I have these two resistors in my schematic, that i want to find footprints for:

i understand that the (Body dimensions = 2.5 x 6.8mm) part is useless. The only info left is (Leads dimensions = 0.6 x 28mm).

I opened the datasheet but this info is not clear to me since i can bend the leads wherever i want.

Also, do i care about the diameter of the leads (0.6mm)? There is no clear P value (in KiCAD's terms) in order to select one.

I opened up the footprint library and there were a ton of THT axial horizontal resistors, but each carried a different prefix like DIN0309, DIN0411 etc. What do these mean and how to select one?


2 Answers 2


Don't over think it - it's not that hard. This one will do:-


Do you see? Your 2.5 mm x 6.8 mm simply has to fit into the box in the footprint viewer. The one I've selected is 2.5 mm x 6.3 mm which is pretty close and will do as the lead pitch is 10.16 mm. So bend leads to fit. Personally, I tend to go long as my boards are pretty childish, so you could go to the next longer pitch (15.24 mm) which gives you wiggle room.

DIN207 is simply an international standard for common parts so that they're easy to fit into designs. Like TO-3 transistor cases. Decide the lead pitch yourself, but of course it can't be less than the body length or you'll be having to shove it in at an angle/vertically/with hammer.

Tip again. Ideally source your parts first so you have confidence of their sizes. Then design PCB and print it out like below at an exact 1:1 scale. Finally test fit by hand...


  • \$\begingroup\$ Vertically mounted resistors aren't uncommon! Densest packing you could get with through-hole components, though of course everything uses surface-mount now. \$\endgroup\$
    – Hearth
    Dec 21, 2020 at 4:07
  • \$\begingroup\$ Thank you. Didn't know that about DIN0207. I guess this is what i should use for the other resistors as well. So i should not take into consideration the leads dimensions parameters. I should only make sure the P is bigger than the component. Now only question is should i take the diameter of the lead? The one is 0.6, the other is 0.45. I guess both will fit through the hole right? \$\endgroup\$ Dec 22, 2020 at 1:34

The footprint for an axial through hole resistor is typically just two through holes and a courtyard/keep-out for the part body and leads. Because you can bend the leads however you want, you have a lot of freedom in how you design the footprint. There is no one-correct way to do it, and there may be many footprints in the library that would actually fit.

The DINxxxx numbers refer to the size of resistor. Beyond that I can't really say specifically what the sizes will be. If you are unsure you can always open the footprint and measure it.

Making a footprint in the Ki-CAD footprint editor is really easy. If you can't find one you want in the libraries, then just make one.

The only restrictions are...

  • The hole diameter is appropriately sized for the leads. Usually this is something like lead-diameter + 10 mills. In your case 0.6mm + 10 mils = 0.825mm holes.
  • Unless you are doing something strange like bending the leads in a U shape, the distance between the holes needs to be at least a far as the part-body length + 2x the allowed lead bend radius. The minimum bend radius is usually something like 3X the lead diameter. Given you body diameter the hole spacing would be 6.8mm + 6 * 0.6mm = 10.4mm.

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