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I'm designing an electrical component into a CAD from the manufacturer drawing. The version I'm interested to is the code 106166-1 (6 positions, last entry in the table).

I'm not sure to understand the drawing about the following dimensions:

  • overall width of the connector
  • distance between any row of pins and the case

Width of the connector I see two different ways to calculate it:

  1. D * 2 + 0.240'' = 0.452''
  2. C + 0.040'' * 2 = 0.412''

But as you can see they are different values. And I don't understand what does "3 places" mean (under 0.040'', on the right of C ref).

Distance between rows and the case I can't see any actual measure of this. Only observing the drawing it seems the inner row is placed adjacent to the case, hence the center of the pins should be at diameter / 2 (0.015'' / 2) from the case.

Would you please help me to understand the right approach to find those values?

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(Links are to a distributor website. I don't have any knowledge of this distributor, but it had pictures of the parts.)

Width of the connector = 0.412"

Based on pictures of the part and other models found here (link), like the 103166-2, I can see that the slot measured by D is not centered, therefore you can't use that to get the width of the part. Instead, C + 0.040" * 2 is the correct measurement. This is verified by taking the width of the edge (0.040") twice, the distance from the edge to the center of the connector (0.066") twice, and B (0.200").

"3 places" This is confusing, but I'm pretty sure it is the width of the left, right, and bottom walls as shown on the middle left view. The top wall seems thicker in the photos and drawing.

Distance between rows and the case. Middle left view.

From the left: The wall is 0.040" thick. Then it is 0.066" to the center of the pin. Look at the top left of the view.

From the bottom: The wall is 0.040" thick (I think). Then it is 0.090" to the center of the pin. Look at the bottom right of the view.

If you mean the right view, the bent pins are 0.157" on-center from the base of the interior of the plug.

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  • $\begingroup$ I meant the right view, I see the pins are bent at 0.157'' from the base of the interior of the plug, but there's no indication about the distance between the tip of the pin and the end of the plug...to calculate the exact position. $\endgroup$
    – Mark
    Oct 20 '18 at 5:14
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As a general rule, there should never be two different ways to find the same dimension from a correctly dimensioned drawing, so if you have two formulas, either you or the drawing have a problem somewhere!

I don't know how you got your first formula D*2 + 0.240.

Looking at the view with dimensions D and E, the length appears to be E + 0.240 + D, assuming the two ends of the component are mirror images - which seems to be the way it is drawn, but there is no guarantee that is the case for all the part numbers in the table, because nothing on the drawings actually says it is symmetrical.

Since dimension E is not given for your part number, you can't use this formula anyway.

In any case, the notation "0.240 typ[ical]" means that the 0.240 is not an accurate dimension. Compare the pins, where the pin diameter is "typical", but the view of the hole pattern gives the definitive size of the holes for the pins.

Your other formula based on a different view, C + 0.040*2, is correct.

For the distance between the end of the pins and the base of the plug: this is probably not accurately controlled, except that the pins don't protrude beyond the base. However, the length of the pins from the tip to the middle of the turned part is 0.318 + 0.157 = .475. The thickness of the pin is 0.025 (from the top drawing), so the pins protrude of .475 - .0125 = .4625 into the plug. The depth of the plug itself is 0.489 so the clearance will be 0.489 - 0.4625 = 0.0265.

@Wercho gives the correct answers to the other questions.

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  • $\begingroup$ About the first formula, my fault was to assume if E is not given, there is only one slot, hence 2D + 0.240. The errors due to tolerances are acceptable for my purpose (I always use typ values with no problem). The last missing information is in my comment to Wercho's answer. $\endgroup$
    – Mark
    Oct 20 '18 at 9:01
  • $\begingroup$ I started updating my answer after seeing your comment on Wercho's. I think you have answers to everything now. $\endgroup$
    – alephzero
    Oct 20 '18 at 9:07
  • $\begingroup$ Sorry, surely I'm blind but I still cannot find the position of the pins related to the front (or back) side of the plug. The horizontal pin's length is 0.318'', the center of the pin is 0.157'' and the plug's width is 0.489''. But what is the distance between the pin center and the front side? $\endgroup$
    – Mark
    Oct 20 '18 at 9:11
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    $\begingroup$ It's half the diameter of the pin, which is 0.025. From the drawing, one row of pins is touching the housing. Even if they aren't quite touching, the minimum distance from the center of the pin to the housing will be 0.025/2 = 0.0125, so the minimum clearance at the other end of the pin will be 0.0265. From the "AMP" on one of the drawings, I guess these are standard Amphenol plugs, so if you are using Amphenol matching plugs and sockets in your design, this detail doesn't really matter anyway. $\endgroup$
    – alephzero
    Oct 20 '18 at 9:19

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