I’m using a 5V DC-DC converter to step-up a coin cell (Nominal Voltage: 3V) to 5V which did not work. I suspected that the high startup current of the converter was causing the Cell voltage to drop, hence the converter never really starting. To confirm this, I used a power supply with 3V voltage and 1 Ohm resistor in series to measure the input current to the DC converter.

As I suspected, the starting current was very high than what the Coin Cell could provide. Below is the plot of voltage drop across the input resistance I got on my oscilloscope. Since the input resistance is 1 Ohm, the current is equal to the Voltage drop e.g. 100 mV = 100 mA. (CurA = Starting Peak Current AND CurB = Input Current after Converter starts up)

[Plot of Input current to DC converter without capacitor, the peaks show the starting current

Someone suggested to put a high capacitor in parallel to the input voltage to the DC converter to reduce the input current. I used a 1500 uF capacitor in parallel but opposite to my intuition, the input current further increased from 292 mA to 384 mA as below:

Input current after using 1500 uF capacitor in parallel to Input Voltage

I’m placing the capacitor in parallel just between the input supply to the DC converter. What is wrong here? My assumptions or values of capacitor?

  • \$\begingroup\$ Your coin cell can not supply the 300 mW you are asking for! Read the datasheet! \$\endgroup\$
    – winny
    Apr 29, 2018 at 11:48
  • \$\begingroup\$ Obviously, I know that the poor coin cell can only provide upto 20 mA surges of current. That’s why the converter never really worked in the first place and that’s not what I’m asking in the original question. \$\endgroup\$
    – Marry35
    Apr 29, 2018 at 11:55
  • \$\begingroup\$ Not clear from your question that you understood that. What’s your load? Where’s your schematic? What’s your soft start time constant? What’s your output capacitance? \$\endgroup\$
    – winny
    Apr 29, 2018 at 12:10
  • 1
    \$\begingroup\$ @UmairRiaz And you expected that the large cap at the input would, somehow, miraculously, provide the current? \$\endgroup\$ Apr 29, 2018 at 12:11


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