1
\$\begingroup\$

I am confused about the input common mode voltage seen by an op amp configured as a differential amplifier for high side current sensing. Assuming that the op-amp and the load for sensing share the same ground reference (fig. 1), it is clear (i think) what the input common mode voltage is.

schematic

simulate this circuit – Schematic created using CircuitLab

(15V+14.9)/2=14.95V

For the majority of op-amps this would violate the input common mode range given a 15V +Vcc. However, what if the op-amp is powered by a separate source (fig. 2) with a separate ground reference? In this case, I initially think that the op amp does not see the large input common mode voltage and instead only sees the potential difference across the shunt resistor. My reasoning is that the amplifier is floating and so from the amplifier's perspective Vcm=(0.1V)/2, but I am not sure if this is true. If not, how would the input common mode voltage be calculated here (fig 2)? I can't seem to wrap my head around this.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ In the first case the op-amp 'sees' 13.63V on the non-inverting input and the same (at balance) on the inverting input. The differential amplifier has a common mode voltage of about 15V and is very sensitive to errors in the resistor ratio matching. You can make the first circuit 'work' for most op-amps just by increasing the positive supply voltage by a few volts, no need for a floating supply which could cause other issues in your second circuit. It's better to use a transistor to level-shift the output and perhaps another buffer op-amp. The requirement for op-amp input voltage range.. \$\endgroup\$ Commented Oct 17, 2023 at 6:41
  • \$\begingroup\$ .. near V+ remains, and there are not many op-amps that are guaranteed to work like that (and they tend to be expensive if they will accept +/-15V supplies). Posted as a comment because the second circuit will likely sort-of work (with or without an actual connection between GND_ref points- but with different outputs) but it's not a good approach. \$\endgroup\$ Commented Oct 17, 2023 at 6:45

2 Answers 2

0
\$\begingroup\$

Your 2nd circuit, the one with a floating GND between the batteries works (assuming it really stays floating) I'm afraid many of us do not believe it stays floating in practice. For ex. V3 may be not constant and there can be some coupling (capacitance, leak, circuits which are added without thinking at first) between the grounds.

Let's assume the GND ref point stays floating. If it happens that the current measurement result is needed in the circuit which has the same ground as V3, you must build an isolated bridge which transfers the result over the barrier between the floating and non-floating sections. That adds some complexity.

My suggestion: Change R1 and R3 bigger or reduce R2 and R4. Then the opamp may well handle the case. Add an other opamp for more gain if that's needed.

As hinted already in a comment, simulate or calculate otherwise how precise R1...R4 you actually need i.e. how much error resistor tolerances may cause. Such analysis is needed to reveal if the circuit is impossible in practice. Sometimes inserting an adjustable part helps, but even it can be sometimes useless.

\$\endgroup\$
0
\$\begingroup\$

One way to avoid the problem of getting too near the high voltage rail is to separate the functionality into two stages. The first stage simply shifts the differential input voltage to a ground reference. The second state amplifies.

Here is an example:

schematic

simulate this circuit – Schematic created using CircuitLab

The common mode voltage at the non-inverting and inverting inputs of OP1 is now 7.5 V. The common mode voltage at the non-inverting and inverting inputs of OP2 is 99.41 mV, and the output voltage is 1.093 V -- all safely within the recommended operating range for probably all op-amps capable of taking +/- 15 V rails.

Some advantages to this approach might include

  • There are commercially available op-amps with 4 same-valued laser trimmed matching resistors in differential-amp configuration. This makes high common mode rejection ratio easier.

  • The additional components (1 op-amp and 2 resistors) is likely to be cheaper and easier to add to your project than another power source

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.