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I have an analogue signal that's 0-4 V that I want to measure using a micro-controller that only measures up to 3.3 V. I want to reduce the analogue signal by 1 V. I've seen designs using voltage splitters, but the voltage reduction is a function of the input voltage. I need something that's a constant 1V.

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  • \$\begingroup\$ Do you need the output voltage to go negative? That will make this more difficult. Can your ADC handle negative inputs in the first place? Most can't. \$\endgroup\$ – Hearth May 24 at 0:34
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    \$\begingroup\$ Use a resistor divider to map 0V-4V to 0V-3.3V(VREF). Because if you instead used say a AA battery to literally subtract 1.5V from the sensor voltage, then the range would be (-1.5V to 2.5V). Not good, your 3.3V single-supply microcontroller won't like getting -1.5V at the analog inputs. Can't exceed common-mode input range. Only practical way is to scale it down. ...or are you saying the signal you're trying to measure is really in the range 1V-4V (not 0V-4V), and you don't need to be able to measure down to 0V? Does the solution need to be passive, or could you use an op-amp? \$\endgroup\$ – MarkU May 24 at 0:46
  • \$\begingroup\$ What if there was a simple circuit that dropped it by 1.25 volts? \$\endgroup\$ – Andy aka May 24 at 10:18
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While a little lazy for me to use 2 op amps when it can be reduced to only 1, this will accomplish what you are after.

It subtracts 1V from your signal, I presume your circuit does not have negative rails, so will clip anything that falls below 1V on the input

The left most op amp is just giving a buffered 1V ouput, the right op amp is a differential amplifier with a gain of 1.

enter image description here

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You're probably better off just using a voltage divider, but if you must, you can use a rail-to-rail input-output op-amp as follows

schematic

simulate this circuit – Schematic created using CircuitLab

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