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Need 5V-Least power dissipation

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Member level 3
Apr 6, 2002
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What's the way of getting 5V from a 9V battery with a minumum power dissipation.What do you recommend:7805,or a voltage divider,or zener , or something else?

I want the battery life to be maximum as it can be.

I would suggest using either a DC-DC converter with integrated LDO regulator, or using a buck-boost DC-DC converter so that as the power output from the 9V battery drops below the threshold for the step-down component of the converter, it changes to step-up mode. This will give you the best performance and lifetime from the battery. National/Maxim/Linear Technology all do parts which can do this.

If you need the cheapest option, then you probably should look a a low-drop out regulator (LDO) with maybe 200mV dropout voltage. This will give you operation from your input 9V downto 5.2V. However when the input voltage is greater than 5V all the power is dissapated as heat in the device.


Least power dissipation

Using 7805 the efficiency you get is very low. For example, if you need 1mA to supply your circuit, then 7805 will ask for 1mA from your battery (let assume that 7805 needs no current). This means that the efficiency is about Pout *100 / Pin = (5*1mA) * 100 / (9V*1mA) = 55%. If you add about 5mA current consumption for 7805 then your battery life will end very soon :)

What i suggest to use is an step-down switching regulator. For example i use max638. Its typicall efficiency is 85% and has low operating current, typically 135uA!!! 85% means that if your circuit needs 1mA, then your battery will supply only 0.65mA plus 0.135mA for max638 means 0.785mA. On the other hand, 7805 needs 1mA plus 5mA for itself, thus 6mA.

I think that 0.785mA is about 7.5 times less than 6mA, so your battery life will be extended 7.5 times.

Have a nice day

If you have any problems don't hesitate to ask

Sorry for my poor english :lol:


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