I'm trying step up the voltage of a size 675 Duracell Zinc Air button cell battery: http://www.microbattery.com/microbat/pdf/duracell/duracell-hearing-aid-battery-008-2014.pdf

from its nominal 1.4 V to 3.3 V using the charge pump TPS60312: http://www.ti.com/lit/ds/symlink/tps60311.pdf

while continuously drawing a current of ~12-14 mA from it. The charge pump should be able to supply 20 mA at 3.3V from its OUT2 pin.

First I checked if the cell is capable of supplying that current by creating a simple circuit with only the cell and a load of a 100 Ohms. The circuit stabilized with the voltage of the cell dropping down to ~1.2V with a current of ~11.5 mA going through the circuit.

Next I connected it to the charge pump, as shown in the schematic on the bottom left of the first page of the charge pumps data sheet linked above (using five size 0402 ceramic 1 uF capacitors, and no resistors as PG is not required), plus applying the same 100 Ohm load to the output as before. Doing this results in the voltage severely dropping across the battery down to ~0.7V, which is below the operating input voltage (0.9V) of the charge pump. The charge pump still generates an output but its below 1V.

I tried some different resistor values and the circuits seems stable with a steady 3.3 V output for lower current draws, but once it goes above ~5 mA the cell voltage starts to drastically drop.

I also tested the charge pump with a power supply instead of a cell, and the same thing happens, the displayed voltage on the power supply drops down to 0.68V once the 100 Ohm resistor is placed into the circuit.

  • What could be the cause of the severe voltage drop?
  • The data sheet mentions one alkaline, NiCd, or NiMH batteries as typical uses for the charge pump, but there is no mention of zinc air batteries. Could this be an issue?
  • I also used physically smaller capacitors (0402) than recommended in the data sheet (0805). Does this matter, and how much does the capacitor physical size actually matter in cases like this (assuming the other properties are the same)?

1 Answer 1


You are severely overloading that cell. Its 600 mAh rating is at a 2 mA current draw. To get your desired 20 mA at 3.3 volts, you will draw at least 47 mA from the cell, assuming 100% efficiency of your charge pump (and it definitely won't be 100% efficient, so will draw even more current.)

  • \$\begingroup\$ Where does that 47 mA value calculated? \$\endgroup\$
    – varkong
    Commented Mar 17, 2017 at 22:33
  • 2
    \$\begingroup\$ Power out of your charge pump is equal to power in, plus losses. Your desired output of 20 mA at 3.3 volts is 66 mW. To deliver 66 mW at 1.4 volts, the cell must deliver 47 mA. \$\endgroup\$ Commented Mar 17, 2017 at 22:37
  • \$\begingroup\$ There are some high current versions of the 675 batteries with current limits of ~30 mA, which just might be enough for 12 mA (which is what I need, not 20 mA). Would implementing Synchronous Boost Converters such as: ti.com/lit/ds/symlink/tps61029-q1.pdf potentially be more efficient? Are there any other methods that would feasible to step up voltage, yet minimising the current draw from the cell? \$\endgroup\$
    – varkong
    Commented Mar 17, 2017 at 23:08
  • \$\begingroup\$ Unless you have a perpetual motion machine on hand, power cannot be created - you will always be limited to "Power Out = Power In Plus Losses". \$\endgroup\$ Commented Mar 17, 2017 at 23:18

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