The advantage of LDOs is primarily their "Low Drop Out" voltage. That
is, they can use input voltages almost as low as the output voltage and
still maintain regulation (the LM1117 is spec'd at 1.5V and graphed to
~1.2V). The good ole 7805  requires upwards of 2.5V differential by
VinMin spec (the graph shows ~1.75-2V) to keep good regulation.

Therefore the LDOs shine when you have multiple voltages at nearly the
same voltage as the input since they will dissipate less heat with a
1,5V differential than a 7805 with 2.5V. Many lower current devices have
dropout voltages as low as 0.2V(typ)!

HOWEVER, if plugged into a system that had the higher voltage required
for the 7805, it will have similar dissipation.

David W. Gulley
Destiny Designs



Pic Dude wrote:

> I finally decided to break out of the 7805 habit for something more
> efficient.  Power consumption is not a major concern since most
> of my gadgets are supplied by an ample source of power (AC or
> automotive DC), but space is almost always an issue, so reduced
> component count, smaller components, etc help.  And smaller
> heatsinks will make a big difference as well.  Hence the LDO kick...
>
> So I've been searching around and came up with this little device
> numbered LM1117-5.0.  Seems like an easy swap from the 7805,
> and costs about a buck each at Digikey.
>
> However, looking at the datasheet
> ( http://www.national.com/ds/LM/LM1117.pdf ),  figure 6 shows the
> power dissipation diagram, with a formula of:
>            Pd = ((Vin - Vout) x Iout) + (Vin x Ignd)
>
> But how do I calculate Ignd?  They state In = Ignd + Iout, but there are
> two unknowns in there.  So far, I haven't found any other info that will
> help me calculate this.
>
> If I look at it another way, Ignd should be Iadj + Ir1 (figure 1), but there
> is nothing that tells me how to calculate Ir1.  Iadj is very small (approx
> -55uA at "normal" temp, so that barely puts a dent in the Pd calc (it's
> almost equal to that of the 7805).  I would expect Ir1 to flow in the
> same direction as Iout, so wouldn't that make Ignd larger, thereby
> making Iin almost equal to Iin?  So what would be the point of using
> this over a 7805?
>
> I'm obviously confused -- can anyone shed some light here please?

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