Hi John,

Actually if the current drain is zero, the regulator consumes only its own
standby current, it doesn't 'syphon off' the excess voltage in any way. It
does carry the same current as the load, so if you draw 1 amp, the excess 2
volts (which will appear across the reg) will waste 2v * 1 amp = 2 watts.
The load gets 5v * 1 amp = 5 watts, giving you some 71% efficiency (only).
This is independent of the load current. You get much better mileage with
one of the modern switching low voltage regs, they have almost 100%
efficiency (well yes, almost...).

Cheers, Jan Didden

----- Original Message -----
From: "John Waters" <john_fm_waters@HOTMAIL.COM>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Sunday, July 22, 2001 6:17 PM
Subject: [OT]: How to reduce power consumption for battery operated
circuits?


> Hi All,
>
> For a battery operated circuit, we want the power consumption to be as low
> as possible, but if the circuit involves a microcontroller, we will need a
> voltage regulator like the 78L05 or 7805 to keep the 5V stable. Since the
> supply voltage is at least 2V higher than the 5V, part of the current will
> be bypassed through the regulator, in other words, the regulator
"consumes"
> some power even when there is no or little current demanding by the
> microcontroller circuit itself.
>
> Is using the regulator unavoidable? What is the usual methods to reduce
> power consumption for a battery operated microcontroller circuit?
>
> Thanks in advance!
>
> John
>
>
>
>
>
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