I’m trying to make a footprint for the 4UCON 10156 connector (same datasheet here or here), but I’m having trouble interpreting the datasheet.
I know that the “fat” part of the plastic is 14.00 mm high, and the “thin” part of the plastic is 7.37 mm high. The distance between the first and fourth row of pins is 5.7 mm. But what I can’t figure out is how the pins are positioned relative to the plastic. In particular, if I knew the size of the little gap that I’ve labeled “?”, then I think I would know everything I need to know.
Yup that datasheet is insufficiently dimensioned (under constrained)
Does the manufacturer offer a 3d model of this part? you could measure the nominal dimension from it.
If not you have no choice but to either contact the manufacturer and ask for clarification or buy the part and measure it from there. (The later is sub optimal as you do not get any information about tolerance ranges)
If it is not critical for your application you could also “trust” the drawing of the suggested footprint. It looks like the 1mm drill just touches the body outline.
A good starting point may be 0.6mm - or 0.635mm .
Probably no less than 0.5mm, the max being 0.7mm.
If your goal is to build a 3D model, this +/- 0.1mm uncertainty is probably fine.
Thanks! I’d been hoping I was just missing something.
As near as I can tell, the manufacturer (4UCON) does not. They seem to only have the datasheet.
There is a 3rd party that seems to have a model, though. I’m not 100% sure it’s the same part (they don’t specify the manufacturer or part number) but it looks like it is.
Yes, I think I’ll just do that. My application is not that critical. It’s just that I wanted to be as accurate as I could, especially if the information was hiding in the datasheet somewhere and I wasn’t seeing it.
That looks about right to me. I have already purchased the part and I’m looking at it. The edge of the pin seems to touch the transition between the two plastic sections. If it doesn’t actually touch, the distance is so small that I’m not sure how to measure it, even with calipers.
Everything you need is on the datasheet. The gap you question is half the width of the pin or 0.15. But given the uncertainty of the other other dimensions you can essentially ignore it.
Third sheet of the drawing gives the pin dimensions of 0.55x0.3. That would be 0.3 for the dimension you are interested in. Looking at the cross-section on the left shows that the part of the body with dimension 14.00 extends to the inside edge of the pin. The dimension of 5.70 is center to center of the pins, so the remaining gap is 0.15 ± 0.10.
That’s the part I’m not sure about. Both the diagram on the left and the diagram on the right seem to show a small gap between the inside edge of the pin, and the part with dimension 14.00.
Yes, but that gap is smaller than the tolerances of the other measurements, so it’s impossible to give a meaningful measurement for it, other than the obvious dimension of half the width of the pin. It’s highly unlikely that you need to be more precise than that.
So the gap between the pin and the main body would be about zero if you treated the pin as 0.55 square
The pitch pin to adjacent pin is 2.54, but row to row is 1.90mm, this means that with 1mm hole, you would have to have a very small annular rig to give track clearance.
I would try to go a little smaller to make it easier to run tracks between pins, 0.7mm is going too far, if any pins are slightly bent, so I would try 0.9mm finished hole size.
The datasheet specifies at least 0.1mm tolerance for the pin size (Measurements given as .XX have 0.1mm tolerance unless specified otherwise) This means the pins could be 0.65x0.4mm.
The smallest hole into which such a pin can fit is ~0.76 If the fab has a hole size tolerance of +/-0.05 you should not go smaller than 0.81mm (so probably already 0.85mm as the fab will not have a 0.81mm drill)
With that we did not yet take into account misalignment of the hole compared to what was designed.
IPC would suggest an increase of the hole diameter by 0.2mm compared to maximum lead size. Would result in 0.96mm (or 1mm) This is what the manufacturer suggests.
I played around a bit. If i use 1mm drill size and use 0.15mm as minimum annular ring i can get 1.24mm clearance between the pads in x direction (should be enough for a trace to leave that area)
This should be more than enough for a trace to get to the inner rows.
Even when i have 0.3mm pad size increase in y direction (for easier soldering) i still have > 0.7mm pad to pad clearance.
This would even be enough for a 0.2mm trace with 0.25mm clearance.
Even with 0.25 min annular ring one can still get 0.7mm pad to pad clearance. But a lot less pad to pad clearance in x direction (might be tricky to get more than one trace fitting.)
Except there are 4 rows of pins, Rene was just showing 2 rows for simplicity. What to do about the two middle rows? As this is a right-angle connector, most use cases would have this connector at the board edge so the middle pins on one of the long sides will have to thread through most of the connector. Even with Rene’s calculations, it may take up to 3 signal layers to fully route this connector.
Two signal layers should be enough even if you can only enter the connector from one side. (This is a likely scenario as the connector will be near the pcb edge)
To show that i use the kicad default min width (0.25mm) and clearance (0.2mm)
With the settings in my example it might be a good idea to make the pads a bit larger in x direction. (2*0.25+3*0.2 = 1.1 So the pads could be 0.14mm wider as the minimum setting i used has a clearance of 1.24mm)
With 0.15mm clearance and 0.2mm trace width one could even get away with one layer.
