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Can someone help me in identifying the component dimensions from this datasheet BD-datasheet

Actually, this datasheet is confusing for me. so someone please help me in identifying the dimensions and fill the following properties by extracting the data from the datasheet:

  1. PAD pitch:
  2. PAD horizontal size:
  3. PAD vertical size:
  4. Drill size:
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  • \$\begingroup\$ It's all there on page 1 - what is your specific problem? \$\endgroup\$
    – Andy aka
    Jun 13, 2017 at 9:52
  • \$\begingroup\$ Yeah, I aware that its available on page1 but am unable to figure out the following values PAD pitch: PAD horizontal size: PAD vertical size: Drill size: \$\endgroup\$
    – kaviarasan
    Jun 13, 2017 at 9:54
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    \$\begingroup\$ The pad pitch is 7.5 mm +/-0.2 mm for all of them. \$\endgroup\$
    – Andy aka
    Jun 13, 2017 at 10:11
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    \$\begingroup\$ @kaviarasan I've added a diagram and please note that at some point you should consider formally accepting the best answer to your question. Here's an example of where you can practice the art of formally accepting answers: electronics.stackexchange.com/questions/304781/… - it's a small price to pay for getting good answers and continued support. \$\endgroup\$
    – Andy aka
    Jun 13, 2017 at 10:45
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    \$\begingroup\$ @Andyaka thx for a reminder. Done \$\endgroup\$
    – kaviarasan
    Jun 13, 2017 at 10:51

2 Answers 2

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Just to expand upon Andy's answer, your drill size should be a bit wider than the maximum dimension marked in red. To be more precise, it should be wider than the diagonal formed by the maximum dimension marked in red (1.1mm) and the maximum dimension on the right-hand diagram showing the depth of the pin (0.8mm). The maximum lead diameter is determined by $$\sqrt{1.1^{2}+0.8^{2}}=1.36mm$$

According to IPC-2222, Level A requires the minimum hole size to be the maximum lead diameter + 0.25mm, so your minimum drill size should be $$1.36mm+0.25mm=1.61mm$$

Levels B and C (tighter fabrication tolerances) add 0.20mm and 0.15mm to the maximum lead diameter, respectively (instead of 0.25mm). Level A is usually your safest bet because it pretty much guarantees that any fab house can handle it.

As for the pitch, the datasheet shows a maximum and a minimum, and you can generally assume that the nominal measurement is right between the two. With a range of 7.3mm minimum to 7.7mm maximum, it is easy to see that the nominal is 7.5mm, with a +/- 2mm tolerance.

Finally, for the pad size, I generally try to make my pads about twice the diameter of the drill size. If the minimum drill diameter is 1.61mm, then your minimum pad diameter should be 3.22mm.

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  • \$\begingroup\$ I would recommend a pad pitch of 0.3 inch or 7.62 mm, a better fit to routing grids of 100 or 50 or 25 or 20 mil. \$\endgroup\$
    – Uwe
    Jun 13, 2017 at 11:48
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    \$\begingroup\$ The user is looking at the metric dimensions, which suggests he is designing his board with a metric grid. Using a 0.5mm grid, or even a 2.5mm grid, is perfectly reasonable and will achieve the 7.5mm dimension very easily. \$\endgroup\$
    – DerStrom8
    Jun 13, 2017 at 12:07
  • \$\begingroup\$ thank you @DerStrom8 you gave me clear understanding this will certainly help me in looking at other datasheets also. \$\endgroup\$
    – kaviarasan
    Jun 14, 2017 at 6:02
  • \$\begingroup\$ @DerStrom8 I use footprint wizard so I won't be using the metric grid. I just enter the properties values. \$\endgroup\$
    – kaviarasan
    Jun 14, 2017 at 7:05
  • \$\begingroup\$ @kaviarasan Doesn't the footprint wizard have a limited number of supported package styles? I don't recall ever seeing this package as an option in the wizard? \$\endgroup\$
    – DerStrom8
    Jun 14, 2017 at 13:13
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enter image description here

  • Pitch between pins in BLUE
  • Drill size in RED - use Pythagoras to get the hole size i.e.: -

Diameter (bare minimum) = \$\sqrt{1.1^2+0.8^2}\$ = 1.36 mm. So make it maybe a 1.5 mm hole.

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    \$\begingroup\$ It might be worth adding that this pad configuration will only do for 2.4A. If the OP needs to go to the full 4A, a heatsink will be required and that might need to be reflected in additional footprinty bits if it's attached to the PCB. \$\endgroup\$
    – Paul Uszak
    Jun 13, 2017 at 11:04
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    \$\begingroup\$ The value of 0.043(1.1) is not the drill size, hole size should be a little larger to fit for the rectangular pins (1.1 by 0.8). Using Pythagoras the hole should be 1.36 mm minimum, so 1.4 to 1.5 mm for the hole size. Pad pitch should be 0.3 inch to fit to common routing grids. \$\endgroup\$
    – Uwe
    Jun 13, 2017 at 11:42
  • \$\begingroup\$ @Uwe fair comment. I shall amend to prevent novice confusion. \$\endgroup\$
    – Andy aka
    Jun 13, 2017 at 12:16
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    \$\begingroup\$ I disagree with Uwe's comment regarding the pad pitch. The OP is presumably not using an imperial grid, so a 2.5mm grid is a better option, which allows 7.5mm spacing. \$\endgroup\$
    – DerStrom8
    Jun 13, 2017 at 13:50

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