# PWM with 50% duty cycle

#### electronicsLearner77

Jul 2, 2015
277
I keep getting questions on how inverter works for different duty cycles. Suppose the inverter has 3 half bridges with 6 switches. If i apply 50% duty with center aligned mode. Then 50% of time in 20KHz the upper switches will be ON and 50% of the time the lower switches and the net result is 0 current through the windings is it correct? Please advise.

#### kellys_eye

Jun 25, 2010
5,684
For an inverter (AC out) both halves of the switching cycle produce an output voltage else you wouldn't get the 'AC' part. At 50% duty cycle you are switching at 100% power (50% each side of the cycle). Anything LESS than 50% will be a power reduction.

Power is produced at the peak of each squarewave cycle whether it is in the positive or negative direction.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,182
the net result is 0 current through the windings
That is only half the truth. Together with the direction of the current the polarity of the voltage also changes. Therefore power dissipation happens in both cycles. For the power it is irrelevant that the net current is zero: net power is not zero.

For the coil, however, it is very well relevant that the mean current is zero. When the on times of the two halves of the full bridge are different from each other, the imbalance will lead to a DC component in the current which in turn will adversely affect the behavior of the inductance. WOrts case: The inductor gets saturated by the DC component and loses (most of) its inductive behavior.

#### crutschow

May 7, 2021
651
Anything LESS than 50% will be a power reduction.
It reduces the RMS voltage which, of course, would reduce the power to a fixed load.

If you reduce the duty-cycle to 25% (each half cycle on for 50% of the time), then the RMS output voltage will equal that of a sinewave with the same peak voltage.

#### crutschow

May 7, 2021
651
If you reduce the duty-cycle to 25% (each half cycle on for 50% of the time), then the RMS output voltage will equal that of a sinewave with the same peak voltage.
I meant to add, if the inverter has that output waveform duty-cycle then, for a given output RMS voltage, it tends to work properly both with resistive type loads, and with electronic loads that rectify the peak voltage to generate DC power.

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