How to design a 3.7v to 5V 2A stepup circuit ?

Discussion in 'General Electronics Chat' started by mebinpm32, Oct 16, 2017.

  1. mebinpm32

    Thread Starter New Member

    Oct 16, 2017
    4
    0
    Hi,

    Sorry for my bad english
    I m working on a project using raspberry pi. I need to power Pi(800mA), 5 inch LCD(300mA), 500mA current for IR leds simultaneously. Total around 2 Amps current i need.I saw many modules available online. But i don't want to buy these modules, i want to design a circuit like this with minimum size.
    Im using 3.7v 4000mAh li-ion battery as power source.

    I saw many circuits.....Can anyone help me please for finding the best circuit.
    Thanks in advance
     
  2. bertus

    Administrator

    Apr 5, 2008
    19,584
    3,996
    Hello,

    Have a look at the LM3478.
    Be aware of the high input current of about 3.2 A, so the battery life is limited

    Bertus
     
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  3. Dodgydave

    AAC Fanatic!

    Jun 22, 2012
    8,103
    1,399
    5V @2amp is 10W, from a 3.7V battery at say 90% efficiency is 3amps, your battery will last about an 1h.30mins..
     
  4. MrAl

    AAC Fanatic!

    Jun 17, 2014
    6,093
    1,311
    Hi,

    There are circuit boards you can use already built. Check Amazon, etc. Also watch efficiency and max current, get a current rating higher than you need by say 50 percent.
     
  5. mebinpm32

    Thread Starter New Member

    Oct 16, 2017
    4
    0
    As per bertus's suggestion, i got one circuit from the datasheet of lm3478. which is attached hereby,
    I think this will works fine.
     
  6. Tonyr1084

    Distinguished Member

    Sep 24, 2015
    3,230
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    At 1.6 amps (calculated from your figures) I'd recommend a supply that can deliver twice that much current. In other words I would want a supply capable of at least 3.2 amps. So 3.2 amps at 5 volts (3.2 X 5 =) 16 watts. If you had a "PERFECT" source of power (no losses due to efficiency) then 3.7 volts delivering 16 watts would require (16 ÷ 3.7 =) 4.3 amps capability. Your LI battery isn't capable of that much current. Now, your circuit (as has been pointed out) (and if it were a "Perfect" source) 1.6 amps (strictly holding to that number) would require (again a perfect source) 5 volts (times) 1.6 amps = 8 watts. 8 watts (divided by) 3.7 volts equals 2.16 amps.

    What we're getting at is that your battery isn't going to deliver the performance you want. And I always question "Why build when you can buy?" OK, I understand, and practice this too - I build even though I can buy but I do so simply for the fun and experience of it all. So if you truly want to build, I understand. But if you're solving a problem then it might be quicker and easier (and cheaper) to buy than to build and blow something up. But as for your power source, that remains in serious question. Unless you only need a little more than an hour of performance.
     
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  7. Tonyr1084

    Distinguished Member

    Sep 24, 2015
    3,230
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    Yes, I'd agree at a bare minimum. but I still like doubling the robustness of my circuitry.
     
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