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I'm wondering, is it possible to provide for example 5V/500mA from a single 1.2V battery and consuming only 50mAh from the battery by using a low input voltage boost converter and a power amplifier ? (L272 and a TPS61200 for example)

I found that it's possile to go from 1.2v to 5v with the TPS61200 and go from 50mA to 500mA with the L272 but what if they're used together ? Will it preserve my battery lifespan ?

Emmanuel Istace
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  • Please read this (http://electronics.stackexchange.com/questions/79279/does-mah-measure-how-long-a-battery-would-last) and then see if you still have questions. – The Photon Nov 22 '13 at 17:00
  • Yes, it's not about how to measure the battery life but if it's possible to use current amplifier and voltage amplifier to longer the battery lifespan. – Emmanuel Istace Nov 22 '13 at 17:11
  • The other question tells you, if you have a 50 mA-hour battery, you can only expect it to last so long when supplying a 0.25 W load, no matter what chips you use. – The Photon Nov 22 '13 at 17:22

2 Answers2

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Consider conservation of energy.

Very roughly, 1.2 V * 50 mAh is about 215 J.

5 V * 500 mA is .25 W.

At a .25 W rate of use, 215 J will be consumed in about 450 seconds, or 7.5 minutes.

Edit

Let's look at the chips you mention.

TPS61200 is a boost converter. It will "boost" the power supply voltage from 1.2 V to some higher voltage --- 5 V in your case. But it will draw more current from the battery than it delivers to the load. This is because it is not magical and it is not able to create energy from nothing.

L272 is an op-amp. It can deliver an output current of up to 700 mA, controlled by a lower power input. But it must get the energy to deliver that output from it's power supply pins. You must provide a +/- 12 V supply to use this amplifier according to its specifications. If have some other source of power to provide the +/- 12 V, then you can certainly use the L272 to control a 5 V / 500 mA signal powered by that supply. But the power will be taken (very inefficiently) from the +/- 12 V supply, not magically created by the L272.

The Photon
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  • Ah, you're taking his 50mAh as literal rather than a typo. I took it to be a typo since, in his penultimate sentence, he uses 50mA. – Alfred Centauri Nov 22 '13 at 16:55
  • @AlfredCentauri, Haha, didn't see that. More likely OP doesn't know the difference so we'll have to wait and see --- and then redirecting to an old question about the difference between power and energy will probably be better than either of our answers. – The Photon Nov 22 '13 at 16:58
  • I'm perhap's not clear. I means, is it possible by using voltage amplifier and current amplifier to preserve battery and so being able to get 5V at 500mA from a battery of 1.2V and consuming only 50mAh from it ? (using the TPS612000 to amplify voltage and the L272 to amplify current). From what I understand the L272 amplify the current, so I can output 500mA from a 50mA without draining 500mA from the battery. In that case with these two components in a circuit ou could be able to get a 5V "battery mAh * current amp gain" from a 1.2V battery. – Emmanuel Istace Nov 22 '13 at 17:08
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    Your system must obey conservation of energy. If you deliver 0.25 W to the load, you must draw 0.25 W (plus extra because no converter is 100% efficient) from your battery or some other power source. – The Photon Nov 22 '13 at 17:11
  • Ok thank you, I was thinking that amplifiers could 'violate' this rule. Sorry for the stupid I'm really fresh in the EE hobby. And thank you for you answer. – Emmanuel Istace Nov 22 '13 at 17:22
  • Conservation of energy is just about the strongest rule there is in physics. It's up there with the speed of light among the laws of physics that we never never never see violated in the physical world. (http://en.wikipedia.org/wiki/Laws_of_thermodynamics) – The Photon Nov 22 '13 at 17:25
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    I find it useful to regularly remind myself that batteries don't store electricity and capacitors don't store charge. Today I spent a happy half hour reading [ELECTRICITY MISCONCEPTIONS](http://amasci.com/miscon/elect.html) – RedGrittyBrick Nov 22 '13 at 17:36
  • @RedGrittyBrick, sorry I don't see the connection between your link and my answer. Is there something I wrote that I could write more clearly? – The Photon Nov 22 '13 at 18:02
  • @ThePhoton: No, sorry, my comment should have been addressed to the OP, I thought the OP might find the link as interesting and useful as I do. – RedGrittyBrick Nov 22 '13 at 18:57
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I found that it's possile to go from 1.2v to 5v with the TPS61200 and go from 50mA to 500mA with the L272 but what if they're used together ? Will it preserve my battery lifespan?

Let's just say the answer is "yes". Say you had a 1.2V battery which can now provide 5V, and where it could provide 50mA, it could now provide 500mA, with no reduction in runtime.

Say I need to power my house. I need roughly 120V and 40A. I could then take this small battery, then cascade these "voltage amplifiers" and "current amplifiers" until I attain the necessary power. No need for the electric utility! Just swap out a couple AAA batteries every few months!

Clearly, this would be a violation of the law of conservation of energy. The battery contains only so much energy, and nothing we connect to it (that doesn't contain stored energy itself) can enable us to get more energy out of it.

How then do boost converters work? Clearly it is possible to get 5V out of a 1.2V battery, but how?

Remember that voltage is a force that works on electric charge, such as electrons. There are analogous machines to boost converters that work on mechanical force, instead of voltage (charge force). For example:

  • levers
  • jacks
  • pulleys
  • ramps

If a mechanical system has no losses, then power in (my work moving the jack handle) must equal power out (increased gravitational potential of the lifted car). Recall (or reference on Wikipedia) that mechanical power is the product of force and velocity:

$$ P = Fv $$

So if I'm lifting a car with a jack, and the car is moving half as fast as I'm moving the jack handle, the force applied to the car must be twice what I'm applying to the handle, otherwise the input and output power would not be equal. This would be a 2:1 mechanical advantage. This is how I'm able to lift a very heavy car with just muscle power: I apply less force, but over a greater distance.

Boost converters (and in fact, all switch-mode power supplies) are essentially the same thing. However, electrical power is the product of voltage and current:

$$ P = IE $$

So say I have a boost converter that gives me 5V from a 1.2V battery, and I measure the load current (on the 5V side) as 50mA, then the battery current must be:

$$ \require{cancel} \begin{align} P_{battery} &= P_{load} \\ I_{battery} \cdot E_{battery} &= I_{load} \cdot E_{load} \\ I_{battery} \cdot 1.2V &= 50mA \cdot 5V \\ I_{battery} &= \frac{50mA \cdot 5\cancel{V}}{1.2\cancel{V}} \\ I_{battery} &\approx 208mA \end{align} $$

This works in the other direction as well: if a buck converter reduces the battery voltage, then less current will be drawn from the battery. Or equivalently, if a device reduces the current load on the battery, in necessarily must reduce the voltage available at the load.

Phil Frost
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