what happens if i disconnect the earth pin on a 78L05 - Page 1
Forum section: electronics / general

author: Simon
author_rank: Unknown
timestamp: December 21, 2011, 12:57:23 pm
content: So say I have a 78l05 with 12v on the input and naturally 5V coming out. What will happen if i disconnect the ground pin ? will current stop flowing or will the reg loose all reference and output a random voltage

author: olsenn
author_rank: Frequent Contributor
timestamp: December 21, 2011, 02:07:30 pm
content: My guess is that all current will stop. The 5 volt output requires a greater than 5 volt input between the GND and IN pins. If the GND pin is floating than than so is your power supply

author: firewalker
author_rank: Super Contributor
timestamp: December 21, 2011, 02:22:29 pm
content: My guess is that the voltage will be slightly bellow 12 votls. Don;t remember clearly the internal logic though.Alexander.

author: Kiriakos-GR
author_rank: Super Contributor
timestamp: December 21, 2011, 02:49:11 pm
content: Quote from: firewalker on December 21, 2011, 02:22:29 pmI agree the 78L05 it would look as an simple resistor, or better it would look similar with a diode that causes 0.7V drop.

author: Simon
author_rank: Unknown
timestamp: December 21, 2011, 03:09:43 pm
content: I know that under 5V it will pass the output and will reach 5+ the dropout voltage until something like 7v is across it. maybe it depends on the regulator type as well ?

author: firewalker
author_rank: Super Contributor
timestamp: December 21, 2011, 07:26:48 pm
content: Just tested it. 12.00 in 11.52 out. A 12 volt 11 watt incandescent lamp functions normally. So, I guess that loosing the ground will fry your circuit.Alexander.

author: Simon
author_rank: Unknown
timestamp: December 21, 2011, 07:27:38 pm
content: hm, thanks for that, was it a standard one or a low drop out one ?

author: firewalker
author_rank: Super Contributor
timestamp: December 21, 2011, 08:11:41 pm
content: A normal one (78T05 though).http://www.alldatasheet.com/datasheet-pdf/pdf/53632/FAIRCHILD/MC78T05CT.htmlAlexander.

author: Unknown
author_rank: Unknown
timestamp: December 21, 2011, 10:09:22 pm
content: Note that unless it's spelled out in the data sheet, you're relying on unspecified behavior that may be different between manufacturers or may change at any point in tie. You may have to do your own verification for this spec. I wouldn't expect much variation in something as simple as an 7805, but you never know.

author: ivan747
author_rank: Super Contributor
timestamp: December 21, 2011, 10:11:15 pm
content: I have just tried it with a 78L12 regulator. The voltage accross the input and the output pin remains relatively constant. I attached a 1.5K resistor as a load and the regulator behaves like a diodes, basically. The voltage drop increases a bit when the input voltage goes up to 15V, goes.up to 0.8V, basically. When applying 7V as input the voltage drop is 0.6 V. This, presumably, is caused by the increased current flow when the voltage goes up. The increased current flow.is caused by the resistor.

author: Simon
author_rank: Unknown
timestamp: December 21, 2011, 10:17:02 pm
content: ok thanks guys, I'll have to rethink this slightly

author: ejeffrey
author_rank: Super Contributor
timestamp: December 21, 2011, 10:44:21 pm
content: I would expect that most any 7805 will behave mostly the same way: the output voltage will float up to the the dropout voltage as long as you have a minimum load current. You could make something that operated differently, but I expect you would have to go out of your way quite a bit. If you want to shut off current flow without mucking about with high-side switching just get a regulator with an enable pin...

author: A Hellene
author_rank: Frequent Contributor
timestamp: December 21, 2011, 11:07:19 pm
content: In that class of voltage regulators, the ground pin serves two purposes: To power the internal circuitry of the regulator and, most importantly, to provide the regulator with the 0V reference in order for the error amplifier of the regulator to know the exact voltage between the output and thesystem ground, which is not necessarily equal to the Earth's potential --it is just the point of the circuitwe declareas the specific circuit's ground with a potential of 0V.In a few words, the '7805 will always be trying to maintain a +5.0V potential diference between its "Vout" and "GND" pins.The regulator's output voltage is given by the following equation: Vout = Vref + Voffset, which means that if you drive the "ground" pin of a 5.0V regulator to, say, 4.0V, the output voltage will become 5V + 4V = 9V. Driving, for example, the "ground" pin to -3.1V the output voltage will become 5.0V + -3.1V = 1.9V. Setting the "ground" pin to 0V will force the regulator output to 5.0V + 0.0V = 5.0V.Now, the output of a '7805 regulator with its voltage-reference pin floating will theoretically be equal to the input voltage of the regulator, since there will never be a +5.0V difference between the output and the reference pins that will force the regulator's output stage to cut off or to shut down. But, since the output power stage of the '7805 is a plain darlington transistor in series to a current monitoring resistor, the output voltage will always be lower to the input voltage by at least 1.5V of the Darlington stage plus the voltage drop on the current monitor resistor of a value of a few hundred milliohms.To understand the 78xx family better, try to find theNational Semiconductordatasheets that include the regulators' schematics.-George