2
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

I am imagining of a half-bridge converter feeding a normal resistive load. In series with each of its switches is an inductor. During the switch "on" instance, there should be zero/negligible current flowing through the switch. Does this mean there will be relatively lossless turn on?

EDIT: Here is a schmatic of what I mean. I forgot the anti-parallel diodes, but they're there.

\$\endgroup\$
  • 1
    \$\begingroup\$ For completeness, add a schematic, there's a schematic entry tool for that when you edit your question. Then think about where losses can occur. Can an (ideal) inductor dissipate power? Which devices would dissipate power if the switching wasn't lossless? \$\endgroup\$ – Bimpelrekkie Jun 5 at 17:33
2
\$\begingroup\$

Does this mean there will be relatively lossless turn on?

From the point of view of the power supply feeding the converter, energy will be delivered to the load and also stored in the inductor. The energy delivered to the load is somewhat delayed by the inductor charging up so, for a short period of time less energy is delivered to the load and this may or may not be a bad thing.

But, energy is being taken-up by the inductor as the current ramps up and, that energy can no longer be used by the load so, when the MOSFET turns off it all gets dissipated in the MOSFET because the inductor tries to maintain that current against the scenario of a rapidly open-circuiting MOSFET.

A side impact can easily be that the MOSFET gets too stretched on inductor back emf and gives up the ghost.

I don't see any net benefit; only problems but, if you have a cunning plan, please enlighten me.

\$\endgroup\$
  • \$\begingroup\$ aha the phantom down voter strikes again or have I said something stupid in my answer? \$\endgroup\$ – Andy aka Jun 5 at 18:14
  • \$\begingroup\$ Hi Andy. I added a schematic of what I mean. I understand that the turn-off losses will probably be worse due to the inductance, but if the inductance is large enough, perhaps the turn-off losses will be quite small? \$\endgroup\$ – Lerbi Jun 5 at 19:01
  • \$\begingroup\$ I understood your circuit from the description but it’s always better to have a circuit. Bigger inductance means longer to turn on the load and more energy to destroy the mosfet when switching off. Why not use a simulator and prove it to yourself. Any EE worth their salt uses sim tools now and specifically, to solve ideas like yours (in the absence of web site help like here!) \$\endgroup\$ – Andy aka Jun 5 at 20:58
0
\$\begingroup\$

To answer your question, with a straight up ideal inductor, it can possibly result in zero current only during switching because the current will be delayed.

To achieve this you need a saturable element, when the switch turns on, this component will hold DC voltage until it saturates, this time until saturation will be the delay between current and voltage. Controlling this delay is difficult and has many issues in itself and will only have clear benefits in large transient voltage that happen during turn on.

As Andy aka explained, this does has a drawback during turn off and now you have an adverse effect when your switch off.

If you fully balance the energy as a complete process over time, from the perspective of the energy source nothing has changed. However, at the instant of turn on the current will be close to 0 and losses during that time are reduced.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.