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Can someone explain why the EMI would cause a Bitflip and not always a high read? Why would a pulse invert the signal that’s read? Don’t the voltages effectivel
by echoangle 2y ago
Can someone explain why the EMI would cause a Bitflip and not always a high read? Why would a pulse invert the signal that’s read? Don’t the voltages effectively get added?
- amenghra 2y agoIt depends on how the analog signal is encoded. In some protocols, a 1 is encoded as high-then-low and 0 is encoded as low-then-high.
- echoangle 2y agoAh good point, I was assuming simple TTL where signal level is the bit that’s transferred, RAM is probably using something more complex
- missinglugnut 2y agoYou need to think of EMI as having a magnitude and a direction. Half the time you are adding a negative voltage.
- echoangle 2y agoSince he’s using a Piezo lighter, shouldn’t it be just a single DC pulse like discharging a capacitor?
- missinglugnut 2y agoI was confused on the lighter type so I deleted that part of my response. I think you're correct but I can't say for sure.
- james_a_craig 2y agoEven your example of discharging a capacitor can end up with a pulse both directions, caused by the inductance of the wires. In this specific situation, there's no common reference level, and so the induced pulse will go both directions. You can think of this as being about the edges of the pulse being the parts that actually cause radio to be transmitted, and there's both a positive-going edge and a negative-going edge on a pulse.
- james_a_craig 2y agoSign matters as well as magnitude. The pulse created will have both a positive and negative part - waveform sort of like --^v-- and so you can get either direction bit flip. It's not equivalent to connecting a battery to the pin; EMI's more like AC in that it goes both directions.