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I have a circuit that has a battery output voltage of about 7V and can also have a charger input that when plugged in will output around 9V. The battery/charger power a microcontroller (regulated down through an LDO) as well as a motor driver (simply shown here as a current limited transistor).

The battery charging is taken care of separately through a multi-cell PMIC, but I need to isolate the battery positive output from the charger input when the charger gets plugged in. I figured I could use a transistor switch to control this, but I thought that an easier way to separate the supplies would be to simply put a Schottky diode from the lower voltage to the junction where they connect. I want to confirm the behavior of the diode and if this design seems reasonable (assuming the diode can handle the load current and that it is low turn on voltage).

I know that when the charger is plugged in, the diode will not conduct b/c it will be reverse biased; however, I am not quite sure on the behavior when the charger is not plugged in. It would seem that the cathode side does not have a defined voltage, so I am not sure if it would be considered forward biased or not in this case. I would imagine the diode would conduct but would like a more rigorous answer.

Also, do not assume the driver (represented by Q1) is turned on.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Do you care about the voltage, or the current? "Conduct" implies the latter. \$\endgroup\$
    – CL.
    Commented May 20, 2015 at 7:13
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    \$\begingroup\$ If there is any load between cathode and ground then, providing the anode voltage is sufficient to overcome the 0.7V barrier, a current will be drawn and the only place it can come from is through the diode. \$\endgroup\$
    – Chu
    Commented May 20, 2015 at 7:21
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    \$\begingroup\$ The other side is not "floating", it's connected to the rest of the circuit. \$\endgroup\$
    – pjc50
    Commented May 20, 2015 at 8:16

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With the charger disabled by SW1 the diode will conduct and produce a voltage on the cathode of about 0.5Volts lower than the 7V battery. Under heavier load conditions this volt-drop might be a little higher. Under lighter load conditions this may be a little smaller.

BTW diodes don't have a "V_cutoff" they have a forward volt drop.

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