Using VOUT33 as main power for a K60

[jpderyck]

Hello,

I would like to know if we can always use the VOUT33 (with VREGIN connected to a 5V source and internal regulator enabled) as the main source of power for the K60 ?

I see that output current of VOUT33 is maximum 120 mA, is there a case when it is not sufficient for the K60, for example when all peripherals are enabled ?

best regards

JP

[adriancano · Accepted solution]

Hi,

The chip can also be powered by the USB bus directly. In this case, VOUT33 is connected to VDD. The USB regulator must be enabled by default to power the MCU, then to power USB transceiver or external sensor. You can check the information in the specific Reference Manual in the section USB bus power supply.

You can refer to the section Power consumption operating behaviors in the specific data sheet. The IDD_RUN value indicates the current consumption of the chip the all peripheral clocks are enabled and when all are disabled.

Hope this information can help you

Best Regards,
Adrian Sanchez Cano
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[mstroven]

As Jean-Paul notes, the VOUT33 pin has 120mA output capability.  However, in the datasheet, the maximum digital supply current is listed at 185mA.  So it would be helpful to know just how much current each internal peripheral requires.  For example, can I run the K60 at 100MHz with the ethernet peripheral running in 100mbit mode at 90% bandwidth?  Would I have enough current available to also power the ethernet PHY?

Has anyone done this without an external 3.3V regulator?