• State Resolved
  • Locked Locked
  • Replies 2 replies
  • Answers 1 answer
  • Subscribers 62 subscribers
  • Views 279 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

BQ25601: Solar Powered Operated WSN Node

Ayla Shiva
Intellectual 680 points
Part Number: BQ25601
Other Parts Discussed in Thread: TPS63070,

hi,

I need to verify my design, that i'm working on, A Remote WSN node which will be powered by a Renewable Energy Source like Solar, and the Solar Panel Specs are as follows

Vmp=+6v

VOC=+11V

Imp=0.5A

Pmax=3.5W

I have selected BQ2601 as the charger IC for my Li+ion Battery 18650 type, requirements are mostly met with ours, And a Boost convertor like TPS63070 is used to power other sensors

Which will work at greater than +3.3V, I'm very keen to know that, Is this the Best approach of mine?, Or any  Best practices are suggestible which can be adopted into my design.Schematic_For-Eval_Sheet-1_20190314114451.pdf

  • 0
    TI__Genius 11850 points
    Hi Ayla,

    Your block diagram looks like it should work correctly.

    However, a couple of things to double check are your SYS_Min bits and your LDO's dropout voltage. If your battery voltage is lower than the value defined by the SYS_Min bits, then the BATFET operates in linear mode, and VSYS will be regulated about 180 mV above the SYS_Min setting. Therefore, make sure that SYS_Min + 0.18 V - LDO dropout voltage > 3.3 V (preferably with some extra margin) in order to avoid problems when the battery voltage is low.

    Another thing to consider is your battery capacity and how quickly you want to charge your battery. Since the max power your solar panel can provide is 3.5 W, the max charge current is limited. The BQ25601 will apply fast charge current to your battery when VBATLOWV < VBAT < VBATREG. The typical value from the datasheet's EC table is VBATLOWV_RISE = 3.12 V, and assuming your battery charge voltage is 4.2 V, then:

    VBATLOWV_RISE = 3.12 V -> I = P/V = 3.5/3.12 = 1.12 A
    VBATREG = 4.2 V -> I = P/V = 3.5/4.2 = 0.83 A

    Therefore, even with only the BQ25601 and the battery connected and no system running, the fast charge current is limited to about 1 A. This could make the charge time long if your battery capacity is high, especially if your system is running concurrently.

    Best regards,
    Angelo
  • 0 in reply to Angelo Zhang87
    Intellectual 680 points
    Angelo Zhang87, Thanks for the insightful answer, I will start integrating it.