Opamp input voltage divider creates wrong voltage - what could be the cause?

Question

I'm using above 1:2 differential opamp, when 72V applied to the 30:1 voltage divider, voltage across R1 is only 1.727V instead of 2.4V, opamp output is around 3.4V.

Opamp is LMC6062IN/NOPB http://www.ti.com/lit/ds/symlink/lmc6062.pdf

I think it's because of bias current, any suggestion on solution?

Follow up question:

Same circuit below but now this time 0V does not connect to circuit main GND, so there is not second voltage divider but i am seeing around 1.5V across R1

^{simulate this circuit – Schematic created using CircuitLab}

Follow up question #2:

^{simulate this circuit}

Switch on both side of V1, top and bottom circuits are identical differential opamp circuit, bottom one is before switches and connect to v1 directly, top one is after two switches.

When SW2 is open, SW1 is closed: voltage across R3 is 1.4v, output 2 is 2.83v. voltage across R23 is 0.532v, output 1 is 1.08v. Seems like R3 is in parallel with a 60K resistor, does that mean two op amps' inverting and non-inverting pin are tied together? I am using a dual op-amp.

Answer 1

... when 72v applied to the 30:1 voltage divider ...

It's not 30:1, it's 31:1. \$ V_{OUT} = V_{IN} \frac {10k}{200k + 100k + 10k} \$.

... voltage across R1 is only 1.727 V instead of 2.4 V.

But you have a second voltage divider in parallel with R1. "R1" is now effectively 7.5 kΩ. The voltage at R1 will be \$ V_{OUT} = V_{IN} \frac {7.5k}{200k + 100k + 7.5k} = 72 \times 2.44\% = 1.75 \; V\$ and the voltage into the non-inverting input will be 2/3 of that = 1.17 V.

Im (sic) thinking its because of bias current ...

I'm thinking it's because of biased thinking ...

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^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. OP's second attempt.

In the standard inverting and non-inverting op-amp amplifier configurations the output will adjust until the difference between the two inputs is very close to zero. For practical analysis they are connected together as shown in Figure 1. Now it is clear that you have 20k in parallel with your 10k of R1 giving 6.67k.

Running the calculation again the voltage at R1 will be \$ V_{OUT} = V_{IN} \frac {6.67k}{200k + 100k + 6.67k} = 72 \times 2.17\% = 1.56 \; V\$

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Your question 2 circuit is a bit of a mess and I don't think I could analyse it for you. I think you are trying to combine a traditional voltage divider on an non-ground referenced source and feed this into a differential amplifier. I doubt that this is necessary.

Figure 2. Just use the differential amplifier itself.

For example, setting R1 and R2 to 100k and Rf and Rg to 1k would give you a differential amplifier with a gain of 1/100. Your 72 V input on the original circuit would then give an op-amp output of 0.72 V (if it can swing that low).

Why complicate things?

Answer 2

Another problem you may run into with your circuit is the input common mode range of the opamp.

For this device the input must not go closer to the positive supply rail than 2.6v to guarantee correct operation. If the supply rail is 5v this is equivalent to 2.4v relative to ground - if your supply rail is slightly low (it won't be exactly 5v) the max voltage may be lower. A typical device will work better up to about 3v.

From the calculations done by @Transistor you do have margin but it is something to be aware of.

There are a number of devices with improved input common mode range - often referred to as "Rail-to-Rail inputs" - I quite like the LMP7701 range that have very good offset voltage, extremely low bias current and can operate with a supply up to 12v.

Answer 3

This is a classic mistake, as you are assuming that your difference amplifier circuit is acting as a high impedance load to your signal source and front end circuitry. It doesn't, there is a (relatively) low impedance path through R4 and R9, as well as through R3 and R8. This is similar to how non-inverting amplifier circuits (such as simple voltage followers) act as high-impedance buffers, but inverting amplifiers do not. Transistor already did the resulting additional voltage divider math for you, but here is another example with an LTspice and Mathcad calculation comparison for an inverting amplifier circuit.

https://eewiki.net/display/Motley/Analog+Bits+-+Analog+Combinator+Circuit+for+Load+Cells#AnalogBits-AnalogCombinatorCircuitforLoadCells-SufferingFromaLackofBuffering

Also, I'm not sure why you are using a difference amplifier at all since you have one side connected to GND anyway. I of course don't know all the details of what you are trying to do and why, but at first glance I would think you could just use a simpler non-inverting amplifier circuit.