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I have a customized circuit using ATMEGA328p with Arduino Boot-loader to develop a Wireless Sensor Network which sends readings from a JSN-SR04 Ultrasonic Sensor to another node via nRF24L01+ wireless modules. Now, I am running the ATMEGA328p in low-power mode so that it can run on as low as 2.4V which is the voltage of coin cells when they run out. nRF24L01+ module is also capable of running of voltages as low as 2.4V so all is good till this part. I have tested and the circuit works without any flaws.

I am stuck on how to power the Ultrasonic sensor which requires a very stable 5V voltage supply and will not work below 4.7V. The only power source I have on the circuit is a CR2450 Coin Cell (640 mAh). Below are the few things that I tried:

1- Used a DC-DC converter to step up the voltage level to 5V but the DC-DC converter requires a huge starting current which the Coin Cell cannot provide and hence it does not work.

2- Used two Coin Cells in series but the voltage goes upto 6.6V since, the voltage of a new coin cell is 3.3V. This might destroy the ATMEGA328p which has a maximum input voltage of 6V.

So what I need is something to step up the voltage of Coin Cell to 5V without requiring a huge starting current and can be disabled so that its quiescent current (normally 1 to 10 mA) can be cut-off to maximize battery life. I can also use two Coin Cells and step down the voltage through something suggested, but efficiency and Battery life should be maintained.

Anyone working with low-power electronics, please suggest me what is the most efficient way of working with Coin Cells to power external Sensors. Thanks!

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  • \$\begingroup\$ Batteries have ESR that rises to 1k in thin coin cells so you Can’t use a toothpick to whip cream and make butter” as Robin Williams once said \$\endgroup\$ Apr 22, 2018 at 13:28
  • \$\begingroup\$ Use a (very) large capacitor to supply the intial current for the DC/DC? \$\endgroup\$ Apr 22, 2018 at 13:31
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    \$\begingroup\$ When working on low power, the first thing to do is identify the exact details of the required tasks to succeed, how often they are performed and their duration, and their estimated power costs. You can get peak power, minimum power, and average power. You've provided none of that information. Start there. \$\endgroup\$
    – jonk
    Apr 22, 2018 at 16:22
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    \$\begingroup\$ 30mA for 2 milliseconds, so, 60 microcoulombs. dQ = C*dV, so dV = dQ/C = 60/47 = 1.27V sag. No wonder it didn't help. Try 1000uF for 60mV sag. \$\endgroup\$
    – user16324
    Apr 22, 2018 at 17:44
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    \$\begingroup\$ @UmairRiaz What you failed to understand about my comment is that you should disclose those details here. The appropriate answer is much easier then and less sweeping in scope. Which saves me time. Without that, I'm not interested. \$\endgroup\$
    – jonk
    Apr 22, 2018 at 20:32

2 Answers 2

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If you can use two coin cells in series then power Arduino from the middle 3.3V and the sensor from 6.6V.

Also using an LDO with enable and reset to bring 6.6 to 5V you can put the sensor to sleep using enable input and benefit from a "power good" signal from the reset output. Here is an example: http://www.ti.com/product/TPS779

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If your goal is maximum battery life do not run cells in series with different loads on each cell as your run time will be limited to the cell that dies first and you will have remaining capacity that is unusable. Instead put 2 cells in parallel so you can have more startup current and then as other have recommended place the largest ceramic capacitors in parallel with these 2x cells that you have room for. Texas Instruments makes boost converters with ultra low quiescent current in the nA range, you should check those out.

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