Optocoupler Output Question

First, the load that these can switch is only 120mA on the output. Also, you need to limit the current on the input side (LED) so you don't burn it out - 10 to 20mA to the LED (coil substitute). It will start to switch with as little as 0.5mA but I'd make sure it's completely on, to keep the mosfet as cool as possible. Typically a properly sized resistor is enough (value depends on the supply voltage you're using.

If you're switching an AC load, use Pins 4 and 6 as a replacement for the switch of a relay. Ignore pin 5.

The device will NOT act as a rectifier as-is. One MOSFET will switch and the other MOSFET will allow current to flow through the "body diode" since it will be flowing opposite in direction for which a MOSFET can act as a switch since the two MOSFETs in the device are in a "tail-to-tail" arrangement. Note that the next AC half-cycle will cause current to flow in the opposite direction and the two MOSFETs will switch function (one will be the switch and one will be the non-functioning as current flows through the body diode).

If you're switching a DC load, use Pin 5 as the lower voltage side of your switch and pin 6 as the higher voltage (connected through your load to positive supply).

In the DC case, again,
Low- side switching - connect pin 5 to ground and pin 6 to your load. Then connect your load to positive supply.
-or-
High side switching - connect pin 6 to your positive supply and pin 5 to your load. The. Connect the other side of your load to ground.

You do not normally make any connection to pin 5 when using AC. On DC, connect pins 4 and 6 together and that will be the positive side and pin 5 will be the negative side.

On AC there will be two MOSFETs in series so the Rds(on) will be doubled. On DC there will be two in parallel so the Rds(on) will be halved.

There are two MOSFETs so that the current will be blocked in either direction when the MOSFETs are turned off if used with AC. One body diode would conduct in on direction, so two are needed in series in opposite directions to block the AC

When the MOSFETs are turned on, they conduct in both directions. Normally there will be a large enough gate voltage and a small enough current that the voltage drop will be so small that the body diodes don't conduct.

In the AD6C112, the Rds(on) is quite large (20Ω) and so at the maximum current of 120 mA, the body diodes will conduct. At lower currents, the voltage drop will be due to the Rds(on) only. There will be no additional voltage drop from the body diodes.

The MOSFETs themselves do not rectify, and they work just as well forwards as reverse. In the reverse direction, if they do not conduct enough and the voltage rises to 0.6 V or so, the body diode will start to take current.

MOSFETs are often used in reverse as a way of making a rectifier with a lower voltage drop than a diode would have. The LM5050 is an IC that is designed to make precision rectifiers using N-MOSFETs in reverse.

https://www.ti.com/lit/ds/symlink/lm5050-1-q1.pdf