# Calculate battery life when using boost converter

How should I calculate the time taken to drain the battery to below 5 V when using a boost converter to convert DC 18 V to 28 V for a load on 20 mA.

• Your question is a little confusing. Are you asking how long it will take to discharge a 9V battery to 5V? Why do you mention 28V? Is that an output voltage? Where is the link to the datasheet for your batteries...there are different kinds of "9V" battery. – Elliot Alderson Apr 19 at 15:19
• @Elliot Alderson i am boosting it from 18V (9V x 2) to 28V. Yes output is 28 V. This the datasheet celltech.fi/fileadmin/user_upload/Celltech/Prod.sheets/… – Mahesh Apr 19 at 15:28
• Lets say the 9V battery has a rating of 210mAh. – Mahesh Apr 19 at 15:42
• Btw, if it is really a 9V battery form factor (PP3), you may want to use a Li-Ion based one. They are two li-ion cells internally, and have about 540 mAh. – anrieff Apr 19 at 16:41

The basic theory is that energy available = VIt = 18 (V) x 0.21 (A) x 1 (h) = 3.8 Wh and the power required is VI = 28 (V) x 0.02 (A) = 0.56 W.

A 3.8 Wh energy source will supply 0.56 W for 3.8 / 0.56 = 6.8 h.

Now multiply that by the efficiency of your system - maybe 80% (I didn't check) and you could expect about 5 h or so.

If you convert everything to WattHour, it is easy to do:

If that runs for one hour that's 560 mWattHour = 0.56 WattHour

Your battery voltage is 2 x 9 V = 18 V.

At 210 mAh that's 18 V x 210mAHour = 3.78 WattHour

For simplicity we start with assuming that the boost converter has no losses, then we get:

3.87 WattHour / 0.56 WattHour = 6.75 discharges of one hour so 6.75 Hour of run time.

If the boost converter has losses that simply decreases accordingly:

At 20 % loss (80% efficiency) that's 80% x 6.75 hours = 5.4 hours

At 40 % loss (60% efficiency) that's 80% x 6.75 hours = 4 hours

• Thankyou both. Yes 60% seems about right. I have lots of stuff downstream before output. I have lm317, pic microprocessor, relay, digipot etc. – Mahesh Apr 19 at 16:26