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Is there any disadvantage to running a Linear Regulator under the minimum voltage stated in the datasheet? For instance, In the datasheet of the LM7812C2T, it says the minimuim voltage in order to to get 12v out, is 14.5v in. However, if i run it at 12V, the actual output is around 11V.

Is there any reason this is a bad idea?

Edit: Datasheet added as suggested ST L78 Regulator Datasheet.

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You might see 11V, but you have no guarantee that it is a regulated 11V

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Whilst I haven't seen this specific datasheet (how about providing a link?), what is almost certainly really specified is only the behavior when supplied with the stated minimum voltage.

A regulator will typically "work" with a lower input voltage, but this is not necessarily guaranteed. Since in practice, its gain will be reduced, it is somewhat unlikely that you get entirely unstable or oscillatory output. But you will likely see a reduced regulation, meaning that the output is less decoupled from the input. Also, consider that the behavior might become particularly dependent on temperature and individual part.

That said, I have seen recommendations (for hobbyist prototype equipment) to simply supply a (low dropout) 3.3V linear regulator with a 3.3V input if one so happens to not need such a hard-to-bypass regulator in the first place---admittedly with a consumer that has a very wide range of acceptable input voltages.

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  • \$\begingroup\$ On page 11, a guarantee for an ouput >=11.5 V is given under conditions that include an input >= 14.8V. Another guarantee, for degraded line regulation, is given for conditions that include an input voltage >= 14.5V. Some misprint seems to be involved in other line regulation conditions for "16 to 12 V" where the analogous conditions for other regulators seem to imply that the second figure should be the higher voltage. \$\endgroup\$ – pyramids Jul 30 '16 at 12:37
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This is not a complete answer, but a correction to some wrong information given in other answers.

When used the way you describe, the regulator chip is not likely to overheat as some answers have suggested.

Why?

The linear regulator normally operates by putting a pass device in the path between the input pin and the output pin. In the the classic designs like the 780x, the pass device is an NPN transistor, operating as a voltage follower. This means the pass device normally operates in forward active mode, not saturation (as suggested in one other answer).

It is when there is not enough input voltage to keep the pass device in active mode and it enters saturation that we enter drop-out operation, and the desired output voltage is no longer maintained. This means that in terms of the pass device, drop-out is actually the mode that consumes the least power.

The total power of any 3-terminal device can be calculated as $$P=v_1i_1 + v_2i_2 + v_3i_3$$ where the currents are all taken as positive when they flow in to the pins. For the series regulator in drop-out mode, this means $$P=v_{in}i_{in}-v_{out}i_{out}-v_{gnd}i_{gnd}$$ if we now use the more common convention that \$i_{out}\$ and \$i_{gnd}\$ are currents flowing out of the device. By Kirchoff's current rule we also know that $$i_{in}=i_{out}+i_{gnd}$$ so, if the ground terminal is actually connected to the circuit reference potential (0 V), $$P=(v_{in}-v_{out})i_{out}+v_{in}i_{gnd}$$

So what does this mean about drop-out? First, in drop-out, \$v_{in}-v_{out}\$ is as low as it can possibly be, and if the load doesn't have negative resistance, \$i_{out}\$ is therefore also lower than it would be under regulation. So the first power term is no larger than it is in regulation.

The power consumption in drop-out can therefore only increase if the ground current increases dramatically in drop-out. Does this happen? The datasheet of your device has anticipated you're wanting to know this, and provided a curve (ground current is also called quiescent current in these devices):

enter image description here

So generally, the ground current (and so the total power consumption) drops in mild drop-out scenarios.

On the other hand, there's no promise that in strong drop-out situations (say \$v_{in}\$= 3 V, for a '7805) that the ground current won't increase. If you really need to know what happens in strong drop-out I'd advise you to measure the behavior the way you're using it. Perhaps someone with time on their hands will be able to analyze the simplified schematic given in the datasheet to make an analytical argument why the ground current would or wouldn't increase in strong drop-out.

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Yes, running a linear regulator with less than specified input voltage is a bad idea.

The problem is that you have no idea what it might do. The output voltage might track the input minus some offset. However, the output could just as well shut down or go low rapidly as the input voltage is lowered. Quite likely the output impedance goes up too, so the output voltage will vary with the load current. There could also be significant device to device variation.

You really don't know what you may get, regardless of what you measure one device doing at one set of conditions.

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At the stated minimum input voltage, the output voltage should be stable up to the rated output current. At lesser input voltages, the output voltage will sag at less than the rated output current. The voltage drop from input to output exists because of forward drops through the regulator's output transistors inherent in the device design and construction.

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The only information from the datasheet is the dropout voltage is typically 2 V with no min and max value. The device you tested had a dropout voltage of 1 V, but you don't know anything about different devices under different conditions of output current and temperature. If you want a low drop regulator, go and search for one with better specifications. If you find one, compare the maximum output current, it might be much lower.

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