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I have a CO2 sensor that output signal values 30-50 mV. I need to translate these voltages to 0-5 V for my microcontroller with the highest resolution. I understand that I can amplify the voltage using a non-inverting op-amp circuit, as shown, to a range of 3-5 V, but is it possible to expand that range to 0-5 V in order to get better resolution of sensor values?

Circuit image

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    \$\begingroup\$ Can you fix the image please. \$\endgroup\$ – Dean Sep 2 '12 at 10:14
  • \$\begingroup\$ Dean - I've updated the resistor value to 101 ohm if that's what you mean. \$\endgroup\$ – neufuture Sep 2 '12 at 20:35
  • \$\begingroup\$ Since posting this question, I've realized that I had mis read the data sheet and that the output range is incorrect. I have also found additional documentation on the particular sensor I am using. I've posted a new questions here. The responses to this post have been useful in starting to understand the concept of instrumentation amplifiers. \$\endgroup\$ – neufuture Sep 2 '12 at 20:56
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You can use a differential amplifier to subtract the 30 mV offset.

enter image description here

When R1 = R2 and R3 = R4 the transfer function is

\$ V_{OUT} = \dfrac{R3}{R1}(V_2 - V_1) \$

So set V1 to 30 mV and choose R3 = 250 \$\times \$ R1.

A problem with differential amplifiers is that R1 will load the resistor divider to get the 30 mV offset, so that you have to recalculate the resistors, and also V2 will have an input impedance which may distort the measurement.

An instrumentation amplifier is the solution.

enter image description here

Most instrumentation amplifiers are differential amplifiers with a buffering input stage. The input stage sets the gain, while the differential stage is usually a \$\times\$1 amplifier. The amplification is then

\$ V_{OUT} = \dfrac{2 R2}{R1}\cdot \dfrac{R4}{R3} (V_2 - V_1) \$

The Microchip MCP6N11 is a suitable device.

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  • \$\begingroup\$ Just to add, if you wanted you could do it with two opamps, because you only care about the input impedance of the microphone input. The other input is just a voltage divider, so you can just adjust your resistances to compensate for the input impedance, and then unity gain buffer the microphone input \$\endgroup\$ – BeB00 Nov 13 '17 at 17:55
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An instrumentation amplifier is what you need here (though an opamp could be used with some attention to detail)
Depending on your supply (single, dual) you need to be careful though. If using a single supply (e.g. 0-5V) you must make sure the InAmp can handle common mode inputs of the level of your input signals, which will be 30-50mV relative to ground (so the input range must include ground)
Also since your output includes ground (and power rail if using a 5V supply) you must make sure the output can swing fully to both rails. Many InAmps don't do either of these things. The LTC2053 is one rail to rail in/out option, as is the MCP6N11 Steven mentions.

EDIT - the LTC2053 will not be suitable as the input impedance is not high enough. The MG811 datasheet specifies the need for an Opamp/Inamp with an input impedance of >100GΩ, so something like the MCP6N11 Steven recommends is needed. This has an input resistance of \$ 10^{13}\Omega \$, which is \$ 10\ T\Omega \$.
I have left the rest of the answer to demonstrate a typical setup, since the principle is the same regardless of the Inamp used.

Anyway, as long as you take care with the above, the setup is pretty simple. Apply 30mV to the inverting input, signal to the non-inverting input and set gain for (5V - 0V) / (50mV-30mV) = 250.

Here is a dual rail (+-5V) example circuit with the LT1789 InAmp:

LT1789

Simulation:

LT1789 Sim

Single supply LTC2053 circuit (simulation not shown as it's the same as above):

LTC2053

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    \$\begingroup\$ +1 for always exposing people to how easy it is to simulate circuits \$\endgroup\$ – justing Sep 2 '12 at 2:44
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Use an instrumentation amplifier like this one.

Since you want to amplify 30-50mV to 0-5V, 5V/(50mV-30mV) = gain of 250. Use the datasheet to select a gain resistor. For my example, G = 1 + (100k/Rg), so Rg = 100k/(G-1) for 402 Ohms. These values need to be pretty exact, and when in doubt make it a little bigger and sacrifice a little span. Since you want 0-5V, you'll want to set the reference voltage to 2.5V since that is the middle of the span. Use a reference diode for that.

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protected by Dave Tweed Jul 18 '14 at 11:18

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