Combined inductor and current sensor

[@popovlist]

Thanks professor. Do you know what kind of sensor this is? Would that be better than a separate winding for current representative measurement?

[@sambenyaakov]

Do you mean winding on same core? This does not measure current.

[@popovlist]

​@@sambenyaakov I see, but let's say there are just a few windings, for that supposed measurement. Could not that be fed in an integrator so the resultant signal is current?

[@sambenyaakov]

@@popovlist Not really

[@اسلامكمال-خ2ذ]

Dear sir, how do I get these slides

[@sambenyaakov]

Sorry unavailable. You can capure screens

[@dor7sh]

Very intersting idea👍

[@sambenyaakov]

🙏😊

[@gsuberland]

What's funny is I suggested exactly this a few months ago to a bunch of EEs and they all shot the idea down - "it wouldn't be reliable", "if this worked someone would've done it already". Glad to see the idea wasn't as silly as I thought.

[@sambenyaakov]

Great minds think alike 🙂

[@pasikavecpruhovany7777]

Neat idea. EI core with air gap in the middle seems a good compromise from construction standpoint.

[@sambenyaakov]

Great point!

[@OdedArlevski]

Great video! Thanks, Prof.!

[@sambenyaakov]

🙏😊

[@tamaseduard5145]

👍🙏❤️

[@sambenyaakov]

🙏😊👍

[@stevesedio1656]

Very nice. An accurate measure of the inductor current (minus the increased leakage). I also like the EI suggested by Pásikavec. A question, at 0.51 minutes you show an RC across the inductor and say it measures the voltage drop of the internal resistance of the inductor. The output of the switch and the cap are both very low impedances, therefore, IMHO, you are just measuring is the voltage across the cap. To confirm, I modeled this, setting the values of LR and RC to get the same waveform. I doubled the inductor resistance, doubling the voltage across it, the voltage across the C remains unchanged. I repeated this but halved the inductor, voltage across the inductor resistance doubled, the voltage across the C remains unchanged. Am I missing something?

[@sambenyaakov]

See kzbin.info/www/bejne/qaK0q3Wum6p2oJY

[@stevesedio1656]

@@sambenyaakov Thanks, that is the analysis I was looking for. You have the voltage across the LRs network changing with current. If it did, then you are right. I see it differently. Because the switch and output cap are very low impedances, the voltage across LRs doesn't change with load current. The increased "DC" drop across Rs reduces the voltage across L. With the right RC network the voltage across C will be the same as the ripple voltage across Rs of the inductor. It won't provide the DC offset that ripple due to load current. The voltage also won't show core saturation. If the voltage changed with current, you could.