Hello,
I have a circuit where the USB provides the 5V and with an LDO I get also 3.3V.
I have a doubt, what happens if I power my circuit with USB (so I have both 5 and 3.3V) but also with an external power source that connects to the 5/3.3V?
Is it a problem? Why?
It would be nice to actually see what you are proposing on paper as in a drawing/schematic. would it be a problem? No, not if done correctly.
Ron
It's not necessarily a problem. The Arduino has a 5V regulator and a 3.3V regulator. Where I work we have a board with 5V. 3.3V and 1.8V.
It can be a problem when mixing & matching logic signals. A 5V chip "talking to" a 3.3V chip or vice-versa.
It would be nice to actually see what you are proposing on paper as in a drawing/schematic. would it be a problem? No, not if done correctly.
Absolutely not a problem, it's not the whole schematic but the meaningful part for this question
The USB 5V cames from J1 then with U2 we get the 3.3V.
From the two headers J2 and J3 we expose the 5V and 3.3V. I may connect external power sources to these two headers (J2 and J3)
So? Do you think that it is ok to connect power from J1 and J2/3 at the same time? Or should I add something in the middle?
I normally cut off the 5v pins from my FTDI connectors. (or at least bent them out of place)
When uploading the new sketch the external power supply would power the controller.
I have a circuit where the USB provides the 5V and with an LDO I get also 3.3V.
I have a doubt, what happens if I power my circuit with USB (so I have both 5 and 3.3V) but also with an external power source that connects to the 5/3.3V?
Is it a problem? Why?
Yes, its a problem any time there's 2 power sources connected in parallel, unless designed with this capability.
Its not clear (to me) what's happening with +3.3V. If U2 provides +3.3V power to J2-14, then this is OK. However if both U2 and J2-14 provide +3.3V power, they shouldn't be connected in parallel.
A simple solution would be to use a Shockley diode for isolation, to prevent external power from back feeding to the USB power. So, it looks like you need:
J1-4 (VUSB) +5V ---->|---- J2-15 +5V
U2-2 +3.3V ---->|---- J2-14 +3.3V
For reference, see section A-1, component D2 from the Nano Every schematic.
Yes, its a problem any time there's 2 power sources connected in parallel, unless designed with this capability.
Thanks dlloyd
However if both U2 and J2-14 provide +3.3V power, they shouldn't be connected in parallel.
Yes, both can be connected in parallel
A simple solution would be to use a Shockley diode for isolation
But wouldn't the forward voltage (Vf) create problem? for example I have a few 1N5819 diodes home and these have a Vf of 0.6V. If I apply it to the 5V I would get 4.4V and if applyed to 3.3V I would get 2.7V. Wouldn't this create problems?
Also from the schematics of the Nano every I can see that U2-5 which provided the 3.3V is not protected with a Shockley and the 3.3V pad is exposed also in the J2-2. So maybe it is not a problem to have two voltage source connected in parallel?
But wouldn't the forward voltage (Vf) create problem? for example I have a few 1N5819 diodes home and these have a Vf of 0.6V. If I apply it to the 5V I would get 4.4V and if applyed to 3.3V I would get 2.7V. Wouldn't this create problems?
With the 1N5819 at 25°C and 500mA load, VF would be about 0.35V. I think the 1N5817 would have improved characteristics.
Also from the schematics of the Nano every I can see that U2-5 which provided the 3.3V is not protected with a Shockley and the 3.3V pad is exposed also in the J2-2. So maybe it is not a problem to have two voltage source connected in parallel?
Yes, they didn't protect the 3.3V. Probably because the Nano Every is a 5V board and they intended the 3.3V to power some additional user circuitry. Protecting the 5V is more important ... the USB power-cycles when it enumerates. For 3.3V, check the regulator's datasheet. See if there's any issue with applying 3.3V power to the output while Vin is at 0V.
Thanks, I am going to protect only the USB too, I am not fit for that Vf fall on the 3.3V output