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I have 3.8V and 1.2Ah battery. I will use a step up circuit that gives 5v and 200mA output using 2v in (from my battery). Then I will use blocking diode to take voltage only(5v) and I don't need current from this circuit. Again I will also use a step down circuit that gives 1v and 2A output using 3.5v in (from my battery). Then I will use another blocking diode to take current(2A) only and I don't need voltage from this circuit.

So can I get 5v and 2A output in this way from a 3.8V and 1.2Ah battery? And now what will happen with my battery?

I need to understand that if I can get more power out of the circuit than comes out of the batteries. To extrapolate what I want to do, imagine I want to power my house with 240V @ 100A (24KW). I want it with a 1V @ 100A (0.1KW) power source combined with a 240V @ 1A (0.24KW) power source.

If to power my house with 1v@100A(taking the current only from here using blocking diode) and with 240v@1A(taking the voltage only from here using another blocking diode) and combine both of this to get 240@100A.

Is it possible? Well if it is possible I will again feed back the 5v and 2A output to battery to recharge it. So before being the battery empty(if) I will be recharging it again by the estimated output from the combined circuits.

Will the battery really be empty after producing separately stepped volt and current or can I get volt(5v) and current(2A) greater than my battery (3.8V,1.2Ah) in this way?

Does my thinking have a hope? My step up and step down circuits are as following as example:

http://www.alibaba.com/product-detail/DC-DC-2-0-5-v_1547670640.html?spm=a2700.7724838.102.7.vDaE3Z

http://www.amazon.com/DROK-Converter-4-5-24V-Rrgulator-Constant/dp/B00C4792T2/ref=sr_1_16?ie=UTF8&qid=1436885493&sr=8-16&keywords=current+regulator

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  • \$\begingroup\$ Are you sure that's a 1.2A battery, not a 1.2Ah battery? You didn't link to it, but a battery with a fixed current output would be a very strange thing... Also we know you need 2A at 1V. But what current do you need on the 5V supply? What does it power? \$\endgroup\$ – Brian Drummond Jul 30 '15 at 10:11
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As Mike DeSimone points out, your battery can't produce 3.8 V and 1.2 A at the same time.

But even if it did, that would only be about 4.6 W.

You want to use it to power something with 5 V and 2 A. That's 10 W.

You can't do this without first overturning the entirety of conventional physics by disproving conservation of energy.

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  • \$\begingroup\$ This is the answer the OP needs to read. He doesn't need to know circuits or how batteries work. He needs to understand that he cannot get more power out of the circuit than comes out of the batteries. To extrapolate what he wants to do, imagine I want to power my house with 240V @ 100A (24KW). I can't do it with a 1V @ 100A (0.1KW) power source combined with a 240V @ 1A (0.24KW) power source. The energy just isn't there. \$\endgroup\$ – DoxyLover Jul 30 '15 at 4:35
  • \$\begingroup\$ @DoxyLover you really understood what i wanted to know.Yes if to power my house with 1v@100A(taking the current only from here using blocking diode) and with 240v@1A(taking the voltage only from here using another blocking diode) and combine both of this to get 240@100A.Is it possible? \$\endgroup\$ – Mijanur Rahman Jul 30 '15 at 5:01
  • \$\begingroup\$ I have a feeling that (a) the asker has a 1.2Ah battery, not a 1.2A one, (b) he needs 2A at 1V (stated) and some unstated but much lower current at 5V, and (c) he's really asking if it's OK to connect regulators, one buck, one boost, to the same battery, for a combined load of 2W + not much. In which case the answer might very well be yes, depending on the current requirement at 5V being small enough, and the battery being able to supply 2W + the 5V power. But it's a lousy question at the moment so who really knows? \$\endgroup\$ – Brian Drummond Jul 30 '15 at 10:18
  • \$\begingroup\$ @BrianDrummond thanks brother for your answer.Your answer is mostly perfect.could you pls read my question again? I have edited again. \$\endgroup\$ – Mijanur Rahman Jul 30 '15 at 14:30
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    \$\begingroup\$ As written, your question still asks if you can get 24 kW from two supplies rated for 0.1 and 0.24 kW respsectively. The answer is still no. Energy is a conserved quantity in all known physics, so if you have a load that needs 24 kW, you must have a supply capable of producing 24 kW (or more for safety margin). \$\endgroup\$ – The Photon Jul 30 '15 at 15:58
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I have 3.8V and 1.2A battery.

