# lead acid battery charger design

Discussion in 'Electronic Basics' started by kell, Sep 5, 2005.

1. ### kellGuest

What kind of circuit will reduce its output voltage as current into the
load subsides? i.e. a battery charger with output 14.5 volts (nominal)
when the battery draws substantial current and decreasing to 13.5 volts
or so, as the current drawn by the battery drops. I'd like to build a
battery charger and maintainer to use on VRLA of about 5 to 7 AH.

2. ### mikeGuest

If you really want to do what you asked, you've got a problem...several
problems. The voltage current relationship for a charging lead-acid
battery is dependent on current conditions and history. Plot the
voltage/current curve that you want then compare that to what the
battery wants at different parts of the charge cycle. You'll probably
find an unstable condition that oscillates.

Not what you asked, but I'd use constant current charge and sense the
voltage. When it hits the determined level, switch to constant voltage
at whatever voltage you think you want. You're gonna need some kind of
digital latch to remember.

I'd probably use a PIC processor. Use the A/D to sense the voltages and
the PWM to set the currents. Fewer parts than most anything else you
could do. Trivial to customize.
mike

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3. ### kellGuest

No. It doesn't work that way.

4. ### ehsjrGuest

You can sense the current through a resistor, use a current
mirror to develop a voltage, use an op-amp comparator to
compare the voltage to a reference, and use the output of the
comparator to electronically control the constant voltage
source for a stepped voltage reduction. There are some nifty
power management/battery charger IC's available that will
avoid all the design work for you.
http://www.analog.com/en/cat/0,2878,766,00.html

Ed

5. ### mikeGuest

Whaddyamean it doesn't work that way???
If you design it that way, it works that way.

Maybe you meant something other than what you said??
If you intend stuffing a voltage source with unlimited current
into a battery, you're asking for trouble.
If you limit the current, you've got what I described.
You can make it more complicated from there if you feel the need.

Measure the dynamic imepdance of the battery, voltage vs. current
for various states of charge.
Plot the family of curves.
If you can draw a straight line that solves your problem for all
conditions, you can make a simple linear circuit to do it.
I'm betting you can't unless you choose a very limited subset
of the set of conditions.
I'm betting that even the straight line represents negative resistance.
And the battery is nonlinear and time variant. You've got an oscillator.

The way out is to define a sequence of states triggered by
the charge conditions. Simplest way to do that is with a processor.

Or you could simplify the requirement.

mike
mike

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6. ### Byron A JeffGuest

A circuit with feedback.
Actually this can be done with a comparator.
Lead Acid has three main charge phases:

1) Bulk. Current limited energy dump until the terminal voltage reaches
the 14.5V

2) Absorption: voltage limited until the current draw reaches C/100

3) Trickle: 13.8V indefinitely.
True. That's one of the reasons why voltage/current control for SLA batteries
typically has only minimal feedback.

That's bulk charge.
The 14.4V mark for 12V batteries.
As would I.
Exactly. Such a charger is on my project list.
agreed.

BAJ

7. ### Byron A JeffGuest

[/QUOTE]
It does. Here's a web page on the subject.

http://www.solarnavigator.net/battery_charging.htm

Bulk charging is current limited. That limit essentially forms a constant
current charge. You do switch at a predetermined level. You do switch to
a constant voltage absorption charge.

I believe it meets all of the criteria.

BAJ

8. ### kellGuest

I won't use unlimited current.
In the original post I asked how to implement the transition from the
absorption stage to float stage. I read your response in that context.
It didn't occur to me you'd be talking about bulk charging!
Regarding the original question, I know I saw a fairly simple circuit
somewhere that used an op-amp.

10. ### mikeGuest

I think we're in heated agreement.
Take a close look at your circuit.
Ignoring the issues when you hook a dead battery up to it. Let's assume
a subset of all possible initial conditions that don't go up in smoke.

U2 is a comparator providing digital latching function.
It's a two state state-machine controlled by S1 and the battery voltage.

