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Do I need a linear regulator in this circuit?

Discussion in 'General Electronics Discussion' started by syedj94, Aug 12, 2013.

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  1. syedj94

    syedj94

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    Aug 12, 2013
    Hi,
    I am trying to build a portable usb charger the outputs 5V when input is 1.5 or 3 Volts. I found this step up converter: goo.gl/RAV9dC
    The spec for the converter state that the input voltage is 1V to 10V, and the out put Voltage is -1.5V to -10V and 3V to 10V. I frankly don't know how to interpret this information.
    It seems to me that by inputting 1.5V, the output will be 10V. So I am wondering if am right, and IF I am, then do I need to put in a 5V voltage regulator in the circuit?
    Thanks!
     
    Last edited: Aug 12, 2013
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
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    Nov 28, 2011
    The MAX1044 is a switched capacitor charge pump. Its output current capability is relatively low and it's not suitable for charging a USB device such as a smartphone.

    You can buy a USB charger that operates from a single AA cell - in fact I have one. I haven't tested it properly yet, but I know that there is not enough energy in an AA cell (carbon-zinc or alkaline) to fully charge even a small cellphone. You would have to replace the cell at least once, possibly several times. So the whole idea is a bit impractical, unless you use a larger cell.

    In that case you would need a "boost converter" to convert ~1.5V to 5V. The converter would include voltage regulation. Efficiency is always a problem, especially at low voltages like 1.5V. Assuming efficiency is 80% and you are drawing 1A from the USB connector, current drain on the battery will be (1 x 5 / 1.5 / 0.8) which is 4.17 amps!

    Personally I would recommend using a 6V "lantern battery" - the rectangular kind with two springs on the top - and a low-dropout (LDO) linear regulator, if you want a portable internally powered USB charger.
     
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