transformer | PartQuest™ Explore

This is a circuit design for a typical wall-plug power adapter. The sinusoidal source represents a typical US 60 Hz, 120V AC wall supply. A transformer steps the voltage down to a manageable range to be rectified for a 5V DC supply, but the waveform is still sinusoidal. The 4 diode bridge rectifier takes this lower amplitude sinusoidal signal and produces a DC biased result, but still with noise resulting from the rectification process.The capacitor before the voltage regulator chip helps smooth some of this noise. Then the regulator chip provides a self-contained, feedback controlled mechanism to keep the output voltage at the specified value. The output capacitor further smooths any output ripple and provides a reservoir of charge to supply any surge in current load that may be encountered.This example includes a load resistor that allows the exploration of the effects of load. The load is varied during the simulation and it can be seen that above a load of 100 mA, the regulation of the output voltage begins to deteriorate.The addition of a few components can make it possible for such a circuit to provide significantly higher power loads. See http://www.systemvision.com/design/ac-dc-power-adapter-current-boost-regulator for an example.

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AC-DC Transformer and Rectifier II

Description

This circuit compares two different ways to specify a two winding transformer. One way is by specifying the inductance of each winding, with a coupling coefficient between the windings.The other way is by specifying the number of windings and the size and permeability of the core on which the windings are wound. For a perfectly coupled transformer, the flux in the core is completely shared by both windings. In the first model, this corresponds to a coupling coefficient of 1.0. In the second model, this corresponds to a single flux path in the magnetic circuit.L = u0*ur*N^2*A/Lu0 = 1.25663706 × 10-6 [m kg s-2 A-2]L_primaryur = 3000 [no units]N = 10 [turns]A = 100.0E-6 [meter^2]L = 100.0E-3 [meter]L_primary =1.25663706E-6 *3000*10^2*1000E-6/100.0e-3 = 3.76991E-04L_secondaryur = 3000 [no units]N = 100 [turns]A = 100.0E-6 [meter^2]L = 100.0E-3 [meter]L_secondary = 1.25663706E-6 *3000*100^2*1000E-6/100.0e-3 = 3.76991E-02The third circuit shows how a magnetic circuit can be used to represent imperfect winding coupling. In this example, the length of the core is divided between the two horizontal core segments. A third (vertical) core segment is also shown. This magnetic path carries flux that is not common to both windings, thus reducing the coupling between the two windings. When the length of this segment is very long, its effect is negligible, approximating the simpler magnetic circuit above it..

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