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I have a signal that is roughly 0.2V + noise fluctuations of order 0.1-2 mV. Ideally I want to amplify this signal such that the mV fluctuations become about 1V. In other words I want to amplify the signal by about 1000x.

However, if I flat out amplify the signal, the total signal becomes 200V + 1V fluctuations, which I can't reasonably read on some bench top DAQ (0-10V range).

Is there some combination of circuit elements that can take my input 0.2V + 1mV signal and spit out only the amplified fluctuations (i.e. 0V + 1V fluctuations)?

edit: I should say that these fluctuations are controlled by me physically squeezing a pressure gauge, so they aren't necessarily high frequency. Basically the signal rises to 0.202V when I squeeze, and 0.200V when I let go. I want to see that excess 0.002V blown up to 1V, but I may be squeezing and letting go slowly in general.

Capacitors block DC and pass AC.

You can use a series capacitor into an opamp with whatever gain you need.

Even better might be a simple RC high-pass filter...One capacitor (series) and one resistor (to ground) in front of your amplifier.

Like this:

^{simulate this circuit – Schematic created using CircuitLab}

R2 and R3 set your gain. C1 and R1 set your low frequency cut-off. The formula you use to find the cutoff is:

$$Ftext(Hz) = frac12 pi R C$$

Use a coupling capacitor prior to the amplifier. The DC signal will be blocked but the fluctuations will pass through.

Digital designer here so I'm not certain, but...

The other answers assume high-frequency fluctuations. Instead you want to subtract the 0.2 V and amplify that. You can use a summing amplifier to subtract the offset, if you've got positive and negative supply voltages. I think you can also use an inverting configuration where the non-inverting input is at 0.2V instead of ground.

Sure, just an ordinary inverting op-amp can do that:

^{simulate this circuit – Schematic created using CircuitLab}

Remember that an op-amp wants to make its inputs the same. So if you put 2V on the non-inverting input, and the signal input is also 2V, the output will be 2V.

But say the signal input is 2.1 V. The op-amp wants to make the non-inverting input also 2V, and will have to drive the output higher than 2V to make that happen due to the voltage divider action of R1 and R2. The selection of these resistors thus sets the gain.

Keep in mind any source impedance will effectively add to R2, so if your sensor doesn't already have a low-impedance output, you may want to buffer it.

You have a couple options for realizing V2, since you probably won't want to find a 2V battery. Since the op-amp's input impedance is quite high, this doesn't need to be a low impedance source, so it could be as simple as a potentiometer across the power supply. Of course this will make the circuit somewhat dependent on the supply voltage, and the small but non-zero input current to the op-amp will introduce some error, so if you require high precision you might find an adjustable voltage regulator more suitable.

Here's something inspired by the first 2 answers. Make a 10-second low pass filter of the input signal and feed that into an op-amp's non-inverting input (+). Then take a 1-second high pass filter of the same input signal, and feed that into the inverting (-) input of the same op-amp.

Fluctuations get subtracted from the average and amplified a lot. If it's too much amplification, a resistor in series with C2 will lower the gain. This also inverts the fluctuation signals. If you want them non-inverted, follow this with a gain of -1 inverting stage.

^{simulate this circuit – Schematic created using CircuitLab}