I have found that the range of VSYS is 1.9V~4.4V and I want to set 3.3V as Curie's main power. Does it mean that I can power Curie with 3.7v Li battery directly and not need a LDO chip to transform 3.7V to 3.3V?
Thanks for your interest in the Intel Curie Platform.
I have checked the datasheet for the Intel Curie Module and it states that it only works with 3.3 V or 1.8V interfaces. Please see Section 4 of this datasheet: https://software.intel.com/en-us/articles/intel-curie-module-datasheet Intel® Curie™ Module Datasheet | Intel® Software .
Could you be so kind to provide me the link where you find this information?
I will be waiting for your reply, have a nice day!
I have found the information from Table 3-1 of this datasheet: https://software.intel.com/en-us/articles/intel-curie-module-datasheet Intel® Curie™ Module Datasheet | Intel® Software .
And from Figure 5 of https://software.intel.com/en-us/articles/intel-curie-module-design-guide Intel® Curie™ Module Design Guide ，it seems like I can connect VBATT to pin VSYS as main power of Curie and there is no need to transform 3.7V of the battery to 3.3V.
Am I right?
Seems like you are right! Reading again Section 4, it is talking about internal voltage regulators. I apologize for the misunderstanding.
Thanks for providing the right information!
Have a nice day!
The USB_3P3 is a dedicated LDO for the USB, if you power the Curie by using the VDD_USB you should be fine. Since the Curie is intended to be supplied by direct USB power, a charging device or a battery.
Hope you find this information useful, have a nice day!
I want to confirm the message you send. If I leave ball L4(VSYS) floating and at the same time ball K4(VDD_USB) is connected to USB power, the chip could still work well and every function is normal?