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EDA => Eagle => Topic started by: TheBorg on May 26, 2014, 04:07:36 am

Title: What drill size should the pads be for this part?
Post by: TheBorg on May 26, 2014, 04:07:36 am
I'm working on a footprint for a part in Eagle (this is probably my third pcb I've designed using the software, so I'm quite new at it).

The part # in question is S 6B-PH-K-S in this datasheet (http://media.digikey.com/pdf/Data%20Sheets/JST%20PDFs/PH%20Series.pdf). The pin width is 0.5mm. So should I set the drill size at 0.5mm in eagle, or something like 0.7mm (which from what I've gathered is a more common size).
Title: Re: What drill size should the pads be for this part?
Post by: kikib on May 26, 2014, 04:24:50 am
Since the pin width is 0.5mm the drill size should be a bit bigger, how else is it going to fit in the hole? You have to account for manufacturing tolerance for both the component and the PCB as they won't be exactly as quoted. Some datasheets will provide you with a tolerance. Also since it's a multi-pin part you will need to also think about any tiny deviations in tolerance which will have a carry through effect on the alignment of the pins with respect to the holes on the PCB.

So for a pin width of 0.5mm you should provide a comfortable hole size and add a bit on top for manufacturing tolerance. Since you have 2mm pin spacing and 0.8mm hole (as example), you will have 0.4mm on each side of the pin and therefore still left with a 1.2mm spacing from drill hole to drill hole. So 0.7 to 0.8mm sounds reasonable but always check your footprint by printing it off and comparing to the physical part and obviously it's better to be able to fit the component than not able to fit it at all. You could then for rev2.0 PCB tighten the tolerances.
Title: Re: What drill size should the pads be for this part?
Post by: TheBorg on May 26, 2014, 04:28:14 am
Sounds good. Thanks!
Title: Re: What drill size should the pads be for this part?
Post by: Monkeh on May 26, 2014, 04:47:02 am
The datasheet actually tells you the recommended drill size.
Title: Re: What drill size should the pads be for this part?
Post by: TheBorg on May 26, 2014, 04:47:00 pm
The datasheet actually tells you the recommended drill size.
Heh. So it does.  :-DD
Thanks again!
Title: Re: What drill size should the pads be for this part?
Post by: NANDBlog on May 26, 2014, 05:33:00 pm
Have you ever tried putting a R lenght square through a circe with the same diameter?
(http://www.supplierlist.com/prod_img/helen777/205085_Shape_sorting_cube_with__chunky__game_puzzle_.jpg)
Title: Re: What drill size should the pads be for this part?
Post by: LukeW on June 02, 2014, 09:01:43 am
Get the datasheet, look at the recommended PCB layout part of the datasheet, not the data on the component itself, and it says 0.7mm hole. So that's what you'd use, right?

But then again if it's an 0.5mm x 0.5mm square pin, then that's 0.7071mm across the hypotenuse, and you might want a little bit more room for error, so maybe 0.75mm (not standard?) or 0.8mm.
Title: Re: What drill size should the pads be for this part?
Post by: TheBorg on June 05, 2014, 04:37:03 am
Get the datasheet, look at the recommended PCB layout part of the datasheet, not the data on the component itself, and it says 0.7mm hole. So that's what you'd use, right?

But then again if it's an 0.5mm x 0.5mm square pin, then that's 0.7071mm across the hypotenuse, and you might want a little bit more room for error, so maybe 0.75mm (not standard?) or 0.8mm.
If there's one thing I've learned here, is that immideately after I post something I know I've made a stupid mistake. In this case, blowing past the PCB layout part of the datasheet while looking for it.  :-DD
Sticking with 0.7mm for now, and I'll double check when I get my parts ordered.
Title: Re: What drill size should the pads be for this part?
Post by: kizzap on June 05, 2014, 06:49:59 am
x = L * sin ?°

where:

x = diagonal of the pin (and thus the minimum diameter)
L = length of the side opposite the corner you are measuring from, and
? = the angle from the adjacent side to the hypotenuse of the effective triangle (the hypotenuse being the diameter we are after), which in a square pin is 45°

alternately: given a square with side lengths of Y:

hypotenuse (min drill diameter) = sqrt(Y² + Y²)

But I guess that is just pandering.