Why these LM2596 based buck converters consume this much current on their own?

I have one of those lm2596 360 mini mini converters and while stepping down 8.2V to 5V it draws 50 mA on its own, and the coil is getting a little bit warm. Every connection is ok, there is no short circuit or anything. I also have one of those bigger ones with the same chip and after removing the LED onboard it still draws around 8mA. I thought this chip is efficient, but this is ridicolous in my oppinion.


I have also tried an LM317T linear regulator board and it only consumes less than 1mA.
What is going on here?

A buck converter can output almost all the power it's getting from its input (less typically 5-15% losses), which means if the output voltage is significantly lower than the input, the output current can be significantly higher.

But I am measuring the current between the power source (2x18650 batteries) and the input of the regulators. Measuring here 50mA means it is sucking my batteries. Why does it do that?

I don't know what's typical for the LM2596 but of course it depends on the circuit design, and switching regulators typically consume more quiescent power than a linear regulator. It's always "running" and oscillating.

It will have maximum efficiency at or near maximum current-output.

My whole circuit is going to consume about 100mA at max, but mostly 10-20 mA, so I guess I am better to go with a linear regulator.

What you pay is what you get.
Things from China could be counterfeit and/or cheap/old technology.
This one, with a spec sheet , has an idle current of 200uA.
Leo..

tosoki_tibor:
Why does it do that?

Who knows ?

What did the supplier of the device say when you suggested they might be a problem with thier product ?

I have some of these too, but I haven't used them. So with just the batteries connected (no wiring on output), you're reading 50mA?? Have you tried another mini360 module?

EDIT - I have the mini360 2A version. I connected it to my power supply at 8.3V and adjusted it to get 5.03V output. Then I measured the current at 3mA without anything connected on the output side. I used a Fluke 27 DMM so I'm pretty confident in the results.

tosoki_tibor:
I also have one of those bigger ones with the same chip and after removing the LED onboard it still draws around 8mA. I thought this chip is efficient, but this is ridicolous in my oppinion.

The datasheet for the LM2596 quotes quiescent power consumption of nominal 5mA, maximum 10mA, so 8mA
is entirely normal (some of that is inductive loss no doubt). Its a 3A chip, 8mA is a fraction of a percent loss.
Not ridiculous at all. A linear regulator at 8V in 5V out and 3A would be losing 9W and be 60% efficient or so.

For 20mA load you'd need a regulator with very low quiescent current, whether linear or switched. A 3A regulator is clearly the wrong choice for 20mA anyway. Something with a max current of 150mA or thereabouts
would make sense.

You wouldn't have justification to complain about fuel consumption if you put a truck engine on a golf cart would you?!

AFAIK all switching regulators need a minimum load on the output.
Can you try to put a 10K resistor on Vout and GND and measure again?
See if it is changing anything

The LM2596 will work with no load. Hard to find this in the datasheet though, but its right at the end of
page 21: http://www.ti.com/lit/ds/symlink/lm2596.pdf?&ts=1589293761359