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| ADC differential input termination |
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| OM222O:
Hey guys I'm using an ADC with 2 differential channels. The capacitors on the input filter stay charged and I still get readings when I disconnect the leads. I really don't want to add an op amp to the front end as the circuit is very sensitive and low offset op amps cost a fortune (about 2 bucks a pop). my next best option is a resistor across either the input or the filter caps (not sure which one is better). again the system is very sensitive (10uV accuracy, 500nV resolution) so I'm not sure if adding a termination resistor will add noise or offset. The input to the differential channel is the voltage across a resistor which is 200kOhm at maximum. even with a 1Gohm termination resistance, the parallel resistance is about 199.96k which is not bad but still not great ... If there isn't a good way to discharge the caps without introducing errors, I think I'll just leave it as it is because accuracy is more important than random readings when the leads are not connected in my application, it's just a minor issue which I hope to solve :-+ |
| ejeffrey:
A 1 G load on a 200k source is .0002. so a 100 mV input will have a 20 uV error. Or said another way that is only about 12 bit accuracy. If your source impedance is constant and known you can calibrate that which I assume you are going to have to do anyway with such a high source impedance. If you have some out of band way to detect input presence I would look into using a switch (CMOS or mechanical) instead of a load resistor. That will avoid the long time constants of big resistors and not load the signal source. |
| OM222O:
It's used as a part of a 4 wire measurement device. I can tell when the the load is connected and disconnected but the test resistor value would range from a few milliohms to 200k, I assumed worst case scenario in that calculation. The ADC is 24 bit which is why I said it's not too bad, but it's not great either. That also means there is some way of calibrating out the error caused by the termination resistance (first calculating the total resistance seen on the input, then reverse engineering the parallel combination because we know the sum and one of the resistors ... although to do this I have to find a way to do floating point operations to an insane level of accuracy as that would probably cause more error than the current through the 1G resistor itself). My reasoning for 1G is also anything above that is basically open circuit and won't solve anything for me ... I also would like to avoid any kind of manual or mechanical switch as that would become the first obvious weak point. a mosfet can be used but it has to work bidirectionally somehow (ADC can read + and - voltages, and you wouldn't want your device to stop working just because you connected the leads backwards). The leakage through a mosfet can also be a lot higher than a high value resistor (usually a few uAs) If you know a way to solve this please let me know. P.S: the ADC has a function to short Ain+ and Ain- to VDD/2 ... I'm not sure if it actually shorts anything physically to VDD/2 or it's just an internal reference node or something. |
| Kleinstein:
The mode to connect the inputs to VDD/2 could be a good option to discharge a capacitor at the input. The SD ADCs tend to be not very high input impedance - so more like in the 100 K range, when actually used. A much higher impedance usually requires an extra buffer - so ADCs have that internal, but not many. If connected to the outside world the ADC input would likely need some protection from ESD or similar. This can be a difficult part for high resistance sources. |
| 2N3055:
H11 photocoupler maybe? |
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