I am trying to design a datalogger that runs off battery power and am trying to maximise runtime.

The logger is infrequently (every 5 minutes) logging 0-30V analog and digital signals.

One of the key current draws at this stage is the voltage dividers required to reduce the signals to the 3.3V or 5V tolerated by the microcontroller. As even loads as high as 1Mohm begin to add up (16 inputs total = approx 0.48 mA).

This seems like a pretty common problem but I haven't been able to turn up an accepted solution.

I have contemplated options such as

  • Reed Relays - Semi impractical due to the required quantity, possibly more beneficial at even lower sample rates.
  • Some kind of buffer IC - Haven't been able to turn up any parts that accept a voltage so far out of the supply range
  • \$\begingroup\$ actuating the reed switch or keeping the buffer alive will definitively make the problem worse. Why not simply a high-Ω voltage divider, switched on only on demand by a mosfet? \$\endgroup\$ – Marcus Müller Feb 20 '17 at 22:33
  • \$\begingroup\$ Would you be able to recommend a method of actuating the high side mosfet switch, I am assuming you would need to have a pair of N and P channel mosfets to allow a low voltage logic signal to switch the signal. \$\endgroup\$ – Hugoagogo Feb 20 '17 at 22:36
  • \$\begingroup\$ a single n-channel would suffice. Don't try to switch on the high side! \$\endgroup\$ – Marcus Müller Feb 20 '17 at 22:37
  • \$\begingroup\$ Wouldn't it still pull almost exactly the same current, due to the path through the input pin / clamping diodes. \$\endgroup\$ – Hugoagogo Feb 20 '17 at 22:38
  • \$\begingroup\$ I see your point, hm. \$\endgroup\$ – Marcus Müller Feb 20 '17 at 22:41

I have had this problem once before . What I did was use 10 K resistors on the bottom of the dividers .This was to keep the source impedance seen by the ADC at always less than 10K. This is normal but do check your ADC just to be sure .I used 100n caps to ground and no buffers .Your caps may be different due to sampling and noise issues.Then I switched the resistive dividers with the smallest cheapest p channel fets that could be googled at the time .The RDS on was extremely small compared to the resistors in the divider so there were no concerns about accuracy.This worked well because the duty cycle of data aquisition was very low.

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  • \$\begingroup\$ It is a bit hard to follow what kind of circuit you are suggesting, any chance you could stick in a schematic. how did you switch the P-channel mosfets, did you just have a single N channel to turn on all of the P-Channels \$\endgroup\$ – Hugoagogo Feb 20 '17 at 23:42
  • \$\begingroup\$ I used one N chan grounded source to switch one High side P chan .There were space issues and voltage clearance issues so that was why I was glad to get rid of the Buffers. \$\endgroup\$ – Autistic Feb 22 '17 at 10:56

use (analog) switches to switch in one divider when measuring and switching out when not.

you will likely need some buffer analog buffer as well, or you have to slow down the measurement.

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  • \$\begingroup\$ Can you recommend any example parts where the contacts are not limited to the supply rails. \$\endgroup\$ – Hugoagogo Feb 20 '17 at 23:40
  • \$\begingroup\$ it is generally difficult to find beyond-the-rail anything. Analog has some isolators that you may want to try - cannot remember the part numbers off my head. a simpler solution is to find a switch and power it from a rail that's sufficiently high vs. your signal to be switched. \$\endgroup\$ – dannyf Feb 21 '17 at 0:07
  • \$\begingroup\$ an alternative solution is to stick a switch (bjt/mosfet) at bottom of that divider and only turn on that switch (from the low side) if it is to be measured. So none of the dividers will be conducting when not measured. During measurement, only one divider is being powered. something like 2n7000/7002 would work for situations like that. \$\endgroup\$ – dannyf Feb 21 '17 at 0:09
  • \$\begingroup\$ @dannyf I recommended that in my question comments, but op had a valid concern: if you cut the lower side connection of the voltage divider, the upper side resistor+ ADC pin will become the new voltage divider, and that means the input voltage would be very high \$\endgroup\$ – Marcus Müller Feb 22 '17 at 10:14

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