3
\$\begingroup\$

Pad layout

It seems like there's something missing. I can't figure out how the rows are spaced relative to each other.

The pitch is 1.257mm. The distance between the centers of 17 of them is 20.112 -- that doesn't really add information. What about the 22.578? It seems to be measured from some arbitrary point within the pad to the same point on the opposing pad. The B diagram doesn't show what that point is. It looks like it might align with the right side of the smaller pad in A. If so, that would mean the pad in B could be considered split into a rectangle of length 1.5 (with the rounded 0.4 bit tacked on the end), and the distance between the rectangles (ignoring the rounded part) is 22.578. That spaces opposing rows, and adjacent rows are placed by centering. But this doesn't seem to fit on a grid -- 22.578/1.257 = madness.

Am I missing anything here? (And why would someone design something like this?)

The module itself looks like this:

MG323-B module

\$\endgroup\$
  • \$\begingroup\$ Please share the datasheet source for the part concerned. I am curious about that do-not-connect pad marked A. \$\endgroup\$ – Anindo Ghosh Aug 1 '13 at 10:35
  • \$\begingroup\$ A is not an NC pad, it's an RF input. download-c.huawei.com/download/… \$\endgroup\$ – Dan Ellis Aug 1 '13 at 21:45
1
\$\begingroup\$

The outer dimension provided (22.578 mm in this case) is the size of the square ceramic or plastic body of the intended part.

The rounded portions of the footprints provide the touch points for the LCC contacts, which extend beneath the body, as well as to the sides.

There is a reason LCC parts will often not fit into a sane inch or mm grid: Typical applications of LCCs use a spring-loaded socket, with either through-hole or surface-mount contacts underneath:

PLCC (Source: Wikipedia)

The socket's own pins are generally either single or double row at each side, usually fitting (in your part's case, for instance) a standard grid, say 50 mils. Hence, to allow for the edges of the chip carrier, the chip itself would need to be marginally smaller, but the sprung contacts take care of it.

This concept breaks down with newer parts, because the drive for ever thinner sides on chip carriers means that someone will invariably crack the materials strength challenge, and come up with something just-a-bit smaller to save board space. This usually doesn't matter much, as a board designer using a PLCC would perhaps also need to deal with 50 mil grid parts, even 1 millimeter grid parts, and occasionally "just because I felt like it" pitch parts, so a bit here and there is taken in stride.

\$\endgroup\$
  • \$\begingroup\$ It's not a chip, it's a module. Essentially just a PCB that gets soldered onto the surface of another PCB, so there is no chip carrier. The measurements in the diagram are literally the ones I need to make the pad stack to on my board. The module itself looks like this: cdn.techship.se/uploads/images/98/… \$\endgroup\$ – Dan Ellis Aug 1 '13 at 21:51
  • \$\begingroup\$ That part looks like it would fit a 68 pin 1.27 mm pitch leadless chip carrier if you don't want to solder directly, so the layout sizing still makes sense. From what I see, the outer dimension is the dimension of the PCB, like it would be in a PLCC part. \$\endgroup\$ – Anindo Ghosh Aug 2 '13 at 4:50
1
\$\begingroup\$

The "B" drawing indicates that the inner end of the pad has an 0.4 mm radius. It appears to me that the 22.578 measurement is between the centers of that radius on the two rows of pads.

\$\endgroup\$
  • \$\begingroup\$ That doesn't seem to point to a more natural grid alignment. 22.578/1.257 = 17.96, so if we assume that they're aligned, the distance between the two pin centers would be 18 * 1.257 = 22.626. That's a difference of 0.048, meaning the center of the pin would be 0.024 in from the center of the circle. Does that seem likely? If so, why wouldn't they specify that instead of leaving the read to work it out in such a long-winded fashion? \$\endgroup\$ – Dan Ellis Aug 1 '13 at 22:34
0
\$\begingroup\$

It's very simple:

All measurements are relative to the center of the radius on the inner R0.4 mm (the dot in sub section B). The pitch is 1.257 mm, the distance from the the end pad measurement point and the center-line of the row /column center-line pads is 1.233 mm (this should have been shown). 20.112 + 2* 1.233 = 22.578.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.