PIC18F14K50 - USB clarification required

[benoitstjean]

Compiler: 5.026
Device: PIC18F14K50

Hi guys, for now, this is mainly a pure PIC question but it'll probably lead to CCS USB questions when I get there.

This is my first try at USB on a PIC so please bear with me. I have a little dev board with a 18F14K50.

I'm going through the PIC documentation and what I want to do is be able to power-up the device from an external source but also via USB. In both instances, the entire circuit will be powered via 5V: the external power will have a 5V regulator otherwise the USB port will provide 5V to the circuit. The USB will also be used to flash the device with a new firmware using a bootloader that will be previously programmed on the device.

My question:

On page 243 section 22.2.2.1 <Internal transceiver>, the third paragraph states "The internal USB transceiver obtains power from the VUSB pin [...] VUSB must be supplied with a voltage source between 3.0V and 3.6V".

In the docs DS40001350F page 262, figure 22-11 <Dual power example>, the VUSB pin is tied to a cap to GND. In the few schematic examples I found online, the USB does not appear to be used.

So why does one page say that VUSB must be supplied with a voltage source but page 262 shows a cap tied to GND?

Any pointers appreciated.

Thanks!

Ben

[temtronic]

Deja vu , I think....
This sounds real familiar and I think Mr. T (?) explained it a year or so ago. Maybe use 'search' and see what comes up but I recall that one of them paragraphs is wrong(maybe copy/pasted) from another device's datasheet ?
Jay

[benoitstjean]

Yeah, I'll continue my research and try to search the forum but that's one problem - the search engine is terrible and anything can come-up.

Thanks.

Ben

[Ttelmah]

In this case it's because the data sheet is for both the 1xK50, and the LF version. The 'standard' version has a built in 3.3v regulator to feed this pin. The LF version does not have this, and you have to supply the pin.

So the pin needs 3.3v, and on chips without the regulator, _you_ have to supply this. On chips with the regulator, you only need to enable the regulator and supply the decoupling on the pin.

Look at figure 22-1, and Note2 below this.

[benoitstjean]

Ok. The circuit I found has an "opened" jumper to VBUS decoupled to GND with a cap. If that jumper is closed, then it is fed 3.3V. I'm just curious as to why the manufacturer put a jumper on that line if 3.3V is mandatory.

So then for the PIC18F14K50: I **must** feed it with 3.3V. This will then activate the internal USB circuitry. Is this correct?

The dev kit I ordered is 1188-1092-ND from DigiKey. The schematics is included.

I think I'm good for now.

Thanks!

Ben

[Ttelmah]

No.

If you have the PIC18LF14K50 (note the L), you must feed the pin with 3.3V. If you have the PIC18F14K50, then all it needs is the capacitor. The board jumper is so it can support both chip versions.

[benoitstjean]

Ah! Ok! Gotcha. I have the 'F' version so *just* the cap and that's it.

Thanks!!

Ben