4
\$\begingroup\$

Is there a very cheap way to measure several voltage differences of 0.1V to 10V (to ~1% accuracy at 10Hz, say), where each pair of measurement terminals may be tens of volts away from the ground of the circuit that needs the measurements?

As far as I can see, one either isolates each analogue signal before it reaches the ADC (using one isolating amplifier per channel - not cheap), or has one serial ADC per channel (not cheap -- I was hoping to multiplex a single ADC), each floated at the appropriate level, with optoisolation of the ADC output.

Are there any other strategies? Isolated DC/DC power supplies are themselves not enormously cheap (~£3/unit) so ideally I would want something that doesn't require a power supply floated at the level of each measurement input. Is this possible?

\$\endgroup\$
4
  • 1
    \$\begingroup\$ How cheap does it need to be? You can get a single-channel serial ADC for $0.78. \$\endgroup\$ – Dave Tweed Oct 22 '12 at 13:39
  • \$\begingroup\$ OK, so the main cost is then the floating power supply. I'm surprised that I can't find isolated DC/DC converters at less than 1W, which seems overkill for that 1mW ADC... \$\endgroup\$ – Chris Johnson Oct 22 '12 at 14:26
  • \$\begingroup\$ You really can't get 1 mW from the circuit being measured? But in any case, at that power level, you can just build a capacitor-coupled supply from discrete components. Should cost much less than $1 per channel, as well. I assume you don't need a lot of isolation, just the ability to float +/- tens of volts. \$\endgroup\$ – Dave Tweed Oct 22 '12 at 14:32
  • \$\begingroup\$ Do you have an example of that capacitor-coupled supply? Does it just look like the rectifier+capacitor+regulator of an AC->DC output stage? \$\endgroup\$ – pjc50 Oct 22 '12 at 14:36
3
\$\begingroup\$

An answer in another question sparked an idea for this.

You maybe could use a high-side current monitor circuit like INA168:

enter image description here

This allows the input voltage to have a common-mode offset of up to 60 V and will output a ground-referenced 1-10 V signal for your inputs, with an appropriately chosen RL. You may even be able to get a gain of less than one, though I haven't read the datasheet closely enough to know if this is guaranteed to work.

If you can reduce the voltage range to be measured to something like 0 - 1 V instead of 0 - 10 V (say with a simple resistor divider), you can use the somewhat similar INA193, which allows up to 80 V offset, but has a fixed gain of 20 V/V.

If you can't reduce your input voltage range, you could try building your own circuit with a similar topology but lower gain. Your accuracy would probably be reduced due to inability to match components as well as TI can do in a single chip.

\$\endgroup\$
0
\$\begingroup\$

Have you looked at In-amps? (Instrumentation Amplifiers) this is exactly what they are designed for. As well as current mode drivers for signalling in noisy/variable voltage environments. An AD8420 is but one example.

\$\endgroup\$
1
  • \$\begingroup\$ "each pair of measurement terminals may be tens of volts away from the ground of the circuit that needs the measurements". AD8420 has maximum input voltage of Vcc - 1.8 V, and maximum Vcc of 36 V. OP needs to be more clear about what range he needs to cover with "several tens of volts" before we have any reason to think this will work. \$\endgroup\$ – The Photon Oct 22 '12 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.