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Given this diagram, where Vcc is a 3.3V supply powering an Arduino, if I were to connect both a USB and a battery at the same time, which will be used to power the board and why?

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    \$\begingroup\$ The cathodes of both diodes always are at the same voltage. Consider what this implies about the current through the diodes. \$\endgroup\$ – CL. Mar 26 '17 at 21:20
  • \$\begingroup\$ Whichever one has a higher voltage will power the board. When the voltages are close, there will be some sharing. In other words, the transition will be gradual, not an instant switch-over. \$\endgroup\$ – mkeith Mar 26 '17 at 21:29
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That depends on the voltage on the battery.

If the battery voltage is greater than VBUS.5V plus a diode forward drop then D2 will be forward biased and D1 will be reverse biased. The battery will be driving the regulator and the USB will be effectively disconnected.

If the battery is less than VBUS.5V minus a diode forward the converse is true. D2 well be reverse biased and D1 will be forward biased. The USB will be driving and the battery will be effectively disconnected. However is will receive the diodes leakage current as a charge current.

Anywhere in between and the current will be divided between them in a ratio that depends on the forward voltage across their respective diodes.

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  • \$\begingroup\$ This is because the higher the voltage the faster the current? Also, what do you mean by forward and reversed biased? \$\endgroup\$ – user7252850 Mar 26 '17 at 21:30
  • \$\begingroup\$ wouldn't current still be going through both diodes? \$\endgroup\$ – user7252850 Mar 26 '17 at 21:39
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    \$\begingroup\$ @user7252850, see this question electronics.stackexchange.com/questions/294974/… \$\endgroup\$ – Trevor_G Mar 26 '17 at 21:59

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