1
\$\begingroup\$

I want to calculate the time for a Li-Po battery (7.4 V, 2200 mAh, 30 C) to discharge. Battery powers up an embedded device, that requires 5 V, 147 mA.

For this reason, I use a voltage regulator, in order to regulate voltage from 7.4 V to 5 V.

I haven't chosen a battery yet and as a result I cannot provide any datasheet. I 'm looking for an approached disharge time and what is the way to calculate from a discharge graph.

\$\endgroup\$
  • \$\begingroup\$ You... look at the discharge graph and interpolate. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 4 '14 at 17:49
  • 1
    \$\begingroup\$ With a simple linear regulator, 2200mAh / 147mA = something like 15 hours. A more complex buck regulator could give you maybe 3 to 5 extra hours. \$\endgroup\$ – Brian Drummond Jul 4 '14 at 17:52
1
\$\begingroup\$

If your load is from an LDO the battery current will be about the same and the load is 2200/150=14.7h or 1.6C

Looking at the voltage profile 1C to 2C, interpolate the average voltage ratio out/in for your regulator design, and remember that cutout rating voltage changes with test , supplier and battery model, which could be 3V, 2.75 or 2.5V per cell (x2 for 7.4 type) This could affect early cutout of regulator you choose or increase losses more rapidly.

Not all LiPo's have the same curves, but typical for 1.6C is 7.15 +/-0.3 from 80% to 20% charge at 20 deg. C. and a bit more at 30 deg.

(Edit) A buck regulator if 100 % efficient at load should reduce battery current. Your output voltage range is fixed at 5V but input is from 8 to 6V, thus the ratio is 5/8 to 5/6 or 63% to 83% (73% avg) which is your efficiency for an "extra low" (<1V) LDO regulator.

A good " buck" may give you 92% initially and increase as input drops on this part. But this rated for full load of 0.5A, so read the curves, but you would expect battery load to be 150mA x 73%= ~ 110mA and thus expect **20h from a 2.2Ah battery at 110mA average or 20h= 1.2C** vs 2200/150=14.7h for an ultra-LDO. http://www.ti.com/product/tps62170

This gives you ~36 % more time from efficiency gain and battery time, assuming your load is constant.

You may choose this regulator and bigger battery if desired operating time is more.

\$\endgroup\$
  • \$\begingroup\$ 5/6 efficiency is 83%, not 63%. \$\endgroup\$ – WhatRoughBeast Jul 4 '14 at 23:59
  • \$\begingroup\$ Ty WRB. Edit. Done \$\endgroup\$ – G.I. Jul 5 '14 at 1:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.