I have a printer motherboard with the LM2596 IC. The older version of the motherboard had one 100uF capacitor at the output pin of LM2596 but LM2596 recommends that 220uF be used. Because a cheap 100uF capacitor was used, there were issues with the printer such as temperature fluctuating.

People that had issues replaced the 100uF with 220uF and the issues were solved. I had the-same issue then opened up the device to replace the capacitor. After making this post, I realized that the motherboard version I have came with a supposed fix for this issue. The new motherboard uses two 100uF in parallel to fix the issue. Those two 100uF parallel capacitors should only give 200uF not the recommended 220uF. Even with this fix, I still get temperature fluctuating sometimes.

Is there a problem if I replace the two 100uF parallel capacitors with two 220uF? Is this going to cause a problem? I mean, that would be 440uF.

If that's going to be a problem, how about removing the two 100uF parallel capacitors and put just one 220uF?

Most post I found also mentioned that the a poor quality capacitor is used for LM2596 IC is used and this should be replace so that's why I feel like I have to replace those caps with a high quality one.

  • 1
    \$\begingroup\$ Keep in mind that electrolytic capacitors, which your 220μF ones almost surely are, have very wide tolerances on them. ±20% is normal, and it's not uncommon to see even wider ones. So a nominally 220μF capacitor could be actually 200μF and still be within spec. \$\endgroup\$
    – Hearth
    Nov 22, 2018 at 17:51
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    \$\begingroup\$ What is the part number on the onboard capacitors. Hard to see a manufacturer changing a design without correcting a known problem. \$\endgroup\$ Nov 22, 2018 at 17:55
  • \$\begingroup\$ @Felthry I am ware of that. \$\endgroup\$
    – Programmer
    Nov 22, 2018 at 17:57
  • \$\begingroup\$ @StainlessSteelRat You can find that on my other post I linked in the question. It says "100 35v UT" on it. The problem is mostly corrected but still happens sometimes. \$\endgroup\$
    – Programmer
    Nov 22, 2018 at 17:58
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    \$\begingroup\$ Two parallel 100μF capacitors is probably a better arrangement than a single 220μF one, because putting them in parallel reduces the ESR. Your circuit would probably be okay with a higher capacitance there, too; power supplies aren't usually too sensitive to capacitance variations. But I'm not sure you're looking in the right place to fix the problem. \$\endgroup\$
    – Hearth
    Nov 22, 2018 at 18:02

1 Answer 1


I assume that the output terminal of which you speak is not merely a DC power supply. I assume that it serves a control function, in addition to powering something.

If so, then it's hard to say what the effect of bulking up to 440 μF might be. At least two modes of failure are possible:

  1. The output signal at 440 μF might be unusably sluggish. The printer thus might not respond. Reducing the capacitance to 200 μF or 220 μF again would fix this problem.
  2. The 440 μF might draw too much current at the output, storing too much energy too fast. This would not harm the capacitor but it might render the board permanently useless.

Fortunately, #2 is unlikely. Even #1 is fairly unlikely.

Let's talk about #1. The output has an output resistance Rout, and the lower, the better; but for this kind of electronics, unfortunately, it's probably fairly high. The off-board load—that is, whatever it is that the output drives—will have a variable impedance, which means that it draws a variable current, which is why you have a capacitor tapped to a point between: the capacitor supplies the variable current so that the board's other electronics do not have to. The bigger the capacitor, the better it can supply the variable current, so that's good. Regrettably, the bigger the capacitor, the longer the period the output shall want to switch from low to high or high to low.

Switching is not instantaneous. The larger the capacitance, the slower the switch.

The time needed to switch will be approximately t = [-ln(Vmargin/VCC)][Rout][C], where probably 1.0 < -ln (Vmargin/VCC) < 1.5 and where C is your capacitance.

Regarding scenario #2, the potential problem there is that the board's electronics supplying the output must generate switching heat in proportion to the capacitance C. The heating rate dQ/dt will not change, but because the time t is longer, extra heat will be generated each time the output switches low to high or high to low.

Fortunately, outputs like these tend to be designed to handle a lot of switching. Even more fortunate (if you wish to call it fortune) is that #1 limits the effective switching rate. So #1 counteracts #2 to a significant degree.

I hope that some of that makes sense. If you wish to make some extremely rough calculations, then you might assume that Rout = 1.0 kilohm. (For better calculations, construct a voltage divider to measure Rout by Thévenin's technique.)

If you want my recommendation, though, I would be inclined to leave the 200 μF alone, unless you just wish to experiment and do not mind the risk. Otherwise, @StainlessSteelRat is right: "Hard to see a manufacturer changing a design without correcting a known problem."

You say that your version of the board has a supposed fix, but isn't it more likely that the supposed fix is just a fix?


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