Burr-Brown 3650 isolation amplifier (optical)
Here's an isolation amplifier, that transmits an analog signal across an electrical isolation barrier. This is useful for industrial equipment (where you may need to sense current through a high-voltage conductor and transmit it to an earth-grounded control system, for example), medical equipment (where EEG & EKG leads need to be heavily isolated to avoid deadly shocks through low-resistance connections), specialized measurements, etc. This particular one uses light, and has a bandwidth of 15 kHz. Let's open it up:


You can see a few bare dies here - the 3 largest ones are op-amps. All identical designs, but I wasn't able to figure out the manufacturer and part number:


There's also two silicon capacitors (no images, as they're boring to look at), and a resistor:

The middle section, with the white rectangle covered by the 2 black lids, is where the magic happens. I removed these, and found the optical parts underneath - the lid structure is flipped over, on the right-hand side, so that you can see the two channels inside it:

Here's a closer look at the main substrate:

All 4 of these dies are photodiodes. There are 2 separate channels here, each in one slot of the lid structure to block light from each other and from outside.
What transmits the light? There's 2 blank spaces you can see in the photo above if you look carefully: it turns out the LEDs were pulled off the board when I removed the lid. They're still stuck to the bottom of the lid structure:


It looks like these channels are filled with a transparent potting material, which the LEDs stuck to. The bondwires also stayed in the potting material. This potting allows light to pass from the LEDs (on the transmitter side) to both photodiodes (one of which is on the receiver side), while still providing some high-voltage electrical insulation (rated at 2 kV).
You can also see the negative shapes left by the photodiodes in the potting material, which is kind of fun.
So, we have 2 optical channels, each holding an LED and 2 photodiodes. Let's talk about how this all works:

Each of the LEDs is coupled to 2 photodiodes: one on the transmitting side, and one on the receiving side. This is for linearization: LED brightness vs. current is not linear at all, and varies a lot with temperature. Photodiodes are better with the linearity of their their light-to-current characteristics (as long as you stay far above the dark current), but still not perfect. So if you use 2 matched photodiodes, you can use a feedback scheme where the transmitting side adjusts the LED current until it sees the correct output from its photodiode; the receiving side will then be seeing the same output from its photodiode too.
This is the popular approach behind the
HCNR201 analog optoisolator IC, and a few similar ones: the theory of operation and example circuit sections of that datasheet are very much worth reading if you're interested in details. I've looked into this approach pretty deeply, back when I worked on power electronics and was frustrated by the lack of good high-side gate drive probing (but didn't get much of anywhere with these off-the-shelf parts because the bandwidth is low and the CMRR is not great).
The reason for the 2 separate channels, is because the input signal is bipolar (can be positive or negative), while each LED/photodiode set can only work with one polarity. D1, D3, and D5 handle the positive signals, while D2, D4, and D6 handle the negative signals.
Even though the output is a voltage, the input is in the form of currents: if you have a voltage input, just connect a resistor in series. The currents are effectively summed at the inputs of U1, with some kind of RC snubber to tune the frequency response I guess. Putting the LEDs in-line with the op-amp power supply pins is a clever approach: this eliminates the need for separate LED-driver circuitry that eliminates the crossover distortion between positive & negative. R1 provides a dummy load on the op-amp's output so that its supply currents will be roughly proportional to its output voltage. Because the positive and negative photo-currents are subtracted on both transmitting and receiving sides, the LED currents also don't have to be precise: the feedback loop corrects for everything.
For a vague idea of how it works on the receiver (output) side:
D6 conducts a photosensitive current through R9 that creates a positive voltage, proportional to the D6 current. U3 buffers this voltage. (Connecting D6's cathode to this buffered voltage, instead of Vcc, is done to maintain 0V across it and therefore mostly remove any bandwidth-limiting effects of D6's capacitance, if I remember correctly). U2 then looks at the difference between D6's "negative output" voltage, and D5's photo-current through R10, and generates the output voltage based on that.