# Using LM2596 step down with LiPo

i'm doing a project with esp32 (3.3v), dht11(3v-5v), soil sensor (3.3v-5v the cheapest Chinese), nrf24l01 (2.6 - 3.6v). I'm using lifepo4 right now (3.32v) but when the battery drop about 3v, soil sensor won't work.

So I'm gonna use lipo 18650 4.2 max, then use step down LM2596 to regulate at 3.3v.
My question is, when i connect battery to 4.2v,2.6Ah --> LM2596 --> 3.3v. do I receive 2.6A at output too? I mean is it always drain 2.6A? or it just drains Operating quiescent current: 5mA.

Sorry if this is a noob question

• The LM2596 is not suitable for this purpose because it has too high a quiescent current at 5-10 milliamps - the regulator itself will waste too much of your battery even when the circuit it is powering is asleep and drawing microamps. Please confine your experiments to safer battery types like a few alkaline AA cells until you've had a chance to develop a more thorough understanding of electrical basics. Jul 21, 2019 at 14:00
• thank you. i will find solution later. but i just confuse, when i run battery with and step down module, for example the load in sleep state is 20mA so the battery will discharge at 20mA + quiescent current (if i use LM2596, then it will be 5mA). so total battery discharge is 25mA ? @ChrisStratton Jul 22, 2019 at 14:15

First of all as per the datasheet of LM2596 needs minimum 4.5V as supply voltage. So I think it will not start to output 3.3V even though you connected a fully charged battery.

As you have mentioned in the question the capacity of the battery is 2.6Ah, for most of the lipo batteries they can output a maximum discharge current of about its own capacity, for your case it will be 2.6A, but the battery will not support to discharge at 1C for long time. You have to check and confirm it in datasheet of the battery. Even though the battery is discharging at 2.6A it is will be input of the LM2596 and there will be loss at the voltage regulator section depending on its efficiency. More over when the battery discharges the voltage starts to decline, which will also reduce the current. For better reliability you have to use much higher capacity battery like 3Ah.

• The circuit in question is not a high power one. The asker is confused about the regulator, but both the regulator and the battery are unsuitable for this use. Jul 21, 2019 at 14:34
• Yes. The regulator need mich highers voltage Jul 21, 2019 at 14:35
• "LM2596" chinese modules often do not have a real LM2596, last one I got ran at 50kHz where true LM2596 runs at 150kHz. So... don't look at LM2596 datasheet! Also the caps are 85°C high ESR garbage, it doesn't deliver the specified current, quiescent is high, it overheats, etc. Just do not buy these things. Jul 21, 2019 at 14:49

As others have pointed out you have some issues here, but to answer your main question:

The regulator will not draw 2.6A at all times. It will draw only what it needs to power your circuit, plus some quiescent current.

It's a switching regulator so the exact draw will depend on your circuits draw, and the regulator efficiency.