[TriEmbed] How to Damage an LDO

[Pete Soper]

On 12/08/2015 05:36 PM, Brian via TriEmbed wrote:
> Hi,
>
> If a linear voltage regulator sees its nominal output voltage across 
> its GND/OUT pins while its IN pin floats, is it going to be damaged?
>
> For example, a 7805 regulating a 9 volt battery, but the battery is 
> disconnected and an AC adaptor is supplying 5 volts.
>
> This seems like a thing I should /just know/, but I can't figure out a 
> good way to google for it, or to divine it from a datasheet.
Dear Brian (not to be confused with either Brian or Brian),
   The scenario you describe is classic and raises an interesting point. 
A lot of semiconductor devices cannot tolerate the situation you 
describe because internal clamping diodes end up conducting more than 
the few milliamperes they're intended to handle. As Carl said, the data 
sheet is the first place to look.
    Most of us have read a thousand times that the limit for a pin such 
as a GPIO pin is (for example) "VCC+.3v". But guess what: if VCC is zero 
*the limit is .3 volts!! *Applying more than this to the GPIO pin causes 
current flow through internal circuits (typically a diode first) that 
can only tolerate a few milliamperes at most. Without a resistor in 
series with the pin to limit current this is a recipe for escape of the 
magic smoke. If the internal diode destroys itself in a shorted state, 
then the next time VCC is applied some other parts of the chip will be 
fried. If the diode ends up as an open circuit then the normal input 
voltage protection provided by the diode is lost (i.e. a "VCC+x volts" 
applied to the pin will be passed into the more sensitive parts of the 
chip instead of being safely clamped to VCC or ground). Again, this 
varies with different devices, but for MSP430 (from memory) the current 
limit is just two milliamperes.
   But returning to linear regulators, on page 8 of an old TI datasheet 
<https://www.sparkfun.com/datasheets/Components/LM7805.pdf> I found is a 
description of the standard precaution for simple regulators (a diode 
"backwards" across the regulator from output to input).
    A more detailed treatment is here 
<http://www.physics.utah.edu/%7Elebohec/P3610/LecNotes/Lecture09.pdf> 
(see page 9-4).
    Carl's solution would be OK if you weren't picky about what voltage 
actually shows up on the other side of the schottky diode (the Vf 
changes a lot with current and temperature!).

-Pete



>
>
>
> Thanks!
> -Brian (one of many)
>
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