I am trying to integrate a Waveshare e-paper display into my design using their published schematic:
schematic BOM It calls for a 68uH iron core inductor, L1. After searching DigiKey for 68uH iron/ferrite core inductors, limiting it to active SMD parts, of the most popular sizes 1210/1812, I'm still left with 136 choices. The datasheet says the max power consumption is 40mW which doesn't screen out any of the inductors.

Can someone help this mechanical engineer understand what the inductor's function in this circuit is and how to further narrow down the list by choosing DC Resistance, Q, frequency, etc.? I've never specified an inductor before...


1 Answer 1


40 mW divided by 3.3 V implies a current of about 13 mA. So choose any size rated for at least twice this amount for margin with low resistance and the cheapest price.


When an inductor exceeds it rated current, it will lose its inductance, like a spring bottoming out so it acts as a noise filter or low pass filter if there is any noise on the 3.3 V supply. Low supply noise is important for the liquid paper display.

Rev A

It is not clear to me why or where an iron core inductor was specified but these tend to have higher permeability and thus less wire and lower resistance. The reason I mention this is when switching the inductor the current rise is due to the ratio of L/R=T time constant. The max current is Vcc/ESR of the total resistance but the current must be only pulsed for a low duty cycle such that the average current is only . The FET is < 0.3 Ohms the series source is 3 Ohms so I suspect the inductor needs to be similar to the R values of iron core parts with 68uH which is < 0.5 Ohms but it does not have to be iron core, per se.

They should have specified part numbers in the BOM.

Here is my Rev A choice https://www.digikey.com/products/en/inductors-coils-chokes/fixed-inductors/71

I have changed my selection based on <=0.5 Ohm rather than >> 13mA so you will find it rated for much higher currents, even tho it will be limited by the total equivalent series resistance (ESR) of R(L1+Q1+R2)= 0.5+0.3+3 = 3.8 Ohms thus current ramp is 68uH/3.8 Ohms = 17 us. (??) and if pulsed for this duration the ramp current rises towards 3.3V/3.8 Ohms = 870 mA To use less than 13 mA average ( assuming that is correct) the pulse duty cycle must be 13/870*100=1.5% hmm

  • \$\begingroup\$ Thanks for your answer! Your Rev A choice link was to the list of all inductors, but based on your DCR discussion, I refiltered the list and SRN6045-680M rose to the top. Checked it on octopart and availability is pretty wide. Would this be a good choice? \$\endgroup\$
    – MattD
    Commented May 9, 2018 at 11:57
  • \$\begingroup\$ That’s good I don’t know what happened to the did you can you link . Truncated... thank you \$\endgroup\$ Commented May 9, 2018 at 19:40
  • \$\begingroup\$ @MattD Did you have any success with integrating the E-Paper display into your design? Was the SRN6045-680M a good choice? \$\endgroup\$
    – James
    Commented Dec 21, 2023 at 11:42
  • \$\begingroup\$ The SRN6045 worked great \$\endgroup\$
    – MattD
    Commented Dec 22, 2023 at 23:28

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