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I want to know if there is a way of increasing a voltage drop by a given battery at a time.

So, I have this hp 50g graphic calculator here which uses 4 AAA batteries and they last like 4 hours of constant use. I've used a battery bank on the calculator usb port but it doesn't last long either because if there is a usb connected to the battery bank, even if the other end of the usb cable is disconnected, a small irritating green led stays on, and the bank lasts like 3 days disconnected from everything.

A solution that I've come up, is to buy a Tp4056 lithium battery charger and use a 3.7v nk18650 battery, then install it at the bowels of the calculator to make it rechargeable by the usb port.

So I need to make 3.7v go to at least 5v, (because the low bat warning starts at 5v), or to simply come up with two or three of the batteries and make the voltage go up 4.2v, because that voltage is the max chargeable voltage of the tp4056.

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  • \$\begingroup\$ Would it not be easier to use NiMh cells, preferably the Low-self-discharge type and make some arrangement so that you can charge the NiMh cells through an adapter. The problem with using a boost converter is that it will never have a low enough quiescent current. \$\endgroup\$
    – Bimpelrekkie
    Commented Sep 21, 2016 at 8:18

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Use boost converter to step up 3.7 V to 5 V. There are various range of input/output voltage of boost converter chip.

You can also simply buy boost converter module, it is quiet cheap on the market.

Below is an example of boost converter: Boost converter

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  • \$\begingroup\$ oh nice, and it works well on low currents? Because the calculator needs some 50uA when in standby so it can retain it's memory \$\endgroup\$
    – Miguel Henrique Salviatti
    Commented Sep 21, 2016 at 3:46
  • \$\begingroup\$ Is there a version of this circuit that provides an output of 14V @ 2A? If not, what would a person need to do to make such a circuit? \$\endgroup\$
    – DIYser
    Commented Sep 21, 2016 at 3:59
  • \$\begingroup\$ I think it will work well on low current, probably will produce ripple voltage but it will not effect to the calculator. \$\endgroup\$
    – Tristanto
    Commented Sep 21, 2016 at 4:17
  • \$\begingroup\$ Quiescent current will be about 300 uA - if this is too much then look for a different regulator. \$\endgroup\$
    – Andy aka
    Commented Sep 21, 2016 at 7:36

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