Annular Ring
Annular Ring Rules & IPC Standards
| Standard / Condition | Requirement | Application Consequence |
|---|---|---|
| IPC Class 2 (Consumer) | 90° Breakout allowed | Via functions, but may fail under severe thermal stress |
| IPC Class 3 (Mil/Aero) | 2.0 mil (0.05mm) minimum | Guarantees 360° intact copper ring, maximum reliability |
| Standard Drill Wander | ± 3 to 5 mils | Requires pads to be 10 mils larger than the drill size |
| Laser Microvia | ± 1 mil wander | Allows ultra-small pads, critical for HDI and mmWave routing |
| Teardropping | Required at trace junction | Prevents breakout from severing the trace connection |
Ring = (Pad Diameter − Finished Hole Diameter) / 2
Example: 20 mil pad, 10 mil hole = 5 mil annular ring (assuming perfect centering).
RF Parasitic Capacitance Impact:
C ≈ (ε · A) / d
As the pad diameter (A) increases, the shunt capacitance to adjacent ground planes increases, lowering the via's characteristic impedance.
Frequently Asked Questions
Why does drill breakout happen?
Mechanical drill bits are not perfect. They deflect, machines have tolerances, and PCB laminates shift during pressing. A standard drill can wander up to 5 mils from center. If your pad isn't large enough to absorb this wander, the hole breaches the edge of the copper, weakening the via connection.
How does it affect RF impedance?
Every via is essentially a coaxial cable passing through the board. The via barrel is the center conductor. The anti-pad in the ground planes is the outer shield. A large annular ring adds bulk copper to the center conductor, acting like a shunt capacitor. This lowers the via impedance, causing VSWR degradation at high frequencies.
What are teardrops?
A teardrop is extra copper added to smooth the junction between a thin trace and a via pad. If a drill wanders exactly toward the incoming trace, it can sever the connection. The teardrop widens the landing area, protecting the electrical connection and reducing mechanical stress concentration.