Saturating inductor
Inductance falls from Lzer toward Linf as the core saturates.
- electrical
- Modelica Standard Library
- 2 ports
- 4 parameters
saturatingInductorLibraryElectricalDescription
An inductor with a smooth saturation curve. At small currents the inductance is Lzer; as the current grows the inductance falls toward Linf. The curve passes through Lnom at the current Inom.
Example
Transformers and a saturating inductor A coupled-inductor and an ideal transformer on a 10 V sine, and an inductor that saturates as its current rises.
Also in this example:Voltage sensorGroundRampResistorTransformer (coupled)Ideal transformerAC voltageControlled current
Ports
Conserving terminals 2
-
+
pPositive pin. Current into p is counted as positive.
-
−
nNegative pin.
Parameters
-
Nominal current
Inom1 A≥ 0
Nominal current, in amperes, at which the inductance equals Lnom.
-
Nominal inductance
Lnom1 H≥ 0
Inductance at Inom (flux linkage divided by current), in henries. Must lie strictly between Linf and Lzer.
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Initial inductance
Lzer2 H≥ 0
Inductance near zero current, in henries. Must be greater than Lnom.
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Saturated inductance
Linf0.5 H≥ 0
Inductance at very large current, in henries. Must be less than Lnom.
Equations
v = p.v − n.v, i = p.i, p.i + n.i = 0 Psi = Linf · i + (Lzer − Linf) · Ipar · atan(i / Ipar) v = dPsi/dt Ipar is solved at the start so that Psi(Inom) / Inom = Lnom
Implementation
Modelica.Electrical.Analog.Basic.SaturatingInductorAssumptions and limitations
- The simulation stops with an error unless Linf < Lnom < Lzer.
- No hysteresis, core losses, or winding resistance. The curve is symmetric in current.
See also
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