OK, in the truck case it was a very low impedance power source that could sink as well as source current. Most supplies are not intended to do that. A direct battery connection is the one common exception.

... interesting... I can easily jump the diode and re-test the circuit. See if it makes any difference.

 

hat's interesting ... L1 and C1 are there for obvious reasons. But the diode is (supposedly) there in case the power supply falters for a moment, so that C1 feeds the load (in your diagram, shown as R1) and does not discharge itself back into the power supply.

Diode should be not incorporated in the middle of the filter, as RF functional, but should be connected before filter.
Such diode already exists - D_PS.

 

Diode should be not incorporated in the middle of the filter, as RF functional, but should be connected before filter.
Such diode already exists - D_PS.
View attachment 185047

You were absolutely right, Danko. I visually inspected the wall wart's circuit (power supply) I'm using, and sure enough, it had a diode and a cap at its output as you've shown. So a diode in my circuit is really of no use, and only causes a detrimental voltage drop. But the capacitor it has is rather small. 470 µF to be precise. And it has no inductor at its output, although it does have a common-mode choke at its input.

So in my circuit, I'm doing away with the diode, but keeping the fat 4,700 µF cap... and the inductor, of course.

Question. I have two different inductors with me. One's rated at 2.7A and 20µH. The other one is 3A and 50µH. My circuit is only drawing 1.5A. I'm currently using the 20µH one and it's been working fine. Which inductor would be best to use? The one with the largest inductance? Or would using a larger inductor run the risk of backfiring in some way?

 

Which inductor would be best to use? The one with the largest inductance? Or would using a larger inductor run the risk of backfiring in some way?

Larger inductance is better (lower ripple) and nothing bad will happen.