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I made a PCB with an arduino clone and a tiny motor driver. The arduino clone itself needs +5 volt power supply, so I decided to use a DC/DC boost converter to boost a 3.7 volt Li-Ion battery to +5 volt. Then, a battery charger comes to my mind. After some thinking time, I wired my circuit like this, sorry for bad lights and an misspelling "converter"... my circuit

I used one-cell Li-Ion charger bq24032a, according its datasheet, it can accept power from USB and/or AC( AC not used in my circuit). It can also provide power to system while charging the battery.

Some facts I discovered:

  1. ONLY USB presented, battery absented. Bq24032a's OUT pin voltage will be +5 volt approximately( with no load).
  2. ONLY Battery presented, USB absented. Bq24032a's OUT pin voltage will equal to battery's voltage approximately...of course...( with no load).
  3. USB and battery presented, Bq24032a's OUT pin voltage will be +5 volt to battery's voltage. The charger is charging the battery, so the OUT voltage just lowered...

I used tps61032 for DC/DC converter, it is dedicated for one-cell Li-Ion battery boost to +5 volt, but accept input voltage for 1.8-5.5 volt.

So I have two question:

  1. Is my circuit's block diagram just looking OK?
  2. When bq24032a's OUT voltage is +5V( ONLY USB present), is it OK for tps61032 to boost a input +5 volt to a +5 volt output? Sounds weird...

By the time I posted, I already have my PCB in hand... But I cannot figure out the two question( and many many tiny questions...). Answers desired... thanks :)

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    \$\begingroup\$ You can probably get away with a lower voltage if you change out the crystal. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Dec 24, 2013 at 11:20

2 Answers 2

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A1: The block diagram is a good start. A schematic would be even better.

A2: 5V is fine. The data sheet shows it is 97% efficient with a 5V input (see figure 7 in http://www.ti.com/lit/ds/slus534e/slus534e.pdf)

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Regarding your first question, the block diagram doesn't really add anything to the discussion and it doesn't have enough detail to say whether it will work or not. For the benefit of readers who pay for their data I would suggest leaving it out, or drawing a proper diagram instead to taking a poor photo of a whiteboard.

As to the second question, the TI data sheet indicates that the converter will operate with input voltages from 1.8V to 5.5V. That suggests that you should be able to use it from 3.0V to 5.0V. If you have the PCB in hand, build it and find out.

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    \$\begingroup\$ Sorry for the block diagram gives little info for you. But from my point of view, block diagram just gives circuit ideas for people for quick view. My PCB worked, and the boost converter worked at +5 volt input also. But I cannot decide if it is reasonable to give a +5 volt input for a +5 volt output boost converter. \$\endgroup\$
    – iouzzr
    Commented Dec 24, 2013 at 15:18

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