I'm using the TPS63050 buck/boost converter in a design. The datasheet basically just says "don't leave it floating" in the table on page 4. I want this enable pin to be under the control of an external circuit, and so I will implement a pull-down resistor on that pin to accommodate the condition where that external circuit would otherwise float the pin. What I can't quite ascertain from the datasheet is, how big a resistor can I get away with for this purpose. I was thinking of populating a 1 MegaOhm resistor to GND, to minimize the static current draw required just to enable the regulator.

Is there any reason I should be concerned, or any likely problem, with using such a large resistance in this context? My gut says it should be fine, but it would make me feel even better to get some positive affirmation or advice to the contrary from my colleagues here on EESE!

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    \$\begingroup\$ Ilkg Input leakage current for the En pin is max 0.1uA. \$\endgroup\$
    – Oldfart
    Nov 19, 2019 at 17:20

2 Answers 2


Since the leakage on that pin is so low (high impedance), you risk the pin being susceptible to interference with such a high value resistor. Something more reasonable like 10 kΩ would probably be better, but still not needed.

This pin is usually just shorted to the power rail. The static current draw will be next to nothing (typically 0.01 µA).

  • \$\begingroup\$ Just for my own education, could you elaborate a little more on why the current being so low (high impedance), makes the pin more prone to interference? Larger value for the pulldown (1M) does create a higher drop (1M*0.1uA)—which could be an issue if the leakage were greater, but what am I missing? \$\endgroup\$
    – Big6
    Nov 19, 2019 at 18:23
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    \$\begingroup\$ Because any current induced will result in a large noise voltage (since it is across a large resistance), possibly enough to flip the state of that pin. \$\endgroup\$ Nov 19, 2019 at 18:25
  • \$\begingroup\$ Got it, makes sense. That would be in addition to the drop caused by the leakage current. \$\endgroup\$
    – Big6
    Nov 19, 2019 at 18:31
  • \$\begingroup\$ Yes, but keep in mind the drop will not be as you calculated because the pin is not a current source. The leakage current given in the datasheet gives you just an idea about its impedance (which is very high). \$\endgroup\$ Nov 19, 2019 at 19:55

You can use a very high-value resistor there, but make sure that the resistor is as close as possible to the EN pin. You want the "high impedance node" (i.e. the metal between the resistor and the EN pin) to be as small as physically possible.

This is because there will be some parasitic capacitance between the switching nodes of the converter and the metal of the EN pin. This will cause some of the voltage switching waveform to appear on the EN pin.

As long as you make sure the metal of the EN node is very small, you can use pretty much any pull-down resistance. This is because the input capacitance of the EN pin is probably of the order 0.1-5pF. As long as the parasitic capacitance is ~10x smaller than this, the switching noise will never be a problem.

So pay close attention to your layout and you should be fine. Don't be afraid to use 0402 resistors.

  • \$\begingroup\$ Unfortunately, I can't get it much closer than 0.3" away from the pin on a 10 mil trace against a pretty solid ground plane below it on a 4-layer board... so I think it's worth rolling the dice, bearing in mind that resistor just needs to bleed the enable pin to GND to shut the system down. \$\endgroup\$
    – vicatcu
    Nov 19, 2019 at 20:48
  • \$\begingroup\$ If you are prototyping it then it would be easy to just pluck out the resistor and put in a smaller one if you run into trouble. \$\endgroup\$ Nov 20, 2019 at 0:13

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