Martin said:
I an running a battery powered circuit that uses a 3.3V LDO regulator.
The drop out voltage is 4.3. I would like to add a LED indicator to
the circuit to show when the battery voltage has fallen below about
4.5V (ie. LED goes on of off). Can anyone please offer a simple
circuit to achieve this?
Martin Evans
One of the cheapest methods is to use a reset chip such as this:
http://www.onsemi.com/pub_link/Collateral/NCP302-D.PDF
They make a 4.5V version if you don't want a resistive divider. These
come in open drain and push-pull versions. That would reduce the whole
effort to one lone part of about 15 cents, so forgive me if I don't draw
a schematic
A reset chip has another advantage in that you can set a time delay. In
other words you can keep it indicating "low" even if the batery just
have a wee fainting spell of short duration because some load came on.
All this feature requires is the investment into a 1-2 cent ceramic cap.
As for drop-out, as Tim wrote, 1V isn't really LDO but marketeers
sometimes call it that because it sounds cool. Be careful with real LDO
regulators. Besides being iffy in stability with low ESR caps they often
have some undocumented issues. Such as going berserk when the source
impedance (your batery) becomes too high.