[SOLVED] DC-DC converters dependence on frequency.

[SherlockBenedict]

In DC-DC I tried to change the frequency of the switch and I get different answers. I can understand why I am getting it because of the fact that the current flowing through the inductor is different for different frequency. I want to know the exact expression of this dependence of output voltage on frequency in DC-DC converters. This is very serious issue because when I used buck converter with a very low frequency, the output voltage is more the input voltage!.

Any help would be appreciated.

Thanks.

 

[BradtheRad]

The lower the frequency, the more time the transistor is on. Hence the greater the amount of time the current can ramp up through the coil.

The steepness is based on the L/R time constant. The flux field intensity is based on current and henry value. Webers = A x L.

Suppose the transistor is left on indefinitely. The coil current will plateau, meaning it will reach its highest limit as based on total DC resistance.

When the transistor shuts off, the field collapses. The coil generates current around the output loop.

Under certain conditions, namely if the energized coil suddenly sees high impedance, the coil will generate a high voltage kick.

A simulator may show extremely high current developing in conditions of zero ohms resistance. This will produce unrealistically high volts.

However, normally a real buck converter should not be able to produce a higher volt level on the load, than the value of the incoming supply V.

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Edit:

Now I am seeing a mode of operation, where a very high charge develops on the output capacitor after I change the load to a high resistance. The capacitor has greater than the supply V for a while .