Fred said:
You're welcome to post a Tech-chat schematic of your brilliant idea.
Anyone can shoot their mouth off with a vague description of "just do
this" or "just do that", and when we get down to it, most of the time
they don't know their butt from a hole in the ground. So let's see a
schematic of your brilliant simplification the rest of us incompetents
overlooked.
Implementation depends on regulator, input voltage, etc.
If the regulator is switching the current could be detected by looking
at duty cycle, frequency, or whatever. That would be teh simplest solution.
For a linear regulator: bypass the pass device with a current regulator
set to the desired end point. Note: current source should have a disable
input.
In the linear regulator, detect when the error amp output hits its
limit. It may well be that this ouput goes to 0 v or negative. Detect
with comparator, spare opamp, discrete transistor with base resistor, or
even tie to logic with a resistor.
This output going 0 or negative has to shut off both the linear
regulator and the current source. Shutting off the current source and
reducing the output set voltage of the power supply is enough: the
supply will stay out of regulation as the setpoint is lower, but
charging has stopped.
Do not make the voltage too low: the supply has to come back into
regulation when an empty battery is connected.
The reduced voltage will keep the supply out of regulation, giving a
schmitt trigger effect.
Kill the current source by pulling a node to ground.
Change the reference or feedback, depending which way you feel, to
reduce teh voltage of the main regulator.
TO do it quick and dirty, one needs:
Three diodes, two resistors and a transistor for a (poor) current source.
Tie its base to ground to shut it down: 2nd transistor. Mayeb a diode in
the ouput for protection.
3rd transistor + resistor is the required inverter that drives this from
the op-amp output. Use the collector of this transistor as the ground of
the feedback divider if you feel like doing things the dirty way
(changing the supply output voltage).
You need some capacitance on the output (probably via a resistor) to
keep everything from oscillating of no battery is connected.
I hope you enjoy your schematic drawing program, I'm looking forward to
the picture.
Thomas