Operation of single-phase half-wave rectifier with active load
Conditions: La = 0, Ld = 0, a=0
The a peak rectified current of the load and a peak thyristor current are

Direct component of a rectified voltage is

Direct component of rectified current and current of the thyristor are

RMS current of the thyristor is equaled to RMS current of a secondary winding of the transformer

A peak inverse voltage

Let's define 1-th harmonic Ud and id


It is an even function.
Then

A RMS rectified voltage

A ripple factor of rectified voltage Ud for 1-th harmonic is

A ripple factor of rectified voltage Ud is

RMS value of EMF of secondary winding is
.
Instantaneous value of i1 we should define from a magnetic balance condition of the transformer by a variable component. Magnetic voltage drop is not taken into account. The direct component is not transferred into a primary winding.
Then

Let’s define rms value of i1

Let’s define the total powers of the transformer’s windings.
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