Hi
I has some questions about BQ25010.
1) I connected the battery voltage pin (3.7~4.2V) to the "EN" (DCDC enable pin).
I don't want the BQ25010RHLR to activate the DCDC when the USB cable is connected.
Is it okay to connect the battery to that pin?
Making that always enable seems a problem. So I connected the DP3T switch to make the VOUT floating.
2) I referenced the BQ25010EVM to create 3.3V as the output.
I used 680K (R13) and 120K (R14) with extra capacitors (C33, C34 and C35).
I also used 22uH (L4) for more load current.
Will that configuration work to create a 3.3V?
Also, should I use bigger inductor since this powers BLE which uses small current?
Although the LED driver will be using the BQ25010's output voltage as the logical voltage to use I2C,
the boost conver and the LED will use most of the current.
3) Just to check, if I want the output to be 3V,
can I replace the resistors to 750K (R13) and 150K(R14)?
Then what will be the value for the capacitors (C33, C34 and C35)?
4) I don't use the AC input.
Then do I need to connect a resistor to ISET1?
5) Can my battery be charged 500 mA charge rate?
I uploaded the battery's datasheet. Check the MS word file.
6) To use the 500mA charge rate, the ISET2 has to be high.
I'm confused that what voltage level is meaning high.
So, connecting the 5V USB voltage to the ISET2 pin configures the charge rate to 500mA?
7) The nRF MCU's GPIO can handle up to 3.6V.
I'm trying to use the STAT1 and 2 pin to read the status state.
However, I'm confused that what voltage level is meaning high.
If STAT1 goes high, what voltage does it shows?
3.3V? Equal to the battery's voltage? Or does it shows 5V when the USB is connected?
8) Can I always tie the FPWM to GND?
I'm not sure if the chip outputs 3~3.3V and the currents I want.
In the "Power-tps51" picture, which uses TPS61089RNRR
9) I copied the schematic from the typical application which creates 9 volts.
So the connector makes the VBOOST (output of TPS61089RNRR) and VLED connected.
VLED is connected to the anode(+) of the white LED.
Thanks