-1
\$\begingroup\$

I'm designing a power supply filter based on inductor - capacitor topology, as shown below.

LC filter

However, when the load is suddenly disconnected, a high current starts flowing through the capacitor, as the inductor tries to keep the current constant. What are the ways to prevent this? I'm thinking about adding a short circuit protection across the capacitor.


Suggested solution: flyback diode across the inductor.

It would be forward biased during periods marked in red below - the current then is not flowing through the diodes.

The current won't flow through any diode - when input voltage is low - say 0, then the inductor is maintaining the current, which flows through the load, capacitor and diode itself.

forward bias

I performed a simulation in Falstad, the diode conducts high current.

enter image description here

And in LTspice, which shows high current.

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ If load is disconnected, the energy from the inductor is transferred to the capacitor resulting in high current flow. \$\endgroup\$ – kubajed Jan 7 at 19:50
  • \$\begingroup\$ The problem, I suppose, is that the LC by itself is underdamped without the load. It may be detrimental to circuit performance, but if you are able to add damping anywhere it may stabilize the voltage response. Could be in series with the cap, or the inductor. shunt resistance could stabilize too, but I assume that is a no-go. Maybe a shunt RC could help. \$\endgroup\$ – mkeith Jan 7 at 20:23
  • \$\begingroup\$ You simulation appears to have a load of 500u\$\Omega\$, and in another shot (same simulation?), you have a sine wave input of 100V with 0 impedance. Are these figures what you want? If it were not for the 100V, I would say you were designing a utility substation. In a more likely scenario, the current through your inductor would be much smaller. Apart from that, one way to dissipate unwanted AC from your filter is to add in parallel with your cap, a second cap in series with a resistor. \$\endgroup\$ – Math Keeps Me Busy Jan 8 at 3:16
  • \$\begingroup\$ Would you mind sharing what is the expected load current and the approximate DC voltage at the capacitor? \$\endgroup\$ – mkeith Jan 8 at 4:43
  • \$\begingroup\$ Load current 3 kA and capacitor voltage about 1,5 V \$\endgroup\$ – kubajed Jan 8 at 7:44
0
\$\begingroup\$

One way to do this is with a TVS diode across the capacitor after the inductor which turns on after a certain voltage is reached. Size the diode higher than the voltage that the load will see (so it doesn't turn on during normal use, and only turns on in the event of an over voltage condition).

enter image description here
Source: http://www.flexautomotive.net/EMCFLEXBLOG/post/2016/09/15/automotive-centralized-load-dump-test-requirments

Another thing you could try is a flyback diode across the inductor enter image description here

The diode conducts nA in simulation and is negligible. ![enter image description here]

\$\endgroup\$
14
  • \$\begingroup\$ Yes, I agree that this could work, but before the voltage across the capacitor reaches the TVS activation voltage the capacitor gets damaged due to overcurrrent. \$\endgroup\$ – kubajed Jan 7 at 19:26
  • \$\begingroup\$ Flyback diode shorts during normal operation \$\endgroup\$ – kubajed Jan 7 at 19:41
  • \$\begingroup\$ I don't see how the flyback diode can become forward biased in normal operation. Maybe I am just not thinking about it correctly. \$\endgroup\$ – mkeith Jan 7 at 19:53
  • 1
    \$\begingroup\$ How can it become biased? the current through L1 can only go from the diodes to the load (D1 and D2 can only conduct if current is flowing to the load). The only time the flyback diode will conduct is during flyback operation where the load is disconnected (or if the breakdown of D1 or D2 should happen which it should never reach that condition in a properly designed circuit). \$\endgroup\$ – Voltage Spike Jan 7 at 20:06
  • \$\begingroup\$ I added a simulation in my original post. The current flowing through the diode is comparable to the load current. \$\endgroup\$ – kubajed Jan 7 at 21:55

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.