On Mon, 12 Oct 2009 05:43:52 -0500, RogerN wrote:
"Too_Many_Tools" wrote ...
On Oct 11, 9:52 pm, "RogerN" wrote: Snip
I've also been disappointed with some of the smart chargers Snip
Seems my auto voltage regulators keep my
batteries happy for years on the car I drive daily. So, maybe if I had
a charger that automatically started charging when the power was on, I
could set a time to charge the battery for maybe an hour a day, to
simulate driving the car an hour a day.
Smart chargers are based on a chip that have the lead acid charge/
decay curves telling it how to act. I suspect it monitors the voltage
of the battery and if that info is faulty then the charger works
erractically.
I too am a believer in settng chargers up on a timer. /
How long to charge? Good question. I just do it a hour a day with 2
amp chargers and it seems to work. YMMV / TMT
That gives me an idea, if I use a 2 amp charger that starts when
connected, wired the output through relays and use a spare PLC, I can
use one charger and automatically switch it to each battery for 1 hour a
day. I can separate the long runs and run 120VAC to an on board
charger.
An alternative to 2 Amps for 1 hour per day is 80mA all the time,
although 25mA - 50mA might be enough. You can make a constant-current
source with a few transistor/resistor/diode parts powered by a 16-20V
wall-wart. Some circuit variations are shown in following refs. For
fixed font, click More Options, Show original.
http://groups.google.com/group/sci.e...b9f1d7a1bb7f7f
High side, PNP (eg 2N3906), w LED voltage ref and indicator. Choose R1
between 100 ohms and 100K ohms so that LED brightness is adequate and there's
at least 100 uA (microamps) to spare to drive the base of the transistor.
Choose R2 = Rs = sense resistor to satisfy Rs * Im + Vbe = Vled, where
Im = desired current, Vbe = typical Vbe drop ~ .7V, Vled = LED voltage ~
1.6V. Eg, for 50mA, Rs * .05 + .7 = 1.6 -- Rs = 18 ohms. Attach battery
to be charged between "Out" and ground. With supply voltage Vs and charge
voltage Vx, there will be Vs - Vx volts across R2 and Q1; eg, if Vs=20, Rs=18,
Vbe=1.6, Im=.05, and Vx=13.6, then V(R2) = 18*.05 = .9V, so Vce=20-0.9-13.6
=5.5V. So 275mW is dissipated in transistor with Vs = 20V; it would be
better to use Vs = 16V, giving Vce=1.5V and power = 75mW.
http://groups.google.com/group/sci.e...dfa74d63ad83af
Low side, NPN, w LED. Use 2N2222, 2N3904, or similar. If supply voltage Vs
is fixed and reasonably stiff, one can also use a resistive divider: Put
R2 = 100K in place of the 150 ohm resistor and R1 = 5K-9K in place of the
LED. Eg, with Vs=20VDC, R1=5.6K, and max current = Im = 60mA, sense resistor
Rs = 4.7 ohm = ((Vs*R1/(R1+R2))-Vbe)/Im. One can also put a 5K trimpot in
place of R1, to make the constant current adjustable from 0mA up to about
40mA. However, with a 10K pot, max current would be about 210mA (which is
over the 200mA absolute maximum rating of 2N2222 and 2N3904 transistors)
unless you increase R2 to say 102K.
http://groups.google.com/group/sci.e...4c255179d3484a
Low side, NPN, as drawn uses one transistor and one resistor per additional
CCS. Advantages of diode voltage-ref over resistor divider ref: Temperature
compensation, if diode and transistor junction temps are the same; and
less dependence on exact value of Vs.
--
jiw