In my project, the current budget is very tight. With less than 1mA, I have to isolate sensor's signal voltage and be able to measure it with ADC on the other side of the isolation barrier. What I have thought of is that using VF conversion and send the digital signal via photocoupler.

I have looked for commercially available VF converters. But they all consume a few mA's.

Can Anyone suggest a good way to isolate an analog voltage signal with limited current budget and board space?

  • \$\begingroup\$ how often to sample? In other words, how often do you need update on the analog signal? \$\endgroup\$
    – Marla
    May 21, 2014 at 3:06
  • 1
    \$\begingroup\$ You need to specify bits required per reading and data rate (sample per second). This makes a vast difference to the power budget. \$\endgroup\$
    – Russell McMahon
    May 21, 2014 at 3:08
  • \$\begingroup\$ How about an ADC on the isolated side, and isolating at the bus? \$\endgroup\$ May 21, 2014 at 3:12
  • \$\begingroup\$ @Marla Sampling rate doesn't have to be very fast. About 10Hz would suffice. \$\endgroup\$
    – Steve
    May 21, 2014 at 3:29
  • \$\begingroup\$ @RussellMcMahon Sampling rate can be as low as 10Hz \$\endgroup\$
    – Steve
    May 21, 2014 at 3:29

3 Answers 3


Please advise required data rate and isolation voltage and any applicable isolation requirements.

Capacitive coupling will work well if your isolation requirements allow it.

Inductive coupling with eg an oscillator allows low power - effectively an "RF" link.

Formal opto couplers tend to require mAs of current.
Some, such as rhe eg IXYS CPC1001N are specified for operation down to0.1 mA and may operate at lower currents. Other opto couplers may also operate at lower than usual current but care is needed in design to not get a "works in some cases" situation. Available data rate may be affected by low currents.

You could make your own opto coupler with eg an LED and a PIN diode and get good results at very low currents.


Extra information: Sampling rate can be as low as 10 Hz.

If you transfer say 20 bits of data per sample and operate at 10 Hz.
Bits/second = 200 Bits.
At 2 Kbps duty cycle = 10%
At 20 kbps duty cycle = 1% At 200 kbps duty cycle = 0.1%

So if you have a say 5 mA optocoupler operating at 20 kbps the mean current is 5 mA x 1% = 50 uA
= 5% of power budget.

Lower mA opto couplers and faster data rates reduce average current proportionally.


A radical solution, and not for the timid.

IF the sensor has a bit of current drive capability (as witnessed by your implication that the sensor could have enough to drive an LED in an opto-coupler). Drive an inductor with many many turns. Creating magnetic field H. H proportion to N (turns multiplied by current : NI)

Use Hall Effect sensor to measure magnetic field.

I like thinking towards infinity and radical.

  • \$\begingroup\$ This would produce a signal output that is the differential (wrt time) of the sensor voltage - is this what the OP wants? \$\endgroup\$
    – Andy aka
    May 21, 2014 at 7:39

You have not specified the resolution you need. That said, I think your idea of using a V/F converter is a good starting point. You ought to be able to do something with a few hundred uA, with an output frequency in the 10 Hz to 1 kHz range. What I'd then do is feed the output to a CMOS one-shot with a short pulse width, like 1 - 10 usec, or even less. Use the one-shot to drive an optocoupler. With the short pulse width and low frequency, the duty cycle of the LED would be less than 1%, so the average current would be low. Short pulse width means you need a fast optocoupler, with attendant higher detector currents, but this all occurs on the other side of the isolation, so it shouldn't be a problem.


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