Electronic – arduino – Voltage drop @24V2A using a MOSFET with Arduino

If you are using a logic IO pin that can drive 3.3V, the current into the opto's diode will be about: -

\$\dfrac{3.3 volt - V_{F}}{330\Omega}\$ where Vf is forward diode voltage of about 1.2V (see opto's data sheet).

This is a current of about 6.4 mA. The current transfer ratio of the opto is a minimum of 50% therefore you can expect to deliver at least 3.2 mA to the emitter of the photo-transistor.

It could be more - according the the opto's data sheet, CTR can be as high as 600%. However, there are other problems to solve. If you look on page 7 of the data sheet you are pretty much limited to only being able to supply 5 mA from the device when the collector-emitter voltage is above 10V and because your supply is 36 volts you are likely to fry the chip if CTR is higher than 100%.

To avoid all of this I'd be tempted to use a voltage regulator from the 36 volts supply - bring it down to 12 volts then use the opto - no need for the zener protection.

You need a commmon ground between the two. Probably your bench supply (-) output is grounded and the other ones are not, and you are getting a ground through the USB cable and your PC, so the grounds are connected through the 3-pin wall plug when you use the bench supply. If you unplug the PC it will probably stop working. You may wish to use a 1K resistor between the two grounds, just in case.

Also (and this is not the cause of your proximate problem), your BUZ11 is really an inappropriate part for that service. It's not a logic-level MOSFET and is only guaranteed to fully turn on with 10V Vgs. You should use a MOSFET that has an acceptable specified Rds(on) with 3.3V or lower Vgs. The particular sample of BUZ11 you happen to have may sort of turn on enough typically, but that is not good engineering at all.