During the first part of the charge cycle,
U2 does absolutely nothing. Vout gets pulled down with prejudice to
almost the battery voltage and tracks up as the battery charges and it's
voltage increases. So, you have initial charge current limited only by
the 0.2 ohm resistor and the current limit in the LM350. Go look up the
GUARANTEED specs on the current limit for this part.
We have some random current limit that's dependent on the LM350 and the
ambient temperature and the capabilities of the heat sink it's mounted
on. Once the system comes out of current limit at some battery voltage,
the
current starts to taper as a function of the vout-vbattery across the
resistor R7. When the current drops to the threshold of u2, it's output
goes low and reduces Vout. But guess what...R7 is a resistor.
Vout is known. So measuring the battery voltage is IDENTICAL in
function (mathematically equivalent) to measuring the current (voltage
across the resistor). You have three variables, knowing any two
determines the third.

So, you get a constant current (of indeterminate value) charge that
terminates when the
battery voltage reaches the threshold. At that point you have a state
transition to lower Vout
and remember that state with a digital storage element, u2 in this design.

We're in heated agreement about the process.

IF only the circuit shown
had predictable characteristics. Just put a splash cover over the
battery so when the LM350
finally gives up and melts itself shorted and the 18V causes the battery
to emit hydrogen that's ignited when R7 catches fire....

But heck, it probably works most of
the time.

Are we having fun yet???
mike

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11. ### ehsjrGuest

That circuit INCREASES voltage into the battery
as current subsides, until the set point is
reached. That does not satisfy what you posted:
"What kind of circuit will reduce its output voltage as
current into the load subsides? "

Prior to the 150 mA set point:
You have a fixed voltage in through a fixed resistance
to produce Vout to the battery.
Vsource-----R7------Vout
Vout = Vsource - Vdrop
Vdrop = I*R7

As current goes down, Vdrop goes down. For example:

Vin I Vdrop Vout
14.5 2 .4 14.1
14.5 1.5 .3 14.2
14.5 1 .2 14.3
14.5 .5 .1 14.4

So, until you reach the set point, voltage will
not decrease as current decreases.

To say nothing of the disasterous design. It is
actually worse that what I'll post below, but
I'll use the numbers posted at the site.

He says it will start charging at 2 amps with
14.5 volts at the output of the LM350. To get
2 amps of current through a .2 ohm resistor,
what does the voltage across the resistor need
to be? E = I * R; E = 2 * .2; therefore E = .4
so a battery that is sitting at 14.1 will be charged.
Not a good thing.

It gets really ugly if you analyze the circuit.

Ed

12. ### not_realGuest

Ignoring the issue of why don't you just go out and buy a charger. I got
a really nice one from Walmart for \$40. The proper way to charge a lead
acid battery is:

1) Use constant current (I believe about 1/20th of the amp-hour capacity)
until the battery reaches a certain voltage (1.4V per cell?).
2) Continue charging at constant voltage until current drops to 1% of
AH capacity
3) Continue charging at constant voltage (1.35V per cell) indefinitely to
maintain charge.

Please don't take those numbers as gospel, do your own research cause
thats from the top of my head from looking into this a few years ago.
This also ignores the issue of desulphation which involves periodically
injecting high frequency AC current pulses into the battery in order to
knock off sulphur crystals from the battery plates. You are trying to
resonate the plate so the exact frequency may be difficult to determine a
square wave might be more useful but you still probably want to sweep
pulse width.

13. ### kellGuest

Ta heck with all these pics and op amps. Real men do it with discrete
components!

pos power
in_____________c e______________________ B+
| \ / |
| \___/ |
|_/\/\/\____| \
| /
| \ R1
\c /
\ |
| |
|_______|
| |
/ /
/e \ R2
| /
| \
| | R3
| |__/\/\/\___
| | |
| | |
| neg power in |
| |
|______________________|__________B-

The pass transistor (top line) should really be a darlington... forgive
me for not wanting to spend all afternoon drawing ASCII. The R1 and R2
voltage divider defines float voltage. R3 is a power resistor on the
order of 0.1 ohms for a small VRLA, or a much smaller value for a car
battery. Current into the battery causes a voltage across R3 that has
the effect of raising the voltage setpoint controlling the the shunt
resistor.

14. ### kellGuest

Current into the battery causes a voltage across R3 that has
Excuse me, I meant "shunt TRANSISTOR"