IR drop
Why your 3.3 V rail reads 3.18 V
Copper has resistance. Push current through it and some of your supply voltage is spent getting to the load instead of arriving at it. The loss is I times R, which is where the name comes from. On a schematic the rail is a single node at 3.3 V. On a board it is a resistor network, and the far end of it is not 3.3 V.
Where the millivolts go
Three places, and only the first is the one people estimate:
- Along a track. A 0.25 mm track in 1 oz copper is about 19 mΩ per centimetre. Ten centimetres at 2 A is about 380 mV one way, and the return leg spends the same again.
- Through a via. A single 0.3 mm via is about 1.5 mΩ, which is nothing until it is the only path. Four amps through one via is a measurable step in the rail and a hot spot at the same time.
- Across a plane. A plane is not an equipotential but a sheet resistor, with a hole in it wherever a via field or a slot interrupts the current's path. Two points 30 mm apart on the same “GND” can differ by tens of millivolts, and a part that reads its reference from one and its signal from the other sees the difference as noise.
When a small drop matters
120 mV out of 3.3 V is under 4%, which sounds survivable, and often is. The cases where it is not:
- A regulator's feedback tapped in the wrong place. It regulates whatever it can see. Sense at its own pin and every millivolt of drop to the load is yours to keep.
- An enable or reset threshold. A pin that needs 0.7 × VDD does not care that the rail is nominally fine.
- An ADC reference. Drop on the reference is gain error on every reading, and it moves with load.
- Anything at 1.8 V or below. The same 120 mV is 7% of a 1.8 V rail and most of the margin on a 1.0 V core.
How to find it without guessing
A per-track hand calculation needs the current in that track, and that is the part you do not know: a rail that splits four ways, a return that takes the low-impedance path instead of the drawn one, a plane carrying most of it. So the problem is circular: you need the solve to know the current, and the current to do the sum.
To break the circle, Amptrace meshes the copper you drew, solves the whole network at once, and gives you the potential at every net and the drop along every track, via and plane. That includes the reading at any point you hover, the one a multimeter would have given you three weeks later. Related: the trace-width calculator for the heating side of the same current, and what else it checks.