John Pearson wrote:
> I have a 5v voltage regulator where the data sheet states the
> absolute maximum input voltage is 9v DC. I want to use transistor
> batteries on it and the voltage that the regulator sees on the input
> side, with the full circuit operating, with a new battery, is about
> 9.2 - 9.3 volts.
>
> What do you think. Am I looking for trouble?

What part of "absolute maximum" don't you understand?

> I don't want to slip-in a diode because I want to get the maximum
> life out of the battery without a voltage drop penalty.

OK, stop typing and trying thinking instead.  Whether there is a diode in
there dropping the input voltage to the regulator or the regulator drops the
additional voltage itself, you still have the same current drawn and
therefore the same efficiency.  At 9.3V in and 5V out, the efficiency is
only 54%, assuming the regulator ground current is minimal compared to the
load current (probably the case).  The only difference the diode will make
is raise the minimum battery voltage that will still produce regulated
output.  Let's say the regulator needs 500mV headroom.  Without a diode,
this voltage would be 5.5V.  With a 700mV diode drop, it would be 6.2V.
6.2V is still "dead" for a 9V battery.

If efficiency is really important, use a switching regulator.


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