# Buck converter quiescent current is too high

My circuit uses a LMR50410Y3FQDBVRQ1 buck converter to convert an input voltage from a LiPo battery to a 3.3 V supply voltage.

The first batch of the PCB uses the 3.3 V fixed output variant (LMR50410Y3FQDBVRQ1), so I don´t assemble R203 and R204. R203 is replaced by a zero ohms resistor and R203 is unassembled. I have connected the PCB with my laboratory power supply and a multimeter for current measurement and my multimeter (and the power supply) displays a ~7 mA supply current.

Why does the design draw that much power without a load? Texas Instruments writes something like a few µA in her datasheet.

• A really long shot but is your inductor saturating? Does it work with some mA of load on the output? May 4 at 12:53
• mmh is there a way to check if the inductor is saturating? I use the type 1239AS-H-3R3M=P2 from Murata. May 4 at 12:59
• No issues there. More long shots: fake component? Something in parallel fooling you to believe the TI circuit is to blame? May 4 at 13:08
• I don´t think so. I have ordered the components from Mouser (or Digikey). May 4 at 13:09
• Super poor layout? May 4 at 13:40

Read spec again (2)

That is the non-switching current spec. 120 µA

What are your design specs? {must haves & nice to haves} idle, load , sleep

I'm not sure how FPWM mode is controlled, but that would explain your idle efficiency.

12V FPWM Graph below shows 5% eff @ 1mA meaning it takes 20mA to supply 1mA

• Indeed the Shutdown quiescent current is at most 10 uA, however that is under the condition that $V_{EN}$ = 0 which is not what you (OP) have. May 4 at 12:12
• Texas Instruments write on Page 6: Input supply current when in regulation = 90 µA @ VIN = 12 V and VOut = 3.3V and RFBT = 1M. So I expect that the idle current of my device is somewhere around a few µA. May 4 at 12:14
• Are you using FPWM Version? May 4 at 13:08
• Yes. I use the version LMR50410Y3FQDBVRQ1. May 4 at 13:17
• that might explain it (-1) May 4 at 13:18

I'm really sorry to ruin your day, but you really need those resistors...

Maybe this will help

• Texas Instruments write in her datasheet: Feedback input to the converter. Connect a resistor divider to set the output voltage. Never short this terminal to ground during operation. For fixed output version, directly connect to output point. So I have to connect FB with SW and leave the ground connection open May 4 at 12:51
• The datasheet you posted does not work for me, and I only found a version where the last part is not mentioned. Try to short FB with SW and check the results. May 4 at 13:02
• FB and SW are shortened because otherwise the DC/DC isn´t regulating (I have accidentally left the connection open last week). May 4 at 13:08
• that's only 50 uA into the resistors not 7 mA.. Not the correct answer May 4 at 14:25

The device you are using does not have the power-saving PFM mode, so it will always run in FPWM (Forced PWM) which explains the high idle current. Solution is to use the device with PFM mode.

Also on your layout none of the caps are connected to ground plane on L2, and the area of the hot loop is not optimized, so expect trouble with noise and emissions.

• I'm glad bob agrees with me May 4 at 14:00
• What do you mean by "none of the caps are connected to the ground plane on L2"? L2 isn´t connected to the ground. And what do you mean by "hot loop"? May 4 at 14:47
• L2 = Layer2, not inductor, sorry about lazy fingers abbreviating lol May 4 at 15:42
• hot loop is slang for "loop where high di/dt AC current flows", it is the most important current path in a DC-DC converter, and it should get top priority for layout. analog.com/en/technical-articles/… May 4 at 15:44