I would like to build a DC-DC converter circuit using EasyEDA that can output about 80V from 7 Nissan Leaf battery modules that are about 55V nominal.

I am looking at the datasheet for the LM5164 and page 16 shows an example 12V converter with a parts list on page 17.

I don't know much about inductors; I think I get their basic principle of magnetics but I have never tried to build a functional circuit using inductors before; until now I have just used linear regulators.

The parts list says to use inductor MSS1246T-683MLB but JLC-PCB do not appear to stock this. How would I know if SWPA6045S680MT is an acceptable equivalent?

I assume that resistor RFB1 and RFB2 are generating the reference voltage that sets the output to 12V. What formula would adjust these 2 values to make the output 8V?

  • \$\begingroup\$ what relevance does the resistance of the inductor have? the datasheet recommends Inductor, 68 μH, 170 mΩ, 1.8A but i cannot find one. I am looking at part SWRB1207S-680MT that clams 68uH 140mΩ 2.1A \$\endgroup\$
    – Jay Dee
    Jul 1, 2022 at 22:23
  • \$\begingroup\$ SWRB1207S-680MT should be OK. Exact resistance should not matter. That inductor probably uses slightly thicker wire to handle slightly higher current, giving slightly lower resistance. Good Luck! \$\endgroup\$
    – emrys57
    Jul 2, 2022 at 11:15
  • \$\begingroup\$ thanks, so far i have made this schematic in EasyEDA link once i have finished it i will print the board and come back here with my findings. im struggling to understand the PGOOD pin at mo, would i be correct in saying it is a sink pin that would connect the negative of an LED via a resistor and only sink power lighting the LED if all is good? \$\endgroup\$
    – Jay Dee
    Jul 3, 2022 at 13:24

1 Answer 1


I built this circuit and got it working, short of having to change the RON resistor as I accidentally had it printed with an 0805 SMD 1 Ω and not a 100 kΩ resistor by mistake.

The SWPA6045S680MT does work fine.

enter image description here

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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