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Topic: USB and external power supply both plugged -> Need some clarification (Read 440 times) previous topic - next topic

fmuller1

Hi,

I am starting to use the Arduino UNO Rev 3, and I was a bit looking around at the schematic to understand how things are laid down. My skills are limited in hardware as I'm more of a firmware/software guy...

So I was trying to understand how the Arduino is behaving when both USB and external power supply are plugged; and I think I well understood the role of the MOSFET openning the USB VCC when the external power supply V1 is above 7V.

Where I am not sure to understand, is how does it behave when the power supply isn't reaching enough voltage to get 3.3V out of the op amp, and feeding the gate of the MOSFET. In that case, the USB VCC will not be disconnected and will be used to feed U2. But since there is still an external power supply connected, we will still have something at the output of U1, and therefore we will still have the +5V line carrying a voltage. It won't be 5V but maybe 3.3V or less depending on V1, but there will still have a voltage. And so at the source of the MOSFET, we will have something coming from U1, as well as the USB VCC.

Am I understanding right or what am I missing? I am just really confused by the 5V node that isn't disconnected even when V1 isn't high enough to disconnect USB VCC through the MOSFET...

Thanks for your help, it really bugs me not to understand properly how this part of the schematic works...

PS: take a look at the picture I attached to understand what +5V node I am talking about  :)

promacjoe

When an external power supply is used, and it is above 7V, The comparator goes high and the P-channel MOS FET is turned off. Now the MOS FET works as a diode.

If the external power supply is less than 7V, the 5V regulator will not function properly. It will either produce less than 5V, or shut down completely. the comparator turns the MOS FET on, the MOS FET will allow voltage to pass through, With very low resistance.

Actually, the MOS FET, could remain off regardless of the external input voltage. It would act as a diode, and the 5V supply would come from whichever source was higher.

fmuller1

Thanks for your reply promacjoe :)

What you said got me actually thinking more about that regulator used  to get the 5V out of V1. And you're right, I checked and if V1 < 6.5V, it won't output any voltage.

I didn't pay attention to that when I posted my question, that is why I was worried that the regulator could output a voltage even with V1 < 6.5V. Which would have led to 2 power supplies set in parallel and supplying at the same time.

Anyway, thanks again for your help :)

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