c> I don't see anything in the Max. Ratings section that would cover "back
> driving" like this.  Would this typically be acceptable for this
> configuration.

Fig 20 seems to cover your situation explicitly.
I read that as meaning that with the regulator set to a nominal 3.8V
output and with Vin = 0 then applying a voltage to the output produces
a linear resistive response of about 80 k.
E&OE YMMV.

Vin = 0 MAY assume hard grounding of Vin, as happens when there is a
solid resistive load sharing the input voltage at Vin and power is
removed. It seems likely that having Vin floating is less damaging
again but again YMMV.

Classic ye older regulator problems happened when Vin was crow barred
and Vout cap discharged through regulator. Death happened. Solution
was a back diode across regulator. It seems unlikely to be needed,
given Fig 20, but adding a reverse Schottky diode across LDO may add
safety. And, if load drops to near zero power may add problems at very
high temperatures as Schottky leakage exponentially rises - BUT also
not usually a problem.

So. Common sense and data sheet seem to suggest it's OK. I don't
recall a fig 20 type diagram in other data sheets - well done ST -
assuming I'm interpreting it correctly.

> The output cap ESR stability range is given in Fig. 21.  The allowable range
> is roughly 0.05 to 2 Ohms for a 2uF output cap.  The caps that I'm going to
> use are ceramic SMD 805 (no name, Chinese).  Will the ceramic cap meet the
> 0.05 Ohms ESR requirement typically?  From what I've been able to see is
> that ceramic caps are < 0.015 ESR therefore I would be below the
> requirement.  Suggestions?

Re cap ESR - I'd agree with you that it seems marginal.
However, 0.1 ohm track etc resistance probably not too hard to dind
with suboptimal placement - shame when components get "too good" for a
circuit :-).

Cap ESr will vary with brand and more.

If this is one off or small quantity you can measure actual ESR with a
VERY simple tester.
Apply a square wave via a small resistor to CAP. At square wave
transition the voltage will step jump in magnitude and then ramp
exponentially as capacitor charges. The step jump is    Vin x ESR/(ESR
+ series_R) or rearranging

ESR = Vstep x Rseries / (Vin - Vstep)
Oscilloscope allows actual Vin at series resistor and Vstep to be measured.

Too small Rseries loads square wave muchly.
Too large Reseries reduces Vstep size muchly.


            Russell McMahon
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