Hi,
I have to make the footprint for a connector but I’m having a little issue. I don’t find which pad diameter and which hole diameter I must set. Can someone help me please?
I was thinking about setting the pad diameter as 1.2mm and the pad hole as 0.4mm. Am I right ?
The dimensions :
The drawing :
Here is what I’ve done already :
Source :
SPT-THR 2,5/ 3-H-5,0 P26 - PCB terminal block - 1135294 | Phoenix Contact
(scroll to commercial data then open data sheets or technical drawings)
The pin has a diameter of 0.4mm, so your hole need to be a bit bigger.
Minimum from my opinion is 0.5mm. 0.6mm could also be OK.
The pad diameter is related ot the PCB. The bigger it is the better, as long as it does allow to route your board correctly.
So 1.2mm should be OK
So in general, when the datasheet shows a pin of Xmm, I must conclude a hole of X±0.2mm must be set?
Why is the bigger is the better?
Ok I will go for 1.2mm
For the pad,
you must respect the minimum pad size compared to the hole size of your pcb manufacturer.
then it must be big enough so you can solder it easily.
but there is no hard defined rule.
On some of my pad I use an oval form to free some space between 2 pads for easier routing, and still keep a pad big enough for easy soldering.
for the hole size, it’s been a while since I’ve defined one, but yes X+0.2mm seems good.
If I check the datasheet for a B03B-XH connector from JST.
The pins are 0.64mm
The hole they recommend is 0.9mm
And in Kicad lib it’s 0.95mm
So in that case we have X+0.3mm
Oh okay thank you !
I downloaded the 3D model and it seems that even with a hole of 0.6mm, the pins seem too big which my cause issues during the soldering, right?
Should I increase the pad and hole size to respectively 1.8mm and 1mm ?
I don’t know if the 3D model is accurate.
But it’s a big connector, so yes you can increase to a 1mm hole, and 1.8mm pad. You will be able to solder it without problem with the bigger hole.
One tip, if you want to secure your board in case you make a mistake, you can add a via next to each pin. Doing that the signal reach the pin by both the TOP and the BOTTOM. And if your hole is too small you can redrill it and it will work.
The issue when you redrill a hole is that you will loose the plating inside, so there will be no connection between the TOP and BOTTOM pad, and you can only solder on the BOTTOM. So if the signal reach you’r pad via the TOP side you are screwed.
Also on you’r datasheet some of the pins are 0.4mm and some are 1.2mm be careful, Check the bottom picture on your screenshot
Based only on picture you showed.
I don’t understand what Φ0.4 means when they do A-A cut at which it looks that pad is 0.5 x 0.75.
And at the footprint (I don’t understand why ‘for information only’) they suggest Φ1.2 hole.
I increased the sizes and here is the result :
It looks better.
Yes this is why I’m confused about the pins.
Now you mentioned it… the pad seems to be 0.5x0.75 indeed
I begun to suppose that may be when it goes through PCB it is 0.5x0.75 and then for the rest of its length it is Φ0.4. The best would be to have that part in hands, or have a big photo of it.
Even I have never seen such things but it will allow that part to be inserted in sockets. Who knows.
How do you solder the PCB? If you do it by hand, then you can handle a hole that is too big when you careful. When you use wave soldering, you may have to worry that the components are not flat on the PCB when the holes are too big (especially when the holes are all in one line, which they are not in your case).
Don’t forget that it is a connector.
The PCB and the copper pads will have to be robust enough to cope with the strain of wires pushed in and pulled out, possibly very many times.
Ideally, all the pads AND the joining tracks should be as large as you are able to create, otherwise the pads may separate from the board and probably break from the tracks.
As you already have seen (on the answers in this thread) footprint definition depends also on the personal habit of the designer. To give you one more data point, I would as a first step:
During assembling of the first prototype the drill-size should be checked:
According to the experience with this first prototype the drill size could be slightly modified (drill increased/decreased) for a final footprint.
This problem exists mainly for single sided board with all holes being NPTH. Many years ago (previous century) we had such problem with terminal blocks. When soldered, the internal terminal block structure (variant with so called wire protector and not lift - it is important - see construction differences) was not pushed to the PCB and then when someone tried to use it pads were separated from PCB and broken from their track. Even we needed only one layer we used 2 layer PCB with holes PTH and problem was solved (since then we also prefer terminal blocks with lift).
According to my experience PTH soldered pads are fixed enough fast not matter how big pad you use (to a reasonable extent).
Hi @Piotr
I accept and like and relate to this comment:
I also do not know what type of PCB or manufacturing process or number of layers is being used. Nor do I know the purpose or use of the board, nor who is assembling.
My comment was from personal experience over too many years to remember.
I’m also a believer in the saying: “I’ve paid for the copper, so I might as well have and use it”. 
About those pins:
In the detail of Cross section A-A you can see the pins have a (semi) rectangular size of 0.5 by 0.75mm. All those things in boxes, such as the 0.4mm are tolerances, and that tolerance is for the location of the pin. So the center of the pins are withing a circle of 0.4mm diameter from the their nominal position.
I supposed that this 0.4 in box can be something else than I think but couldn’t get the idea what it can be as graphic with two arrows suggests diameter. Probably it is some mechanical drawing standard convention that all people simply knows (except electronics 
).
I even supposed tolerances, but at A-A there are tolerances written 0.1 so assumed it is not this.
It’s common stuff for mechanical engineering folks and as explained in the wikipedia article, those are tolerances. There are also some measurements enclosed in a rectangular box, and those are the measurements that those tolerances. So there is a grid of 5mm by 8.2mm and the center of all all 8 pins are within 0.2mm (radius) on that grid.
Hi look closely on the datasheet- or usually says somewhere a suggested PCB hole size
