How to estimate the DC error for this input and opamp?

Question

I need to amplify a 0 to 2.5 V range DAC output to obtain a -5 to +5 V range final output and I have some of these opamps to use. In my case the DAC output to be amplified looks as follows:

Looking above plot, it means the DAC output will be in stairstep fashion stepping with min 100 to 500 μs intervals and with 2.5 mV increase at each step.

I might also need -2.5V to +2.5V output. So my question is not about the circuitry but minimizing the error causes by the opamp input offset voltage.

It says in its datasheet it is a precision opamp with max 100 μV input offset voltage and can be used for DAC output amplifier in the first page. But how would this opamp input offset voltage affect the output signal in my case (2.5 mV input increase). How much error should I expect between consecutive steps(assuming the DAC output DNL is perfect)?

Answer 1

Think of an op-amp's input offset voltage as being an extra voltage source in series with the one you're trying to amplify. Whatever the closed-loop gain of your amplifier, that gain will apply to the sum of your signal and this offset.

Since you wish to map the input range of 0.0V...+2.5V to an output of -5.0V...+5.0V, the relationship between input and output will be:

$$ V_{OUT} = 4 \times V_{IN} - 5V $$

The op-amp's input offset voltage is effectively a source in series with your DAC voltage source, so you can modify that relationship to include the offset as follows:

$$ \begin{aligned} V_{OUT} &= 4 \times (V_{IN} + V_{OFS}) - 5V \\ \\ &= 4 \times V_{IN} + \overbrace{4 \times \underbrace{V_{OFS}}_{\text{in offset}} - 5V}^{\text{Total out offset}} \\ \\ \end{aligned} $$

Your amplifier will have a gain of +4, and you should expect the input offset voltage \$V_{OFS}\$ of the op-amp to be multiplied by this value, also. For example, if your op-amp's input offset voltage is a constant -50μV, this would appear as a constant 200μV downward shift in output potential, regardless of the input.

Since you will somehow have to add -5.0V to the output, you could take the opportunity to make this (deliberate) offset adjustable, to compensate for any offset the op-amp itself is responsible for. Here's a circuit with the input/output relationship you require, with an artificially induced input offset voltage of 200μV (simulating op-amp input offset voltage, large enough to be visible on a graph, but larger than your OPx84 will exhibit):

^{simulate this circuit – Schematic created using CircuitLab}

The actual input/output relationship of this circuit is:

$$ V_{OUT} = 4\times (V_{IN} + V_{OFS}) - 3 \times V_{ADJ} $$

You can alter \$V_{ADJ}\$ to vary the output offset, and simultaneously compensate for the op-amp's input offset voltage. I've set \$V_{ADJ} = {5 \over 3} V \$ here, with no attempt to compensate for \$V_{OFS}\$, because I want you to see the vertical shift of output due to \$V_{OFS}\$. Here are the graphs of input and output:

As you can see, each step in the output is greater than the desired value by an amount \$4 \times V_{OFS} = 4 \times 200\mu V = 800 \mu V\$.

Probably the worst problem you have to overcome is the variance of \$V_{OFS}\$ with temperature. If you knew what temperature the op-amp was to operate at, and that this temperature will not change, then \$V_{OFS}\$ will remain more or less stable and fixed over a long period of time. However, unless the device claims to have "temperature compensation" (or some other form of compensation such as chopping), no op-amp's input offset voltage remains fixed over a range of temperatures.

ADDENDUM - Input Bias Current

I feel I should also mention another source of offset which you will encounter using this OPx84 device. It has a rather high input bias current, perhaps due to it having a BJT input stage. Well, I say high, at 60nA it's not too bad, but it's huge compared to FET input devices.

Essentially that is the current that each input (inverting and non-inverting) draws, and that current has to come from whatever source is providing that input. In the case of the inverting input, the potential there is provided by R1 and R2. What the inverting input "sees" looking at that source is the Thevenin equivalent resistance, which in this case is the combined resistance of R1 and R2 if they were connected in parallel:

$$ \begin{aligned} R_{TH} &= R_1 \parallel R_2 \\ \\ &= {{R_1 R_2} \over {R_1 + R_2}} \\ \\ &= {{30k\Omega \times 10k\Omega} \over {30k\Omega + 10k\Omega}} \\ \\ &= 7.5k\Omega \end{aligned} $$

While sucking 60nA out of that source, a voltage develops across that "effective" 7.5kΩ which introduces another source of offset error to the system. By Ohm's law, this additional offset will be approximately:

$$ \begin{aligned} V_{IOFS} &= 60nA \times 7.5k\Omega \\ \\ &= 450\mu V \end{aligned} $$

Thankfully, the solution is simple enough; add that same source resistance in the path of the non-inverting input, too. Symmetry of the op-amp's "modified long tailed pair" input stage means that both inputs will draw roughly the same current, and by equalising the source resistance into both inputs, both inputs will suffer the same voltage offset. The net effect is a cancellation of any input offset due to input bias currents:

^{simulate this circuit}

Answer 2

Supposing that you're feeding the DAC output through the non-inverting input of the Op Amp, your input offset will be amplified by your gain.

To get a \$[-5V,+5V]\$ range from \$[0V,+2.5V]\$ you're going to need a gain of 4. So your offset seen at the output will be \$400µV\$ (max) no matter your input voltage. (Resp. offset by \$200µV\$ for an output in the range of \$[-2.5V,+2.5V]\$ ).

At any gain, your output between consecutive steps will be offset by \$ 4\%=\frac{2.5}{0.1} \$ in a worst case scenario.

This is all true as long as the temperature stays at room temperature (\$25°C \$). If the temperature was to vary, your offset will drift a bit. This is defined by the offset voltage drift defined bellow :

In the worst case, your output drift will drift by \$ 2\% = \frac{2}{100} \$ each 1°C.

However this offset voltage drift seems to be generalised for all the 4 Op Amps cited in the datasheet. As the first one (which is the one you're using) has only a maximum of \$ 200µV \$ input offset at 125°C :

So, to me, it's safe to assume that your maximum offset voltage drift is \$1µV/°C\$. So your output drift will drift by \$ 1\% \$ each 1°C.

As Antonio51 commented, you can always use an AZA : Auto Zero Amplifier (or an Opamp where you have access to the Offset Null terminals)