Hi, i'm working with the controlStick f28069, I need to use the controlStick in stand-alone without connecting it to an usb port, and powering it by adding 5v at the pin 16.
In order to do this I must shortcircuit the R15, and remove the resistor R9 so the controlStick will boot from flash.
My question is: if I remove de R9 and shortcircuit R15, would I be able to debug with CCS thru the USB port? If the answer is yes, what was the point of adding the R9 resistor?
Rodrigo,Yes, if you remove R9 and short-circuit R15 you will still be able to debug the controlSTICK via USB and CCS.The purpose of not having R15 was to reduce any risk of a user connecting 5V to a GPIO pin accidentally. We had a request to not have it connect directly by some of our workshop training groups. A pull-up or pull-down on R9 serves to change the F28069 MCU's boot mode. If GPIO34 is pulled high (or disconnected) and TDO is pulled high (or disconnected), the F28069 should boot from flash. So whether R9 is populated or not shouldn't make any difference with the F2806x controlSTICK.C:\TI\controlSUITE\development_kits\F28069 controlSTICK\~F28069controlSTICK_HWdevPkg\R1\a resistor at R9 will only make the design more robust.Thanks,Brett
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