I'm trying to understand this mechanical drawing so I can create a layout for it in gEDA PCB.

Specifically on page 4. I've already managed to place GND 1 and GND 2; those were fairly simple to work out. But GND 3 confuses me, because it's apparent from close inspection of the drawing that it doesn't quite line up with GND 1 and the offset doesn't appear to be given. The same is true for GND 4 which doesn't quite line up.

Also on a semi-related note it is unclear whether the pins are in the order as they are on the SD card or if they are numbered arbitrarily - that is, does pin 1 correspond to pin 1 on the SD card?

While this isn't strictly electronics, I think it passes as an appropriate question because it is an electronic connector and it requires a PCB layout.

  • \$\begingroup\$ In my opinion, it doesn't just pass as an appropriate question, this is an example of a great question! \$\endgroup\$ – Kevin Vermeer Nov 3 '10 at 16:32

If an offset isn't given from the center-line or some other point, it should be symmetrical about the c.l. (always has been in my experience). So 4.7 mm between the inside of GND3/GND4 would mean they are each 2.35 mm away from the c.l., likewise with the 8.3 mm outer to outer.

The pins appear to correspond to the SD card pinout I looked up. It appears backwards on the footprint because the actual SD terminals would be on the right side, but the contacts lead to the left to connect to the board.

  • \$\begingroup\$ +1, the entire left side of that drawing is referenced on the center line defined on the right side (for vertical positions). So start with GND1/2, find the center line, then do the left side. \$\endgroup\$ – Mark Nov 3 '10 at 16:45

Converting drawings to actual library packages is always fun. You have to think like a mechanical engineer. :-)

You have a centerline and dimensions from the center line; you have to work all the math from that point. Specifically, GND3/4 start 2.35mm from the center line and are (8.3-4.7)/2=1.8mm wide. You can see that they are 13.55mm center-to-center to GND1/2.

If there is no mapping given, it's relatively safe to assume that the pins line up with the SD card. I would ohm it out myself or contact the manufacturer though to be sure.

  • \$\begingroup\$ definitely contact the manufacturer for more details, or even better buy one sample in and measure it!!!! \$\endgroup\$ – smashtastic Nov 3 '10 at 7:17

I think it might be easier to put down the numbered pins before GND3 and GND4, then justify GND3 and GND4 to them instead of trying to justify them to the GND1 and GND2 pins.

But if I give it a shot....

You can determine the distance between the horizontal centers of GND1 and GND3 - it's all the way at the bottom and has a value of 13.55 +/- .05 mm.

A good helper is the dotted center line that runs through the middle of the footprint. If we assume GND3 and GND4 are symmetric about that line and they're the same size then we can use two measurements (8.3mm and 4.7mm) give us a height for each of 1.8mm. This means they should be the same height as GND1 and GND2. We can use this information to determine the vertical center of GND3 with respect to GND1.

The top of GND3 is 4.7mm/2 from the center line which is 2.35mm. Its center is .9mm from the edge, so it's 3.15mm from the center line. The center of GND1 is 2.4+.9mm = 3.3mm from the center line, so the vertical center of GND3 is .15mm higher than GND1.



Consider this approach: http://www.penguin.cz/~utx/pstoedit-pcb

  • \$\begingroup\$ Very cool. Looks much easier than typing numbers into a text editor! \$\endgroup\$ – Thomas O Nov 3 '10 at 15:11
  • \$\begingroup\$ Neat! I've not seen that utility before, and imagined that subtle errors (height/width ratio, line thickness, not-to-scale etc) would be present with such an approach. Has that worked well for you in the pas? \$\endgroup\$ – Kevin Vermeer Nov 3 '10 at 16:39
  • \$\begingroup\$ I must confess, I've only tried this approach once. I've got a Python script to write out footprints. I write a little program for each footprint, which generates the pads based on the drawings dimensions. This is not quite as insane as it sounds -- I'm prone to stupid arithmetic errors, so having a program do all the math lets me verify it. \$\endgroup\$ – markrages Nov 3 '10 at 16:55
  • \$\begingroup\$ I have a question. How does the utility know that the pads are X mm's wide when the drawing may be made at any scale? \$\endgroup\$ – Thomas O Nov 3 '10 at 17:15

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