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I'm designing a circuit based on the MC34063A switching regulator to step up 3.7V to 8V.

The 3.7V for Vin comes from a 3.7V Li Ion battery and so I'm trying to calculate the battery life. I'm a bit confused on some of the parameters from the datasheet though. It states that Icc (supply current) is typically 2.5mA. This seems a bit too low to use as a value for the total current consumption. However, I've calculated that Ipk for my application (8V @ 200mA) will be around 1A. I know it's a switching regulator though and so the the waveform for the Ipk value looks more like a triangular waveform. Would an average of Ipk give a better indication of power consumption?

I'm just wondering if anyone would point me in the right direction with regards to calculating total power consumption so that I can calculate my battery life.

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    \$\begingroup\$ The supply current is the current the regulator needs to run its internal circuitry. \$\endgroup\$
    – Hearth
    Jan 17, 2021 at 20:00
  • \$\begingroup\$ @Hearth So would I be able to use 2.5mA as the current consumption when calculating the battery life? \$\endgroup\$
    – RM429
    Jan 17, 2021 at 20:11
  • \$\begingroup\$ Only if there's no load on the circuit. The 2.5 mA is only the current drawn by the regulator itself, there will be additional current drawn by the load, and in a properly designed system the load current will be much greater than the supply current--if it's not, you either chose the wrong controller or you have very strict requirements on other things. \$\endgroup\$
    – Hearth
    Jan 17, 2021 at 20:14
  • \$\begingroup\$ @Hearth Perfect! Thanks. Just to clarify, would I simply add my load current to the 2.5mA to get the total current consumption? My load is 8V @ 200mA. \$\endgroup\$
    – RM429
    Jan 17, 2021 at 20:18
  • \$\begingroup\$ No. Remember, a step-up converter will always have higher average input current than output current (and likewise, a step-down converter will have lower average input current than output current). \$\endgroup\$
    – Hearth
    Jan 17, 2021 at 20:20

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It may be easier to think if it in terms of power. The input power will be the output power divided by the efficiency (say70%) plus the power of the controlling circuitry.

So for 8V @ 200mA the output power is 1600mW. At an efficiency of 70% (not very good but you may not get much better for this circuitry) this is 2280mW. Add the power for the controller (3.7*2.5 = ~10) to get 2290mW.

Divide by the input voltage to get the input current 2290/3.7 = ~620mA. This is subject to a fair amount of variation depending upon the exact circuit used.

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  • \$\begingroup\$ Perfect! Thank you for the clear explanation. I was struggling to understand it in terms of power consumption. \$\endgroup\$
    – RM429
    Jan 17, 2021 at 21:31

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