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I'm trying to learn by myself the principles of a voltage boost converter.

enter image description here

**BATTERY IS 1.2v@4300mAh

I noticed that many sources* suggest not putting a resistor in series with the capacitor (as in the picture).

But doesn't it make an infinite current draw by the capacitor? (R=0)

Thank you.

*for example: http://www.electronoobs.com/eng_circuitos_tut10_1.php
http://www.ti.com/lit/an/slva372c/slva372c.pdf

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  • \$\begingroup\$ But doesn't it make an infinite current draw by the capacitor? No, why would it? And if the capacitor would want to draw as much current as possible, would that current be infinite or would something limit the current? \$\endgroup\$ – Bimpelrekkie Feb 25 at 12:39
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No way. The choke gets filled when transistor conducts, the current start to rise from 0 to some level. Then you switch off the transistor. What happens with the current? It still flows with the same magnitude for the very first moment, into a capacitor of course, and then starts decaying.

The current through inductor is continuous.

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  • \$\begingroup\$ OK I see, and what if I connect in parallel a load to the capacitor, will the load get all the current while capacitor's charging time? \$\endgroup\$ – Danaro Feb 25 at 13:05
  • \$\begingroup\$ @Danaro The load will always get I_load=V_cap/R_load and the rest goes in the cap. \$\endgroup\$ – Marko Buršič Feb 25 at 13:51

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