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"...
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:
- ONLY USB presented, battery absented. Bq24032a's OUT pin voltage will be +5 volt approximately( with no load).
- ONLY Battery presented, USB absented. Bq24032a's OUT pin voltage will equal to battery's voltage approximately...of course...( with no load).
- 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:
- Is my circuit's block diagram just looking OK?
- 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 :)
Answer
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)
No comments:
Post a Comment