Hello dear friends. I'm confused with PF1550 datasheet. There is information about power path but I can't understand how it works. If I correct:
1. My analog scheme I should feed from VSYS;
2. VSYS is power path and its output current depends on equality: VBUS_CURRENT (1500 mA) - CHARGING_CURRENT (1000 mA) = 500 mA
Maybe I'm mistaken and VSYS output current is anything else. Maybe VSYS is not power path...
Sorry, but my English is not good enough to understand so big document, please help me.
Perfectly, I want to charging my battery with 1A current minimum (1.5A recommended) and drive my analog scheme from power path when I need 3.7V and 1.5A too. Is it really with PF1550 or should I use something enough to provide such a power path for my analog scheme?
CPU: i.MX7ULP with LPDDR3.
Solved! Go to Solution.
Hello Vladislav,
I apologize for a late reply. I have just received an answer from an application engineer I have contacted regarding your question. Please see the description.
DESCRIPTION
I think the customer want to know if they can use VSYS to supply power to external .
Yes , he can use VSYS in this way.
If VSYS load is heavy , It can work on Supplement mode , and Battery stop charging , or change to dicharge if VSYS < Vbat.
The max current capability VSYS can power to external is:
VBUS_LIN_ILIM(Max 1500 mA) minus VSYS current to power PMIC (3BUCK+3LDO ).
Customer can ignore the charging current to battery here.
With Best Regards,
Jozef
Hello Vladislav,
I apologize for a late reply. I have just received an answer from an application engineer I have contacted regarding your question. Please see the description.
DESCRIPTION
I think the customer want to know if they can use VSYS to supply power to external .
Yes , he can use VSYS in this way.
If VSYS load is heavy , It can work on Supplement mode , and Battery stop charging , or change to dicharge if VSYS < Vbat.
The max current capability VSYS can power to external is:
VBUS_LIN_ILIM(Max 1500 mA) minus VSYS current to power PMIC (3BUCK+3LDO ).
Customer can ignore the charging current to battery here.
With Best Regards,
Jozef
Thank you very much!