0
\$\begingroup\$

I want to send my gerbers to a PCB manufacturer but its manufacturing capabilities force diferrent minimum anular ring for vias in different layers. For outer layers, its 0.13 mm and for internal layers is 0.175 mm but I altium I can only set a hole size and a total diameter for all layers. How can I change this? I can't found the option. enter image description here

enter image description here

There are no options for each layer.

\$\endgroup\$
5
  • \$\begingroup\$ Think you want one of these: altium.com/documentation/altium-designer/… \$\endgroup\$ Commented Feb 15 at 6:46
  • \$\begingroup\$ That would cause some changes in all the vias that are already placed in my project, my PCB is finished. How could I change this without the libraries? Maybe selecting all similar objects and changing manually in the second picture? Will in this way the clearance apply to 0.41 in external layers and to 0.5 in internal ones? I dont want it to apply to 0.5 in external ones because it is not true \$\endgroup\$ Commented Feb 15 at 7:32
  • \$\begingroup\$ You can edit existing vias with filter queries and Properties. You will need to check collision and move things as needed. \$\endgroup\$ Commented Feb 15 at 8:02
  • \$\begingroup\$ Viewing them in 2D I will notice just the bigger anular ring, right? The 0.175mm ones, even if there is a 0.13 one on them. That was the clearance thing I mentioned, too \$\endgroup\$ Commented Feb 15 at 10:40
  • \$\begingroup\$ This link will take you, even without a valid license, to an excellent (and official) Altium forum: forum.live.altium.com (where all you need to do is register for an account). \$\endgroup\$ Commented Feb 15 at 13:15

1 Answer 1

1
\$\begingroup\$

In your second screenshot, you see where "Simple" is selected? Instead, select "Top-Middle-Bottom" or "Full Stack". This will allow you to set different values for pads on different layers.

enter image description here

\$\endgroup\$

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