:[23]

${\displaystyle \sigma =1-k^{2}=1-{\frac {M^{2}}{L_{P}L_{S}}}=1-{\frac {a^{2}M^{2}}{L_{P}a^{2}L_{S}}}=1-{\frac {L_{M}^{2}}{L_{P}L_{S}^{\prime }}}=1-{\frac {1}{{\frac {L_{P}}{L_{M}}}.{\frac {L_{S}^{\prime }}{L_{M}}}}}=1-{\frac {1}{(1+\sigma _{P})(1+\sigma _{S})}}}$ ------ (Eq. 3.7).

## Leakage inductance in practice

Leakage inductance can be an undesirable property, as it causes the voltage to change with loading.

In many cases it is useful. Leakage inductance has the useful effect of limiting the current flows in a transformer (and load) without itself dissipating power (excepting the usual non-ideal transformer losses). Transformers are generally designed to have a specific value of leakage inductance such that the leakage reactance created by this inductance is a specific value at the desired frequency of operation.

High leakage transformer

Commercial transformers are usually designed with a short-circuit leakage reactance impedance of between 3% and 10%. If the load is resistive and the leakage reactance is small (<10%) the output voltage will not drop by more than 0.5% at full load, ignoring other resistances and losses.

High leakage reactance transformers are used for some negative resistance applications, such as neon signs, where a voltage amplification (transformer action) is required as well as current limiting. In this case the leakage reactance is usually 100% of full load impedance, so even if the transformer is shorted out it will not be damaged. Without the leakage inductance, the negative resistance characteristic of these gas discharge lamps would cause them to conduct excessive current and be destroyed.

Transformers with variable leakage inductance are used to control the current in arc welding sets. In these cases, the leakage inductance limits the current flow to the desired magnitude.

### Measurement of inductive coupling factor

The inductive coupling factor is derived from the inductance value measured across one winding with the other winding short-circuited according to the following:[10][24][25]

Per Eq. 2.7,
${\displaystyle L_{sc}^{pri}=L_{S}\cdot {(1-k^{2})}}$ and ${\displaystyle L_{sc}^{sec}=L_{P}\cdot {(1-k^{2})}}$.
Therefore, the inductive coupling factor is given by
${\displaystyle k={\sqrt {1-{\frac {L_{sc}^{pri}}{L_{S}}}}}={\sqrt {1-{\frac {L_{sc}^{sec}}{L_{P}}}}}}$. ------ (Eq. 4.1)