Let's say I have a 12V / 1Ah rechargeable battery pack. A device that consumes 0.1A will run for 10 hours. When the battery drops to 8V, the voltage will be insufficient and the device won't work anymore.

Most of the current flows through a transistor that sends pulses to a coil (of a metal detector). The pulse width gets shorter as the voltage increases so I guess, the consumption remains the same not matter which is the battery voltage.

Can I increase the number of operating hours for that device WITHOUT increasing the capacity (Ah) of the battery? More exactly by adding extra batteries (series) to the pack and thus increasing the voltage from 12V to 14V? (We assume that the device can operate also to 14V).

My explanation/thoughts:
The capacity of a 1Ah of a battery means that it can provide 1A current for one hour without dropping its output voltage under 75%. If the device stops working at 8V, it means that the battery can be used from 12V downto 8V. This is a 4V difference.

By adding another battery, the total capacity is still 1Ah. But the total voltage is now 14V. So, the battery pack can be used from 14V down to 8V. This is a 6V difference! Therefore, by adding another battery, it takes longer until the voltage drops from max to min. HOWEVER, the load will "burn" more current because now it operates at 14V instead of 12V (I = U/R).

  • \$\begingroup\$ You could add another 12V/1Ah battery pack in parallel - which would double your operating lifespan however when adding batteries this way, the voltages should match so that the drain on each pack is equal. \$\endgroup\$ Apr 27, 2015 at 8:57
  • \$\begingroup\$ Good 2011 discussion related to GSR in my answer here: techref.massmind.org/techref/… \$\endgroup\$
    – Russell McMahon
    Apr 27, 2015 at 12:49

3 Answers 3


12V,1Ah means you've got 12Wh. 14V,1Ah will get you 14Wh.

So yes - you've got more "power" (and you did increase the capacity of the batteries by that...)

Now it depends on how your circuit will use that energy. When it needs less than 0.1Ah @ 14V, it will work longer. If it "burns" the excess voltage, and still need 0.1Ah, regardless of whether it is running at 8,12 or 14V, it will run for the same duration

  • \$\begingroup\$ Most of the current flows through a transistor that sends pulses to a coil. So, I guess, the power consumption will raise as I increase the voltage. \$\endgroup\$
    – IceCold
    Apr 27, 2015 at 8:17

All you need is a Joule Thief

basically it can use nearly all of the energy in a single-cell electric battery, even far below the voltage where other circuits consider the battery fully discharged (or "dead"). The output voltage is increased at the expense of higher current draw on the input

  • 2
    \$\begingroup\$ The Joule Thief type converters have their place and are worth knowing about but are not usually intended for buck-boost operation and efficiencies are usually abysmal due to lack of (in most cases) formal design or even designability. They can be fun and useful but as a battery life prolonger for serious loads are liable to not meet theneed. \$\endgroup\$
    – Russell McMahon
    Apr 27, 2015 at 12:45
  • \$\begingroup\$ I have to agree with Russell. Not quite a good idea. Especially for rechargeable batteries. \$\endgroup\$
    – IceCold
    Apr 29, 2015 at 10:15

If a lower voltage is acceptable a low quiescent current & reasonable efficiency buck regulator will help. If as you suspect, Ioperate is constant with falling voltage then time duration will increase as 1/V. I lower V is provided by a buck converter with efficiency Z ( 0 < Z < 1).
Typically Z = 0.8 to 0.9)

Then new duration ~= Old duration x 12/Vnew x Z
eg For Z = 0.85 and Voperate = 8.5V then Tnew = Told x 12/8.5 x 0.85 = 1.2
= 20% gain.

The maximum possible gain is at 8V (JUST operate) and 100% efficiency (not achievable)
= 12/8 x 1 = 1.5 = 50% increase.

It MAY be possible to operate the device with a 12V pulse of lowered net duration - whether this is possible depends on aspects that you have not told us about. A proper description of the problem and equipment is needed. It is possible that a MUCH longer life is possible but we need complete information.

Current drain is expressed in amp(ere)s (A)
A x H = Ah.

G_G's Joule thief is well intentioned BUT as a general rule, most circuits with that name are built by people who are not aware of what they are doing and the results are not usually marvellous. (The Wikipedia generic example is a typically sub-optimum one). This does not HAVE to be the case, but often is. A modern (or ancient) buck regulator OC (or discreet circuit properly designed will usually be vastly superior.

Here is an example from Elliot Sound Products, of a discrete component buck converter that works well and constant current version here

I know it works well because: Here is a remarkably similar design based on a circuit provided to me by God in 2001 (but, that's another story* :-)). This uses a PFET as the high side switch but this can equally be a small bipolar transistor in this case. The design is remarkably similar to the ESP circuit. As mine appeared on the web some while before his he may have copied the GSR (which anyone was and is freely allowed to do) or he may have received inspiration from another source. It's a "hysteretic converter - which was a well enough known idea at the time - but one I'd not until then heard of. (Despite some extremely smart people saying it has no hysteresis, it does have. The ripple on CBUK2 provides hysteresis and is required for operation. The output is "chaotic" and spreads the EMI signature nicely. For Vin < Vout_design is is well behaved with high side switch on. Switching starts seamlessly as Vout reaches design point.

enter image description here


*. Note specific time and date on circuit - to nearest minute :-). Ask off-list if of interest. An unknown number of these devices were built thousands and possibly 10's of thousands, and they filled a need very well. [I was meant to get royalties - this was a small part part of a larger design but 'things went wrong with payments' as they do][Several paid visits to Taiwan resulted so it was fun overall.] As used the circuit converted Vin of from 12V to 200V+ down to 12V out. Efficiency in the normal 12V-50V range was acceptably good and while only about 50% efficient at 200 Vin, that is vastly superior to the 12/200 = 6% efficiency of a linear regulator. At eg 5W out energy loss is 5W at 200V with this regulator and 78+ Watts with a linear regulator. Most sources say that a buck regulator with a 200/12 = 16:1 + input voltage range is not practical, but God knows why. [ :-)].


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