LP2981A linear regulator datasheet says it needs a resistor in series with C(out)?

Dear Engineers,

We are using the LP2981A linear regulator (5V input, 3V3 output)

The load is a STM32F042K6T6 micro and a TCAN332 CAN transceiver.
The LP2981A datasheet on page 8 says that a ceramic output cap must have a 1R resistor in series with it….

….On our board, which just got sent off for manufacture, we don’t have this. We believe that since our load is only some 10mA, (nowhere near the 100mA full load), we should be OK with the 1uF Ceramic cap that we have on its output? (we didn’t place a series resistor)

We’ve tried to change to a SOT23-5 linear regulator that is able to drive ceramic output caps (without needing a series resistor) but there’s none with the same footprint.

Do you think we’ll be OK due to our light-ish load?

LP2981A-33DBVR datasheet
https://www.ti.com/lit/ds/symlink/lp2981a.pdf?ts=1598302002553&ref_url=https%3A%2F%2Fwww.ti.com%2Fproduct%2FLP2981A

Hi,

according to Fig. 7 on page 11 you might be lucky. As you can see a low ESR is rquired for the low output current. Nevertheless, as you can see by comparing Fig. 7 with Fig. 8 by lowering the output capaciance also the required ESR is increased. Thus for a 1 µF capacitor an even a higher ESR might be required (Fig. 7 is for 3.3 µF).

(i) Check the ESR of your current installed capacitor. Is it high enough, let's say in the range of ~50 m\[\Omega\] (it's hard to estimate and scale a value due to the "strange" log-subdivision) it might be a "reliable" solution.
(ii) Better solution: Increase the output capacitor to 3.3 µF or 10 µF and chose one with an ESR falling in the shown range. As the datasheet provides data for this case, you are on the save side.

BR

Would it solve things if we solder a 1R resistor in series with the output of the LP2981A?

Hi,

not in series with your output. This resistor has to be placed in series with your output capacitor. Thus it has to be installed with one pin connected to the output and the second pin to the capacitor, where the second pin of the capacitor is connected to ground.

BR

what am i missing?
Figures 5 through 8 page 11 each show two curves
The scales look the same on all four graphs (0 to 100 mA, and 0.01 to 100 ohms, ESR, logarithmic)
what does the upper curve represent? (range about 100 to about 10 ohm)
what does the lower curve represent? (range about 0.01 ohm to about 0.1 ohm)
where does the data sheet specify these representation?

not in series with your output.

Click to expand...

Thanks, there are multiple 0603 output caps. (five).......so can we say have a 1r in series with the 2u2 one...then the four 100n caps can do without this resistor?

Hi
MIC5233-3.3YM5 seems to be ok with any Cout >2u2......i believe we will change to this.

Hi,

wwfeldman said:

what am i missing?
Figures 5 through 8 page 11 each show two curves
The scales look the same on all four graphs (0 to 100 mA, and 0.01 to 100 ohms, ESR, logarithmic)
what does the upper curve represent? (range about 100 to about 10 ohm)
what does the lower curve represent? (range about 0.01 ohm to about 0.1 ohm)
where does the data sheet specify these representation?

Click to expand...

the curves are not identical in the four graphs. To ensure a proper operation of the LDO, an output capacitor with an ESR is required, which lies between the upper and lower boundery shown in the corresponding figure.

The ESR has to lie within those boundaries for stability reasons, see [1].

[1] https://www.ti.com/lit/pdf/slva115

BR

Hi,

zenerbjt said:

MIC5233-3.3YM5 seems to be ok with any Cout >2u2......i believe we will change to this.

Click to expand...

pinout and footprint are looking good. Enable pin function is identical. This LDO is also able to be operated with a low ESR ceramic capacitor (>= 2.2 µF). So if there are no price issues (TI part is about three times cheaper @ Digikey), the MIC5233-3.3YM5 should be a suitable repleacement (I have not compared dropout voltage, noise, quisient current, ... ).

BR

stenzer said:

the curves are not identical in the four graphs. To ensure a proper operation of the LDO, an output capacitor with an ESR is required, which lies between the upper and lower boundery shown in the corresponding figure.

The ESR has to lie within those boundaries for stability reasons, see [1].

Click to expand...

thank you