I am in the process of designing a +5V power supply for a hand-held 2xAAA powered device.

When using TI's webench simulator tool to simulate a switcher power supply that I am considering, I noticed on the default schematic for the TPS61099 device that there is a very large capacitor right at the battery input (labeled Cin). It is 150 uF, physically HUGE, and EXPENSIVE ($2.59!). This cap is in addition to the usual moderately sized cap on the input of the switcher (15 uF)

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This is my first battery-powered design, so I must be missing something. I'm used to designing products which run off of a 12 or 24V industrial DC bus, and there was never any need for such a cap on the input.

So why is the cap required? It seems that TI is not the only one to suggest this as a reference design, since this answer shows an LT datasheet that shows a similar setup, also for a battery powered design.

Can I safely leave that cap out of my design? Or is there a particular aspect of battery-powered design that requires enormous input power caps? Why can't I rely on Cinx (15 uF) as the only input cap?

  • \$\begingroup\$ Input capacitors are usually a good idea, regardless of your power source. I would look for a cheaper cap though. \$\endgroup\$
    – Hearth
    Jan 31, 2019 at 3:03
  • \$\begingroup\$ @Hearth I've often used input caps to deal with incoming voltage ripple or ESD or other such transients, but I figured such things are unneeded in a battery powered design, and those caps were much smaller anyway. I would expect that the "Cinx" cap, which is 15 uF, should be enough for most concerns in such a low-power device. \$\endgroup\$
    – rothloup
    Jan 31, 2019 at 3:06
  • 3
    \$\begingroup\$ Some switchers exhibit Negative Input impedance, and oscillate. The big input cap may prevent that. Continue your reading. \$\endgroup\$ Jan 31, 2019 at 3:11
  • \$\begingroup\$ The data sheet shows recommended conditions and a typical application circuit with only 1 input cap of 10µF and 2 output of 10µF each. Odd that web bench recommended 150 & 10 in and 10 out. What options did you choose? \$\endgroup\$
    – Passerby
    Jan 31, 2019 at 3:35
  • \$\begingroup\$ @Passerby I noticed that also. I chose the default options (i.e TPS61099, 0.8 - 4.8 VIN, 5V @ 70 mA out). I thought maybe the application diagram in the datasheet simply presumed that whatever concern the large cap addresses was already addressed to the left of the diagram. \$\endgroup\$
    – rothloup
    Jan 31, 2019 at 3:45

2 Answers 2


The large input capacitor is required for battery operation because of the high pulse current drawn by the boost convertor.

With reference to the datasheet for the TPS61099x you can see that the cutoff current will be between 0.8A and 1.25A.

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Your AAA batteries are extremely unlikely to supply even the minimum of 0.8A when fresh.

Even a fresh Alkaline E92 is likely to see a terminal voltage drop close to 0.2V worst case. As the batteries discharge this will obviously get worse.

The large capacitor allows the battery to recharge it during the switcher off time, and the majority of the on pulse current then comes from the capacitor. You need a really good quality low ESR cap to ensure that it can provide the high pulse current during the on time.

  • \$\begingroup\$ This makes sense. The industrial supplies I was used to designing for were generally low impedance, so that's why I never used to see such large caps on the input. \$\endgroup\$
    – rothloup
    Jan 31, 2019 at 18:37
  • 1
    \$\begingroup\$ @rothloup A big gotcha with most of these chip level SMPS solutions is the input pulse current is a fixed value and there is no way to reduce it. \$\endgroup\$ Jan 31, 2019 at 19:19

The converter input will have negative impedance (it must have, to provide constant power to the output). This negative impedance effect will get worse as the battery voltage diminishes.

A dry cell has an output impedance that increases as it discharges. When the cell output impedance exceeds the amplitude of the converter input impedance, the battery voltage will go unstable and be pulled low.

You need a cap that is big enough to hold the input voltage up when the battery is discharged, both when the device is switched on, and during any transients that may occur during operation.

I can't tell you exactly what resistance to design for, but as an initial guess I'd calculate the battery current at 0.9V per cell, then calculate the resistance to drop 0.6V per cell at that current (0.6V - 1.5V - 0.9V). Then I'd simulate my circuit with that resistance in series with a 1.5V/cell voltage source, and see if it survives the turn-on transient and any transients from normal use.

  • \$\begingroup\$ As an aside, this issue of battery resistance going up with discharged caused a condition called motorboating in old battery powered radios. The radio would come on and pull current, the battery would die, the radio's oscillators would quit, the battery would perk up, go back to start and repeat until someone hits the "off" switch. It would sound like a one-cylinder motorboat engine. \$\endgroup\$
    – TimWescott
    Jan 31, 2019 at 4:59
  • \$\begingroup\$ +1 To a first approximation the cell voltage stays the same and resistance increases. Though if you google "motorboating" you'll get something a bit different. \$\endgroup\$ Jan 31, 2019 at 11:19
  • \$\begingroup\$ @TimWescott +1 for internal resistance measurement method. \$\endgroup\$
    – rothloup
    Jan 31, 2019 at 16:43
  • \$\begingroup\$ @SpehroPefhany for dry cells, yes, a good approximation is that the cell voltage stays constant and resistance increases. This seems to hold true for NiCd and NiMH cells, even when they're mostly dead. The resting charge voltage of Lithium Polymer cells seems to be much more dependent on the state of charge, however. I'm not sure about lead-acid, to the point I'm not even going to guess without checking the web or actually (gasp!) doing measurements. \$\endgroup\$
    – TimWescott
    Jan 31, 2019 at 19:42

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