For a battery-powered design (3 to 3.6V), I'd like to use a sensor that outputs a small, low-power AC voltage signal. To sample this signal, an ADC is used (this ADC is also powered by the battery), where currently the voltage is boosted by a DC source to prevent the sensor's output signal from going below 0V.

To replace this DC source, I figured a voltage regulator that's about 1/2 of the supply voltage (~1.6V) would work. However, there's a concern that as the device's battery drains from 3.6V to 3V, the ADC may read different values due to a fixed regulated voltage used to boost the sensor's AC signal.

Would it be possible or even a good idea to have a regulated DC voltage that is always half of whatever the supply voltage is (i.e.: half of 3.6V, half of 3.3V, half of 3V, etc.)?


The following solution is what would have fully answered my question:

enter image description here


1) By feeding multiple batteries in series to a 3.6V voltage regulator, there would be less concern about the ADC reading different values due to a draining battery supply.

2) A decoupling capacitor / high-pass-filter (from Harry Svensson's response), would remove any inherent DC components in the AC signal.

3) A voltage divider would bias the decoupled AC signal to guarantee that the AC signal would always be centered at half the voltage supply. I think the lower resistance also affects the high-pass filter cutoff frequency.

4) A voltage reference would ensure that the ADC's measurements are less affected by any possible fluctuations in the V+ voltage supply.


closed as unclear what you're asking by Brian Carlton, pipe, Dmitry Grigoryev, Autistic, uint128_t Apr 22 '17 at 0:16

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    \$\begingroup\$ please share a schematic of your current circuit. ' where currently the voltage is boosted by a DC source to prevent the sensor's output signal from going below 0V' how can the signal go below 0 if it is battery powered - that is unclear to me. \$\endgroup\$ – vrleboss Apr 19 '17 at 0:22
  • \$\begingroup\$ Also, what's a "DC source", if not a voltage regulator? \$\endgroup\$ – Phil Frost Apr 19 '17 at 0:27
  • \$\begingroup\$ Just use a voltage divider to halve the voltage and use it as the input to your ADC reference. Is there some reason you need to use a regulator? \$\endgroup\$ – Samuel Apr 19 '17 at 0:30
  • \$\begingroup\$ @HervéGrabas I think the OP just means they're adding a DC offset to the signal. \$\endgroup\$ – Samuel Apr 19 '17 at 0:31
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    \$\begingroup\$ You can make a circuit that outputs 1/2 of the supply voltage (and can supply considerable current). But you wouldn't call it a regulator any more, because varying in response to input voltage changes is one of the main things a regulator is trying not to do. \$\endgroup\$ – The Photon Apr 19 '17 at 0:39

What you want to do is restore a DC component to an AC signal, that can be done like this:

enter image description here

Here's the link if you want to play around, which I strongly advice you to do.

And if you want to boost a signal, then don't bother, there's usually a VREF pin on your board that you can lower which effectively boosts the signal you're trying to measure.

If Vref is 5 volt and you measure 2.5 volt and got 10 bit resolution, then you'll read 512.

If Vref is 2.5 volt and you measure 2.5 volt and got 10 bit resolution, then you'll read 1024.

If Vref is 1 volt and you measure 2.5 volt and got 10 bit resolution, then you'll read 1024.

If Vref is 5 volt and you measure -2.5 volt and got 10 bit resolution, then you'll read 0.

And if you want to reduce then voltage divide with 2 resistors.

If you however want the Voltage Peak to Peak value, then apply a diode and then a capacitor in parallel with a resistor, the resistor will help make the capacitor bleed out. Otherwise you'll only get the maximum value ever reached which will slooooowly bleed out because of real life parasitic resistances. That looks like this:

enter image description here

Here's the link for the picture above.

And here's what you probably want, with filters n stuff. enter image description here

If this ain't what you need, then write something in the comment to this post and I'll update the post.

  • \$\begingroup\$ Wow, that's an amazing circuit simulator thanks for letting me know. One thing about your suggested circuit though, isn't the capacitor followed by a resistor combination a high pass filter? In my application, some of the low frequency components would be needed for analysis. \$\endgroup\$ – plu Apr 19 '17 at 3:05
  • \$\begingroup\$ Look at the graph the voltage outputs, does that look like a high pass filter to you? - It's not a high pass, bandpass, lowpass, bandstop filter. It's simply a "Sample and Hold" circuit that bleeds out reasonably fast. But hey, just click the link and connect it like a high pass filter (connect the 10uf between the diode and the right resistor) and see what happens. It won't turn out good at all. Because you're pumping DC into a capacitor (diode is one way). So you'll get a couple of mV, not what you want. \$\endgroup\$ – Harry Svensson Apr 19 '17 at 23:11
  • \$\begingroup\$ According to here, sim.okawa-denshi.jp/en/CRhikeisan.htm, putting in the 5 uF + 5 k-ohm values into a HPF calculator shows the cutoff frequency to be ~6.3 Hz, so the 500 Hz would hardly show any difference in output. But based on that calculator's Magnitude vs. frequency plot, 1 Hz would show ~-20 dB difference; changing that circuit's frequency to 1 Hz, I observed that the peak of the output is a lot less compared to 500 Hz., so yes, it really is a high-pass filter. But regardless, this can also be a solution as long as the R and C values are chosen with a suitable cutoff frequency. \$\endgroup\$ – plu Apr 20 '17 at 0:32
  • \$\begingroup\$ For the "DC restoration" part, would it also be possible to somehow boost the DC higher to avoid the little bit that goes below 0V? \$\endgroup\$ – plu Apr 20 '17 at 0:44
  • \$\begingroup\$ No, because the negative value is the forward voltage of the diode, and there is no diode with 0 Vf. You could however use a bjt transistor with a 500ohm resistor to its base from vdd instead of a diode, but man... make a new question because it's so unclear to what you want. \$\endgroup\$ – Harry Svensson May 19 '17 at 17:48

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