Am I misunderstanding how Buck converters function?

[cbc02009]

So I built a breadboard prototype of a buck converter I will be using as part of a larger project, but I'm not sure if the results I'm getting make sense, and I was hoping someone could sanity check for me, as the buck converter is stepping down the voltage just fine, but doesn't seem to be stepping up the current.

The converter schematic is attached below. Both the VCC and the V_in are coming from a USB 5V supply. The PWM is supplied by an MCU.

I think it has something to do with the load, since it's just a plain resistor, and I = V/R, so the load essentially dictates the current coming out of the converter. I get that the buck converter is useful because the output voltage will always be duty_cycle*V_in, regardless of the load (within reason), but I feel like either I made a mistake in the design for the efficiency to be so low, or there's just something fundamental that I'm missing.

LTspice simulations match what I'm getting for results, as well, which makes me think I'm just not getting something. Apologies if it's something blatantly obvious that I'm missing.

[Benta]

At only 4.6 mA out, the current used to just keep the circuit running is a large part of the input current. The efficiencies look OK to me. Try with a smaller load resistor and see what happens.

[cbc02009]

At only 4.6 mA out, the current used to just keep the circuit running is a large part of the input current. The efficiencies look OK to me. Try with a smaller load resistor and see what happens.

Thank you, I reduced the load to 10ohms and the results are much closer to what I expected them to be.I definitely need a better power supply though, because you can see the voltage gets dragged down a lot with the increasing duty cycle.

So why does the reduced load make so much difference? is it because it's closer to the internal resistance of the converter? Or is it something else?

[Benta]

It's simply because the IC and MOSFET gates need power independent of the output current. If you remove the load resistor, your efficiency would drop to zero, but the circuit would still consume power.

[cbc02009]

It's simply because the IC and MOSFET gates need power independent of the output current. If you remove the load resistor, your efficiency would drop to zero, but the circuit would still consume power.

Sorry, I should have clarified, I'm only measuring the current going into V_in in both tests, not the current going into the chip.