LarryD:
The Arduino has a built in power switch.
Your 7 volt to the power jack takes precedent over the USB 5v from the USB cable..
That mean that I can directly connect the arduino to the computer, while the arduino is fed into a source of 7v? (see image)
Maybe - the autoswitch circuit looks for Vin/2 to be greater than 3.3V to use power from the barrel jack.
There is a diode between the barrel jack and Vin.
At light current loads, the voltage across the diode might be small enough that 7V - Vdiode is > 6.6V; or it may not, and Vin/2 will be less than 3.3V and USB power will be selected.
You could measure Vin at the power header and see what voltage level you are getting.
@ OP
So you didn't read/understand the posts after that.
Just use 7.5volt on the DC jack, not 7volt.
7volt might not always override USB supply, 7.5volt will.
Leo..
If you are able to adjust the power supply potentiometer to get 7.5v, do so, then the Arduino power switching circuit will work properly as mentioned by everyone above.
.
new99:
(see image)
Well the ends of the cable are wrong, have you seen what sort of connector attaches to the Arduino?
Thank you very much, the truth has made it difficult for me to understand them for what I translate from English to Spanish and the translation sometimes is not correct.
As mentioned by CrossRoads, there is a diode between the barrel jack and the Vin port.
That means if I feed the arduino in port Vin/barrel jack with 8V. Since 8v is greater than 6.6 (Vdiode is > 6.6V), then the arduino will automatically select the power supply through the port Vin/barrel jack.
You are feeding 8V to the barrel jack, or to Vin? Both will result in Vin/2 of more than 3.3V.
CrossRoads:
You are feeding 8V to the barrel jack, or to Vin? Both will result in Vin/2 of more than 3.3V.
To the port Vin with 8v
