DC/DC converters extend battery life

I don't know much about DC/DC converters, or 'pucks' or whatever, but the following link might prove interesting to some.

DC/DC converters from TI extend battery life

DC/DC converters allow the use of energy storage devices (inductors & caps) to convert power from one voltage to another. A buck converter like this can only produce an output voltage that's lower than the input voltage. If the design includes fast, low resistance power transistors, large inductors, and synchronous rectification, it's possible to make them quite efficient, which is what this article is refering to by the phrase "extend battery life".

Seems like a decent enough part, based on a very quick perusal of the data sheet.
http://www-s.ti.com/sc/psheets/slvs432b/slvs432b.pdf

Steve Kurt

Hi PhotonBoy,

I do not know how indepth your electronics knowledge is, but I will try and explain things a little for you is simply and with as much depth as I can ...

you have two uses of dc-dc convertor, one that provides a higher voltage (booster) and one that provides a lower one (buck).

for ease of explanation I will assume an exact voltages for the batteries that do not get lower as the battery runs down


for example with a buck circuit using this component....

suppose you have a pp3 battery (small 9 volt battery) and you wish to illuminate a white led that needs 3.6 volts with 20ma. You now have 9 - 3.6 = 5.4 volts too much that you need to lose, the simplest way of doing this is with a series resistor. to calculate the value you use ohms law

i (current) = v (voltage) / r (resistance)

we know we vant the current to be 20 mA i.e. 0.020 Amps,
the excess voltage we wish to lose is 5.4 volts.

re-arranging the abpve formula we have r = v / i

r = 5.4 / 0.020 = 270 ohms

so we now get the nearest value resistor to 270 ohms and put it in series with the led and power it from the battery.

as 20 mA is flowing through the resistor then 20 mA is thus flowing through the led.

the battery is supplying 20 mA at 9 volts so to work out the power it is providing can be worked out with the formula below

p (power in watts) = i x v

p = 0.02 x 9 = 0.18 watts

yet the led is only using...

p = 0.02 x 3.6 = 0.072 watts

so 0.18 - 0.072 = 0.108 watts is 'wasted ' by the resistor.

note: a simple series regulator will be 'wasting' this energy in the same way as the resistor.

now we will suppose that you use this component to convery this 9 volts down to 4 volts at 95% efficiency.

at 4 volts the resistor you would need is

r = v / i = 4 / 0.02 = 200 ohms

power consumed at 4 volts is

p = i x v = 0.02 x 4 = 0.08 watts

as this is only 95 % efficient the power thus drawn from the battery is 0.08 x 0.95 = 0.8421 watts

by rearranging the p = i x v formula we can get

i = p / v

so we can now substitute in the values for this power consumption and the 9 volt supply

i = 0.8421 / 9 = 0.009356667 mA

note: this is less than half the current taken when using a resistor alone so the battery should (in theory) last twice as long.

the power consumption in the led still stays at 0.072 watts but the overall circuit is now using 0.8421 watts.

The circuit is thus not achieving miracles buy causing the battery to last longer it is simply using the batteries power in a more efficient and less wasteful way.



on the other hand for example with a boost circuit such as the MAX757 component....

with a boost circuit one voltage is taken and raised to a higher one.

note: when you are using a battery the voltage will lower (due to increased internal resistance) as the battery uses up its useful charge.

suppose you now had the same 3.6 volt led that you wished to power from two 1.5 volt alkaline cells.

you could try and illuminate your led from this 3 volt supply but depending on the led you will probably only get a very dim glow if any at all.

what you need to do is increase this voltage to 3.6 volts or more using a boost circuit.

so we shall assume you choose to step this up to 4 volts and provide 20 mA, to power the led with a 200 ohm series resistor, the power needed will be 0.08 watts (as calculated earlier).

given that 85% is a reasonable value for the efficieny of a boost circuit these the battery will need to supply
0.08 / 0.85 = 0.9412 watts of energy.

at 3 volts this comes to ( use i = p / v )

0.9412 / 3 = 31.37 mA

when the battery runs down and a lower voltage e.g. 2 volts is seen then the current drawn is 0.9412 / 2 = 47 ma

and so on..

note: the MAX757 will keep operating until the supply voltage falls below 0.7 volts.

these boost circuits are thus ideal for people who can lay their hands on a regular supply of 'dead' batteries; as batteries that have been used in other high current drain appliances (such as electric toothbrushes / toy racing cars etc) might seem too weak to run them yet will provide in some cases a few hours of use in a boost circuit led torch.

note : batteries from a low drain appliance such as a remote control are usually of no use for these circuits as the internal resistance of the battery will not cause much voltage depletion with the very low current these devices require. when the batteries in a remote control show 'dead' they are usually only fit for the rubbish/recycling bin.

hope this is of interest, and I haven't confused you by going too deep, I have tries to explain it as simply as I can but if you are having difficulty please try re-reading it slowly and working out the formulas for yourself, and remember you would probably take a few days of an electronics course to cover all these fourmuae and principles to a level you will be competent with.


Please excuse any 'spelling errors', some may be typo's, some will be due to the fact that in the UK we spell different to in the US the rest will be down to me !

Thanks very much for the detailed explanation. I'm going to have to digest it carefully tonight when I've got more time to concentrate; right now, I've got to run to a meeting... later...