OK, what that means is that the battery has a nominal output voltage on 3.8 V when the terminals are open and can push 1.2 A of current when the terminals are shorted. These conditions are not true at the same time.

In other words, there's some V/I graph that describes the relationship of voltage to current (given some charge level; it falls off as the battery discharges) that is monotonically decreasing, and all you know are that the points (3.8 V, 0 A) and (0 V, 1.2 A) are on that curve.

I will use a step up circuit that gives 5v and some current output using 2v in(from my battery).

This sounds like you're confused. A step-up switching power converter will need to take 3.8 V in (nominal) and output 5 V in this case. The output current maximum will depend on the power available from the battery. Remember that V/I graph I mentioned? the circuit will settle at some combination of V and I, and the power delivered (P_IN = V_IN * I_IN) must be greater than the power out (P_OUT = V_OUT * I_OUT).

Then i will use diode(say zener) to take voltage only(5v).

A Zener diode, properly connected, simply sinks enough current through itself to bring the voltage across its terminals down to its Zener voltage. If there isn't that much voltage across it in the first place, it will act like an open circuit. If forward biased, it acts like a regular diode with a small voltage drop.

I will also use a step down circuit that gives 1v and 2A output using 3.5v in(from my battery).

So this is in parallel with your step-up circuit? Again, 1 V is the nominal output and 2 A is the maximum the converter can deliver. Further, the power from the battery needs to be enough for both converters; one converter can starve the other, and if that isn't enough power, it'll pull the battery voltage down to UVLO level.

Then i will use another blocking diode to take current(2A) only.

Okay, at this point I have no idea what you're doing and I'm pretty sure you don't either. Can you draw a schematic? i have no idea how that alibaba thing hooks up.

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  • \$\begingroup\$ ok not zener diode say if i use blocking diode(current block) here then will it work as i want? \$\endgroup\$ – Mijanur Rahman Jul 30 '15 at 3:48
  • \$\begingroup\$ It doesn't block all the current; a few microamps will get through. And even then, it needs to have a reverse voltage high enough to not go into breakdown. But still, you get the same effect as an open circuit. What's the point? You really need to draw a schematic and post it. Your text description is not sufficient. \$\endgroup\$ – Mike DeSimone Jul 30 '15 at 3:50
  • \$\begingroup\$ I can not draw good and i am new in electronics.Just suggest me can i get 5v and 2A output in this way?I have another purpose to use this output volt and current greater than battery volt+current. \$\endgroup\$ – Mijanur Rahman Jul 30 '15 at 3:52
  • \$\begingroup\$ 2 A @ 5 V = 10 W. 1.2 A @ 3.8 V = 4.56 W. Your output power requirement is greater than the battery could supply even if it were perfect, which it is not. So no, you can't get that output. You'll need a bigger battery or more of them. If you want to pursue this further, make sure to get the datasheet for whatever batteries you use. They will tell you what kind of practical power levels you can expect from them. \$\endgroup\$ – Mike DeSimone Jul 30 '15 at 14:03
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Attempting to connect supplies with different voltages will cause current to flow directly between them, possibly resulting in damage or death; do not attempt to connect them together.

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  • \$\begingroup\$ I will not connect them together just suggest me can i get 5v and 2A output in this way?I have another purpose to use the output volt and current that is greater than battery volt+current. \$\endgroup\$ – Mijanur Rahman Jul 30 '15 at 3:50

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