Double wound solenoid inductance

QuickField simulation example

DC voltage is applied to the low inductance solenoid with double winding. Transient process of electric current vs. time is analyzed.

Engineering question
How to find double-wound solenoid inductance

Engineering answer
Set up an axisymmetric QuickField DC Magnetics problem for a double-wound solenoid and evaluate inductance from computed field results.

Typical applications
dual-winding solenoids, sensor excitation coils, electromagnetic actuator coils

• Download simulation files (files may be viewed using any QuickField Edition).

Problem Type
Axisymmetric multiphysics problem of Transient magnetics coupled to Heat Transfer.

Geometry
Conductor diameter 0.4 mm.

Given
Copper electric conductivity σ = 56 MS/m.
Voltage U = 0.1 V.
Ambient air temperature T0 = 20°C, convection coefficient average value α = 5 W/K-m².

Task
DC voltage is applied to the coil. Calculate the inductance, resistance and steady-state temperature.

Solution
Transient magnetic formulation in QuickField allows assign electric conductivity to blocks and connect blocks using the external electric circuit. The coil resistance can be calculated using circuit-simulation results: R = V / I.
The coil inductance can be calculated using the field simulation results: L = 2*Energy / I ².
The coil temperature is calculated in a heat transfer problem that is linked to the transient magnetics problem.

Result
Double wound solenoid current I = 1.786 A, Joule heat P = 0.1786 W. Resistance R = 0.1 / 1.786 = 0.056 Ohm.

Magnetic field energy W = 0.187 μJ, inductance L = 2* 0.187 / 1.786² = 0.117 uH.

Solenoid time constant L/R = 0.117 / 0.056 = 2.09 us.

Steady-state temperature of the coil is 50°C.

• Video: Double wound solenoid inductance. Watch on Youtube
  
• Download simulation files (files may be viewed using any QuickField Edition).