I have some trouble with choosing an NTC to limit the inrush current, where I want to bypass the NTC with a relay after a certain time.

The choosen NTC should be one of these here: https://www.mouser.ch/ds/2/18/AAS-920-325D-Thermometrics-NTC-Inrush-031814-web-850596.pdf

There is a total input capacitance of 5mF after the NTC and the Power Supply delivers 24V with a maximum of 30A.

The following schematic depicts the circuit, except that there is no transformer as load, but an input capacitance of 5mF.

enter image description here

The differntial equation system seems pretty complicated, since there are non-linearities from the ntc, which should be taken into account. Somehow I need to know the resistance over time, such that I can calculate the voltage at the input capacitance over time and so the current through the NTC. Has anyone an idea how to get a differential equation relation of the current and the NTC resistance?

  • \$\begingroup\$ If you are using a relay then why bother with an NTC. Alternatively, if you are using an NTC, why bother with a relay? \$\endgroup\$ – Andy aka Mar 15 '18 at 13:48
  • \$\begingroup\$ You may want a time delay on relay \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 15 '18 at 13:54
  • \$\begingroup\$ @Andy Best of both worlds! Although I prefer straight R instead of NTC and bypass with relay. \$\endgroup\$ – winny Mar 15 '18 at 14:01
  • \$\begingroup\$ Well, the delay for the relay comes from the input capacitance, such that the relay will close at around 15V. A straight R may be a possibility too, but I think normally a thermistor is used for such an application. \$\endgroup\$ – HansPeterLoft Mar 15 '18 at 14:26
  • \$\begingroup\$ Define your max I desired, E for load (1.725J) and the R25 is 1.4*120V/IMAX if you do not use ZCS triac to turn on... and E for PTC must be > E load with an R25/R150 ratio >10 or compare input Effective ESR to PTC ESR vs E(t) . What are your unstated V(t), I(t) ESR variables?? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 15 '18 at 17:09

Your Answer

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

Browse other questions tagged or ask your own question.