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« on: January 02, 2012, 06:41:18 pm » |
I'm designing a remote with an Arduino, and I'm wondering about the best way to give it the 5v it needs.
My initial thought was to power it with 6xAAA batteries + linear regulator. But I just read that a pack of batteries in series has the same capacity as a single battery (this was a bit counterintuitive to me), so for 6x1200mAh I'd still only get 1200mAh. What a waste.
So I'm wondering: if I instead put the 6 batteries in parallel (getting 1.3v-1.5v) and then use a charge pump to go up to 5v, will I get both 5v and 6 times the capacity? (ignore charge pump losses)
Or am I breaking some law of physics here, and both methods give the same capacity?
Thanks!
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« Reply #1 on: January 02, 2012, 07:07:34 pm » |
So I'm wondering: if I instead put the 6 batteries in parallel (getting 1.3v-1.5v) and then use a charge pump to go up to 5v, will I get both 5v and 6 times the capacity? (ignore charge pump losses)
Or am I breaking some law of physics here, and both methods give the same capacity?
That is how it works, MAH capacity in series is same as single cell, in parallel the sum of their individual MAH capacity. There are some very inexpensive step-up switching voltage regulators avalible from Asian sellers on E-bay like this one: http://www.ebay.com/itm/Step-up-500mA-Module-DC-DC-Voltage-Converter-/350519982224?pt=LH_DefaultDomain_0&hash=item519c9e7890There are others avalible for higher output current ratings. Many don't work well at one cell voltage input so those would work better using your six cells wired a 3 parallel/2 series cells for a nominal 3 volt output with 3 times the MAH of a single cell. Good luck Lefty
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« Last Edit: January 02, 2012, 07:09:30 pm by retrolefty »
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« Reply #2 on: January 02, 2012, 07:40:59 pm » |
My initial thought was to power it with 6xAAA batteries + linear regulator. A linear regulator is also quite inefficient. If you look at power conservation designs they will use switching regulators, either a buck or boost or buck/boost design of some sort. The boost converter retrolefty linked to would be a good bet.
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« Reply #3 on: January 02, 2012, 07:46:31 pm » |
That is how it works, MAH capacity in series is same as single cell, in parallel the sum of their individual MAH capacity. That doesn't seem right - I think you are confusing the battery's energy capacity (mAh) and the current that can be delivered. If you have, for example, 6 1.5V 1200mAh batteries you will get a total of 7200mAh out of them no matter how they're connected. If they are in series, you get 9V at whatever current a single battery can deliver. If they are all in parallel, you get 1.5V but you can draw six times as much current. Either way, the total energy in the batteries is the same and when they're dead that's how much energy you will have extracted. Pete
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« Reply #4 on: January 02, 2012, 07:50:51 pm » |
Either way, the total energy in the batteries is the same and when they're dead that's how much energy you will have extracted. This was my reasoning, initially. But then I saw this in a website: Hooking the batteries in series will give you a voltage equal to the total voltage of all the batteries, but the Ah and current carrying capacity of only one. For example, three 12v 20 Ah batteries in series will give you 36v 20 Ah. If each battery could put out 200 amps max, then three in series will put out only 200 amps max. http://www.gizmology.net/batteries.htmPerhaps we're thinking in Amperes/hour that the battery can deliver, when we should be thinking in Watts/hour?
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« Reply #5 on: January 02, 2012, 07:55:49 pm » |
That is how it works, MAH capacity in series is same as single cell, in parallel the sum of their individual MAH capacity. That doesn't seem right - I think you are confusing the battery's energy capacity (mAh) and the current that can be delivered. If you have, for example, 6 1.5V 1200mAh batteries you will get a total of 7200mAh out of them no matter how they're connected. If they are in series, you get 9V at whatever current a single battery can deliver. If they are all in parallel, you get 1.5V but you can draw six times as much current. Either way, the total energy in the batteries is the same and when they're dead that's how much energy you will have extracted. Pete Your confusing power capacity with MAH capacity, they are not the same. Series connected cells will discharge at the same rate as a single cell if at the same load current draw, even though the series cells are delivering more total power because power = current X voltage, whereas MAH is just current delivered over a time period. MAH is not a total energy measurement, that would be watt/hour rating, where watts is of course current X voltage. So six series connected cells can supply the same watt/hours as six parallel cells, but each configuration has a differen voltage and current value for it's total capacity. Lefty
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« Last Edit: January 02, 2012, 08:00:19 pm by retrolefty »
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« Reply #6 on: January 02, 2012, 08:05:15 pm » |
Thanks Lefty. That's right. I was confusing power and mAH.
Pete
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« Reply #7 on: January 02, 2012, 10:03:04 pm » |
The higher voltage string and a buck converter will give you slightly longer battery live than a boost converter from an undervoltage string. AH ratings on batteries inprove at lower discharge rates due to lowered internal heating and longer times for chemistries to diffuse evenly during discharge.
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« Reply #8 on: January 02, 2012, 10:20:03 pm » |
Or am I breaking some law of physics here, and both methods give the same capacity?
Yes, you are suggesting a breach of conservation laws. Batteries, no matter how connected, give the same capacity. Energetic capacity is measured in watt-hours. BTW, if you connect the cells in parallel, the stronger ones will discharge into the weaker ones. Regardless of how new they are, there are slight differences in them and this is a recipe for rapid failure. Other options: 1) You could use 5 cells and connect to the DC jack 2) Use 4 cells and attach them to Vin. This avoids the polarity diode so the 117 regulator should be okay with 6V. 3) Use 4 cells and a 1N4001 diode and attach to the 5V input. This will deliver about 5V-5.2V to the board which should be fine.
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« Reply #9 on: January 03, 2012, 08:46:16 pm » |
Hi,
I was interested in this thread because I'm thinking of a solar-powered arduino application. If you had a small solar panel with variable output voltage something like this seems ideal. Does the variable resistor tune the output based on a fixed/known input like your pack of cells, or does it tune the output based on an internal reference voltage it achieves therefore making it suitable for random acts of sunlight?
My initial thought was to have the solar panel charge a battery, which in turn runs the Arduino & sensors, but with this could I avoid the middle-man (for daylight operations only obviously) ?
Thanks
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Measurement changes behavior
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« Reply #10 on: January 03, 2012, 09:08:37 pm » |
Hi,
I was interested in this thread because I'm thinking of a solar-powered arduino application. If you had a small solar panel with variable output voltage something like this seems ideal. Does the variable resistor tune the output based on a fixed/known input like your pack of cells, or does it tune the output based on an internal reference voltage it achieves therefore making it suitable for random acts of sunlight?
My initial thought was to have the solar panel charge a battery, which in turn runs the Arduino & sensors, but with this could I avoid the middle-man (for daylight operations only obviously) ?
Thanks
The variable resistor on that switching regulator sets only the specific regulated output voltage you wish to have, regardless of the input voltage to the module as long as the input voltage in not outside the minimum and maximum voltage limits of the module. So it's a set it once and forget it adjustment. So yes, the regulator module has an internal fixed voltage reference standard it utilizes to control the output voltage. Lefty
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« Last Edit: January 03, 2012, 09:10:11 pm by retrolefty »
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