LM7805 Backfeed Protection

[NateT94]

Hi,
I am a beginner in circuit design and had a question about back-feed protection for an LM7805 voltage regulator.
The circuit I am designing contains multiple IC's and an Arduino that are all powered from a 5-volt rail that comes from an LM7805.
I would also like to add a USB connection so that I can program the Arduino. The 7805 will not be supplying power while the USB cable is attached, the 5-volt rail will be powered through the USB cable.

My question is, what is the best method for protecting the LM7805 from back feed voltage when the 5v rail is being powered over the USB cable.

My first thought was to simply put a diode on the 7805's output, but that would drop the regulated voltage so it's not optimal.

Thanks.

[Zero999]

Put a diode between the LM7805's output and input.

[NateT94]

The 7805 would normally be powered by a 12v wall wart, however, if the 12V supply is disconnected while the USB is connected, then where exactly would the diode be directing the voltage?

I don't really understand how that would protect it.

[bobbydazzler]

I'm not sure the diode's needed unless you're running high output voltage(which 5v isn't) and a 10uf+ output capacitor.  Not sure if that recommendation extends to external 5v power while the 7805 is still connected but off though.

[ledtester]

The 7805 would normally be powered by a 12v wall wart, however, if the 12V supply is disconnected while the USB is connected, then where exactly would the diode be directing the voltage?

I don't really understand how that would protect it.

The diode is used as described in sections 8.1.1 and 8.1.2 on page 13 of the TI's datasheet:

https://www.ti.com/lit/ds/symlink/lm7800.pdf

8.1.1 Shorting the Regulator Input

When using large capacitors at the output of these regulators, a protection diode connected input to output (Figure 15) may be required if the input is shorted to ground. Without the protection diode, an input short causes the input to rapidly approach ground potential, while the output remains near the initial VOUT because of the stored charge in the large output capacitor. The capacitor will then discharge through a large internal input to output diode and parasitic transistors. If the energy released by the capacitor is large enough, this diode, low current metal, and the regulator are destroyed. The fast diode in Figure 15 shunts most of the capacitors discharge current around the regulator. Generally no protection diode is required for values of output capacitance ≤ 10 μF.

8.1.2 Raising the Output Voltage Above the Input Voltage

Because the output of the device does not sink current, forcing the output high can cause damage to internal low current paths in a manner similar to that just described in Shorting the Regulator Input.

[Zero999]

The 7805 would normally be powered by a 12v wall wart, however, if the 12V supply is disconnected while the USB is connected, then where exactly would the diode be directing the voltage?

I don't really understand how that would protect it.

The current would be directed to the LM7805's input. It won't do anything, except for draw a small current.

[Terry Bites]

The data sheet tells all!

[NateT94]

That makes sense. Thanks everyone.
It seems so obvious now haha.

[Noise0]

Hi,
I am a beginner in circuit design and had a question about back-feed protection for an LM7805 voltage regulator.
The circuit I am designing contains multiple IC's and an Arduino that are all powered from a 5-volt rail that comes from an LM7805.
I would also like to add a USB connection so that I can program the Arduino. The 7805 will not be supplying power while the USB cable is attached, the 5-volt rail will be powered through the USB cable.

My question is, what is the best method for protecting the LM7805 from back feed voltage when the 5v rail is being powered over the USB cable.

My first thought was to simply put a diode on the 7805's output, but that would drop the regulated voltage so it's not optimal.

Thanks.

Usually i dont disconnect external power when usb is connected.
Usb connection is usefull not only to program it but even to get some serial print for debug or logging.
This way you don't even have to worry about back feed issues.