Flywheel diodes in buck converters

[ringo888]

Morning all,

Right, i have a basic understanding of the standard buck converter. That it charges the inductor when the FET is switched on and when it switches off the flywheel diode is used as a "closed loop" to allow the current in the inductor to flow to the load. But my question is what would happen if the flywheel diode wasn't there? reason i'm asking is that even without it energy is still stored in the inductor so it must still flow to the load, any explanation?

Thanks,
Ringo

 

[ppdr123]

can u give schematic. the diode may be used to balance residual energy in inductor. but it can be explained more afficiently if you provide schematic.

 

[ringo888]

Sure, see link below for pic:

https://upload.wikimedia.org/wikipe...onventions.svg/800px-Buck_conventions.svg.png

Could it be to do with the inductor voltage polarity jumping from +ve to -ve during the switch "on" time?

 

[FvM]

what would happen if the flywheel diode wasn't there?

Assuming an almost ideal inductor, the energy will be dissipated in the switch, generating any
voltage the switch allows. In case of a mechanical switch, the energy is burned in an arc, with an
electronic switch, it creates either excessive "regular" switching losses or complete breakdown.

With a real device, part of the excessive losses will go to the inductor. In any case, the converter
efficiency would decrease to a linear regulator's behaviour without the diode.

 

[snafflekid]

What happens is the inductor has the flux stored in it and that causes the current to attempt to keep flowing after the switch turns off. The inductor is will attempt to pull current out of the high impedance node it is connected to. The voltage on this node will continue to drop, to supply the current to the inductor until it becomes so much lower than ground that a parasitic diode somewhere turns on or breaks down and creates a current path.

Or ultimately the node goes to negative infinity voltage. How exciting!

 

[ppdr123]

here, the diode provides he current path when the switch is made off.
what happens is, when u make the switch ON, current flows in the forward direction(i.e. IL) providing the load current and charging the capacitor. some amount of energy is also stored in inductor. when u make the switch OFF,inductor do not allow the current to reduce to zero.but since the source is cut off from the circuit, charged capacitor provides some amount of current through this diode. also, energy stored in inductor freewheels through this diode providing current till the effect of inductor is nullified.

 

[ringo888]

OK I understand in principle what your saying, I just don't understand the reason for this action, surely the easiest route for the current path to take (from the energy stored in the inductor and cap) would be through the load (RL)?