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Questions about the BatchPCB service

Moderator: robacarp

User avatar
By sparky
#2996
Minimum hole size is 20mil.

Minimum is 8mil but I would recommend atleast 10mil for a total minimum annular of 30mil.

-Nathan
By Lawrence
#3003
Thanks! That's exactly what I needed to know.
By dhouston
#3628
sparky wrote:Minimum hole size is 20mil.

Minimum is 8mil but I would recommend atleast 10mil for a total minimum annular of 30mil.

-Nathan
That's not how I understand the meaning of annular ring. If the hole diameter is 20 and the annular ring is 10, the pad is 40 not 30.
User avatar
By sparky
#3676
Lemme rephrase :


Image

-Nathan
By dhouston
#3679
Nathan,

Thanks for clarifying.

I just wish it had come 15 minutes earlier. I just did my own step & repeat to combine 10 boards in one order. I'll have to do it over now with 30 mil pads instead of 40.
User avatar
By sparky
#3693
You're fine!

These are just the minimums. You can make vias any way you wish - larger than 20/30.

-Nathan
By dhouston
#3696
sparky wrote: These are just the minimums. You can make vias any way you wish - larger than 20/30.
I understand that but I need every mil I can squeeze out of this board.
By murphyr
#3810
sparky wrote:Minimum hole size is 20mil.

Minimum is 8mil but I would recommend atleast 10mil for a total minimum annular of 30mil.

-Nathan
Have you all had boards made with vias like this?

Just want to confirm before I make my board.
User avatar
By sparky
#3828
We've had many made this way.
By murphyr
#3843
sparky wrote:We've had many made this way.
Cool, thanks for reply.
By Steve
#3870
Sparky, I agree with dhouston, above. Annular ring thickness refers to the distance between the edge of the hole and the edge of the pad. The example you gave is a 20mil hole and a 30mil pad. That would result in an annular ring of 5mil.

5mil is probably OK, though. The worst case scenario is to have more than 5mil misalignment and a 'breakout' of the hole in the direction of a connecting track. That might result in the track connecting directly with the hole plating without the 'support' of the pad. This makes failure at the junction of track and pad more likely.

Steve.
By Jay3
#4228
Here is infos from Olimex page:
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The pad - drill hole clearence must be > 0,406 mm (>16 mils) i.e. the resulting annual ring to be 0,203 mm (8 mils) - i.e. is 0.9 mm drill size is used it MUST have pad/via diameter = min. 1.306 mm
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Following mm/mils conv, they clearly states that a 8 mils border from hole, resulting in a diameter of 35(hole)+2*8=51mils.

According to APCircuits:
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Annular Ring: The width of the conductor surrounding a hole through a Printed Circuit Pad.
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My interpretation is that it is the width of the trace surrounding the hole (so Pad=Hole+2*AR).


Here from PCBFabExpress:
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For plated holes, the Annular Ring should be AT LEAST 7.5 mils wide. For example: If your finished hole size is 125 mils diameter, with annular ring, the pad diameter should be AT LEAST 140 mils (which is, 125 + 7.5*2 ).
--------------

PLEASE!!!! WHAT IS THE TRUE, FINAL ANSWER!

This is of a critical importance, I think. It should be added to the tech specs you list on the main 2.50$ page.

-Jay3
By Steve
#4230
PLEASE!!!! WHAT IS THE TRUE, FINAL ANSWER!
I can't give the final, true answer but Sparky's example was very clear.

20 mil diameter hole
30 mil diameter pad

resulting in annular ring of 5 mil.

In my opinion bigger is always better with clearances, trace widths and annular rings, so my recommendation would be to use the biggest annular ring your board allows (up to about 10 mil, anyway).

Steve.
By SOI_Sentinel
#4232
And the answer is....

Depends on the manufacturer. As these are going through Gold Phoenix, none of the other examples apply except what Sparky's put up.