I would like to use LM2596 converter. Datasheet says: 'If the selected capacitor's ESR is extremely low, there is a possibility of an unstable feedback loop, resulting in an oscillation at the output.'

Why? Why not good a too low ESR value?


4 Answers 4


The ESR of the capacitor forms a zero at frequency fz = \$\frac{1}{2\pi\cdot\text{ESR} \cdot C}\$, which tends to reduce the stability excessively if the ESR is too low. The ESR also should not be too high, of course, so usually there is a recommended range. Older parts that were designed before high-capacitance ceramic parts were ubiquitous may omit specifying the lower limit. This is equally true of LDO regulators.

You can either add some series resistance to the capacitor (maybe a few hundred m\$\Omega\$) to keep it within the recommended range, or use another type of regulator that is designed to be stable with ceramic capacitors. To confirm that you've got it right, you can measure the phase margin, which ideally might be 60 or 70 degrees, for example by injecting a signal and measuring the response.

If you roll the dice and just throw the cap in there, hoping for the best, you may find that it oscillates at temperature extremes or it will have such a low phase margin that it exhibits excessive overshoot/undershoot.

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    \$\begingroup\$ The zero is located at \$f_z=\frac{1}{2\pi r_CC}\$, please correct your expression as the numerator is missing. Thank you. \$\endgroup\$ Aug 28, 2017 at 5:24

As it seems, the dampening effect of a series resistance (remember what ESR stands for!) is required to make the control loop, that a DC/DC switching converter controller actually is, stable.

Without that, charge might just be stored in the capacitor without significant slew, and then the control might (probably) overshoot quite a bit, leading to a suddenly increasing voltage, leading to swift shutoff, leading to aforementioned oscillations.

I wouldn't worry about it until it actually happens. If it happens, add additional trace length, thin your trace, or use a cheaper capacitor.

Think about it as a minimum resistive loading.

If you're generally worried about that: Pick a different SimpleSwitcher. TI's product page lists viable alternatives right at the top. Rule of thumb: Higher switching frequencies mean lower measurable oscillations after an output filter, so that might be advantageous.


This simple regulator demands that both input and output cap are low ESR for low ripple out. This means ESR*C < 20us. This breakpoint effectively attenuates the ripple but also reduces the AC feedback signal which is amplified with high gain needed to regulate PWM on each cycle.

When this happens, the comparator used on the feedback signal ends up squaring an AC ripple feedback signal with more phase noise used to regulate the PWM output switch. Effectively feedback become a low pass filter with high Q peaking from the high Q or ultralow ESR output cap.

When the ESR*C < 5 us reduces to "ultra"-low ESR , now shorter than the typical switching period, the capacitor becomes an ideal integrator (-90 deg). When you add the phase shift in the internal gain block (GB) and comparator, you dont have much margin left at unity gain where the 180 deg Barkhausen criteria is satisfied for oscillation. (an unsatisfactory result ;)

To solve this , a small shunt cap is used on the feedback R to give a phase lead and high frequency boost to feedback ripple. Then you improve step response ringing from better loop phase margin.

Hope this hand-waving description is understood.


No proof given, but there's also some inductance in the capacitors. When the ESR is low enough, the resonant circuit property caused by the parasitic inductance can jump to remarkable effect. High ESR make the resonant circuit too lossy to cause the unwanted effects.

This story could be inspected by simulating.


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