Submitted 7 hours ago by iliketurtiles@programming.dev to askelectronics@discuss.tchncs.de
https://programming.dev/pictrs/image/73178513-32bb-47a8-9a27-b9ace3fd256a.jpeg
This is from a section on why decoupling capacitors should be attached to CMOS chips. It shows current spikes during transitions. Which then because of the inductance of traces connecting power to the chip, will cause the power rail voltage to droop.
But why is the ground voltage also shown to rise? What does it even mean for ground voltage to rise when ground is what voltage is measured against?
Because of voltage drop, if you have a length of wire and run current through it, there is a drop in voltage, so the voltage is higher on one side and lower on the other.
So your voltage on the gnd pin of the IC relative to the other end where its connected to your supply will have a voltage difference that increases with current.
So ground on any implemented circuit is not constant and will have some variation physically where you probe. The schematic abstraction assumes every node is connected with zero ohm, zero length connections but this is of course not the case.
I think what this is showing is that the ground node for this device is noisy and will fluctuate as it is trying to deal with the current spikes. It’s probably relative to the expected ground.
I’ve not ever seen this drawn like this before, but have read about the phenomenon in a few different texts.
This is just my 2c as a fledgling hobbyist and in no way an expert…
but ground isn’t magic. It’s just more of the circuit that’s not intended to have an elevated voltage. The voltage spike is induced on the line just like any other inductance issue. It’s important to be aware of what actually happens in a circuit instead of working off of odd assumptions, like “ground is always 0v”.
If you’re measuring ground at the supply, you’ll have a voltage drop on both the supply line and the return line. It’s much less on a PCB, but if you are switching fast enough with large big voltages, even the inductive load of vias to ground can ruin your day.
there’s voltage drop on both terminals, Makes it clear that there’s impedance on both sides of the switching device
ooterness@lemmy.world 7 hours ago
The voltage in the chart represents the “ground” available inside the chip, measured against some better and more stable ground available elsewhere.