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Can I connect USB voltage after 7805 (at Vout) and left Vin disconnected?

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K33rg4t3

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Hey.
I am making a simple circuit that should be powered from 8V+ unregulatred through the 7805 regulator OR through the USB.
I know that I never should connected two power supplies at once.
My question is: Can I connect USB voltage after 7805 (at Vout) and left Vin disconnected? Will it somehow affect the circuit in the bad way, or the 7805 will behave "safe", as diode or something?

Here is the simple schematic, I am aking about this particular case. Is this safe or no, and, why?
powersupplyquesrion.png
PS: I know that the connecting two power supplise at once is bad, I am asking only about this particular case (Vin left disconnected, only 5V from USB connected)
 

Pay attention to the application hints in the 7805 datasheet about "input short", in case of doubt install the suggested diode.
 

Yes, reverse bias protection. According to the new (TI originated) datasheet, up to 7V reverse bias won't be a problem. I didn't notice this before.

For other regulators, read the datasheets.
 

The word "reverse" nor the "bias" does not exist in this datasheet.
**broken link removed**
Any suggestions?
 

Forbids in sense "reverse bias never occur so its safe to connect" or in sense "reverse bias always occur so never connect it in reverse"? I am having troubles with translating/interpretating that information
 

Hi. It means that it can occur. It can be caused by kickback from a motor, or a large value capacitor - if I remember well - discharging back into the OUT pin of the voltage regulator, ... or a USB power line connected on the same line I imagine, so protect the 7805 with the diode as the datasheets show: from OUT to IN, cathode connected to OUT, anode connected to IN, to be sure you won't back-feed the 7805.
...Or back-feed your USB with the 7805, looking at the schematic you posted, if that's the complete plan, then at a guess I'd put a protection diode in series with, or in parallel to, the output of the USB if it's possible as well.
 

If the regulator is a 7805, I would not recommend it. There are many variations of 7805 and some may be tolerant, some not. If the regulator is a 1117 as shown in the diagram, they CAN tolerate higher voltage on their output than input but if you really want to be safe, and I stress I do not recommend it anyway, you can wire a diode across the input and output pins with the cathode at the input pin. That will ensure the input stays relatively close to the output when USB powered. I caution you that wiring it that way will protect the regulator but it will also conduct USB power to anything else wired before the regulator, including possibly a large reservoir capacitor that will overload the USB while charging.

The only safe method is diodes in series with each supply, you will suffer some voltage drop though. Using Schottky diodes will minimize the drop but not eliminate it. You can compensate for the drop when the regulator is providing power by lifting it's ground pin up with another diode of the same type but you cant do it to the USB source.

Brian.
 

Hi,

If I remember right, then LT1121 is specified for this.
As well as when input is open or short.

Klaus
 

...protect the 7805 with the diode as the datasheets show: from OUT to IN, cathode connected to OUT, anode connected to IN, to be sure you won't back-feed the 7805.

...Sorry, bad mental lapse there last night, I know what I meant but somehow wrote nonsense, my apologies: it is anode to OUT cathode to IN. As Betwixt said: " you can wire a diode across the input and output pins with the cathode at the input pin"

...

- - - Updated - - -

...also, I was thinking, could you maybe get around having both power lines connected together by using either a SPST or DPST/DPDT (you'd need to look at what type would work) pcb mount relay to completely disconnect the voltage regulator when it is not in use? Or something like a MOSFET High Side Switch?
 

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