I'm doing a course in which I need to connect a wheatstone bridge (which contains a light sensor) to an op amp.

I'm using the same devices and voltage supply (9V) as the course, still, when I measure the voltage in the nodes I'm supposed to connect to the op amp I obtain a voltage that (as far as I understand) is too high for the op amp, let me explain:

This is the circuit schema:

enter image description here

The op. amp. datasheet (OPA344PA) is here op. amp. datasheet

From the datasheet,I can see that the input voltage range (VCM) must be between -0.3 and -1.8 V, right?

Now, according to this document Input and output limitations of an op. amp., VCM is (VIN+ + VIN-) / 2.

VIN+ and VIN- are the two nodes of Vout in the whatstone bridge. One of them is fixed, the other comes from a sensor, so it varies.

I have measured both of them with respect to ground (in the case of the sensor, I have taking the maximum voltage that I can get with the current illumination). These lectures are: 1.651V and 2V.

Applying the formula above, I obtain 1.8075. First of all, that value is positive, when the datasheet states that the input voltage should be negative. Even ignoring that, that value is a bit higher than 1.8V, so I guess it's not safe to connect the current circuit to the op amp (if some light is turned on and the sensor voltage becomes higher, it will get much worse)

My question is: Is the above reasoning right? If so, I guess I should use smaller resistors for R2 and R4, so the voltage drop is greater through R1 and R3. Is this correct?

  • \$\begingroup\$ Where are you getting the max Vcm of -1.8 V? I see the -0.3 V on the datasheet, but I don't see -1.8 V. The value in the table under the Max column is (V+) + 0.3. \$\endgroup\$
    – Justin
    Feb 20, 2020 at 15:13
  • 3
    \$\begingroup\$ the input voltage range (VCM) must be between -0.3 and -1.8 V, right? No, that does not sound right, that would mean that for example +2.5 V (which is "in the middle" when using a +5 V supply) would not be allowed. What makes sense is: -0.3 V to Vsupply - 1.8 V. That means, when Vsupply = 5 V, a range of -0.3 V to +3.2 V. Also be sure you distinguish between Maximum ratings (exceed these and you might damage the chip) and operatings conditions. \$\endgroup\$ Feb 20, 2020 at 15:17
  • \$\begingroup\$ @Justin in the datasheet, page 2, INPUT VOLTAGE RANGE, CONDITION, first line. \$\endgroup\$
    – Martel
    Feb 20, 2020 at 15:28
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    \$\begingroup\$ @Martel - The datasheet is actually more row by row. That is the test condition for one of the CMRR measurements. It is saying, "If you run with Vs=5.5V and Vcm between -0.3V and 3.7V the min CMRR is 76 dB, typically 92 dB." It still doesn't have -1.8V, but V+ - 1.8V, which is 3.7V. It is not the limits on Vcm though (unless you require 76 dB of CMRR). Those are on the row above, looking at the Min and Max columns. \$\endgroup\$
    – Justin
    Feb 20, 2020 at 15:32

1 Answer 1


In my opinion there are no problems here as your maximum output is 2V and the opamp is able to handle up to v+(2.5V

The OPA344 and OPA345 are optimized to operate on a single supply from 2.5V and up to 5.5V with an input common-mode voltage range that extends 300mV beyond the supplies.

When the datasheet states that VCM must be between -0.3 and -1.8 are always conditions to other things and are between -0.3 and V+ - 1.8V.

Datasheet section about VCM


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