I have a circuit:

enter image description here

How should I choose the value for the capacitor in offset? How large should it be?


2 Answers 2


I'll refer to this following circuit, since yours has no labels for the components in question:


simulate this circuit – Schematic created using CircuitLab

With the values shown, the DC operating point (quiescent state) of X is:

$$ \begin{aligned} V_X &= 3.3\frac{R_1}{R_1 + R_2} \\ \\ &= 3.3\frac{3.9k}{10k + 3.9k} \\ \\ &= 0.93\rm V \end{aligned} $$

Note: I haven't considered diode "dark current", which will influence the average output level of the amplifier somewhat, but the principles which follow are still valid. Keep in mind that values for R1 and R2 will be chosen to obtain some "average" output level, a level which should also be adjusted to account for any permanent output offset due to the diode's dark current through Rf.

The larger C1 is, the longer it takes for the output of the entire system to arrive at its DC operating point, where \$V_X=0.93V\$, since C1 must first charge to that voltage, via R1 and R2.

To know how long, we can use Thevenin's theorem to derive a simple (single-resistor) equivalent to replace R1, R2 and the source 3.3V. This "Thevenin resistance" will be the the value of R1 and R2 as if they were connected in parallel:

$$ \begin{aligned} \frac{1}{R_{TH}} &= \frac{1}{R_1} + \frac{1}{R_2} \\ \\ R_{TH} &= \frac{R_1R_2}{R_1 + R_2} \\ \\ &= \frac{3.9k \times 10k}{3.9k + 10k} \\ \\ &= 2.8k\rm \Omega \end{aligned} $$

Now the equivalent circuit, which behaves exactly as the original did, looks like this:


simulate this circuit

It becomes clear that Rth and C1 form a low pass filter, and have a time constant (which determines the rate of charge of C1) of:

$$ \begin{aligned} \tau &= R_{TH} \times C_1 \\ \\ &= 2.8k \times 1\mu \\ \\ &= 2.8\rm ms \end{aligned} $$

A rule of thumb is that the time it will take C1 to charge to within 1% of some voltage will be about five times the time constant:

$$ \begin{aligned} T_{99\%} &= 5 \tau \\ \\ &= 14\rm ms \end{aligned} $$

What this means is that the signal at OUT cannot be considered "stable" until at least 14ms has elapsed (after power is applied and has itself settled). That's how long it will take C1 to charge to 99% of its final operating point of 0.93V. Here's plot of \$V_X\$ vs. time, during the first few milliseconds following power-on:

enter image description here

The purpose of C1 is to stabilise the potential at X, to remove as much noise as possible caused by R1, R2 and the 3.3V power supply. It can do this because C1 and Rth form a low-pass filter with a cut-off frequency \$f\$:

$$ \begin{aligned} f &= \frac{1}{2\pi R_{TH}C_1} \\ \\ &= 180 \rm Hz \end{aligned} $$

As long as this cut-off frequency as significantly lower than any unacceptable noise you expect to find at X without C1 present, then it will do its job just fine. Since \$f\$ decreases as C1 increases, there's really no upper limit to C1 in this respect.

For instance, if the 3.3V power source is provided by a switching regulator, you can expect it to contain some rather nasty noise in the tens or hundreds of kilohertz, but as we calculated here, the cut-off frequency of 180Hz will ensure that very little of that noise will make it to node X.

Therefore, cut-off frequency is of little concern here, and our choice of value for C1 is influenced much more by our need for the circuit to settle at its operating point within a reasonable time.

If you are using an ADC to sample the signal, for instance, you shouldn't begin sampling (or at least you should disregard samples) until at least 14ms has elapsed since power was applied and stabilised.

In conclusion, since the larger C1 is, the more noise it filters out (or rather, the lower the cut-off frequency, and the wider the band of attenuation of potential noise), you should aim to make C1 as large as possible, while maintaining an acceptably short settling time.


This is used to reduce the noise on the + pin; this in turn prevents the opamp from amplifying the noise.

There is no precise answer and no component values listed, but a time constant of 10's of ms might be reasonable; use 10 uF.


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