Multi Battery Charger for Lead Acid

[Chris W]

I have a 0 - 15V, 30 amp power supply that I want to use to charge
several small SLA batteries at the same time. The problem is the
batteries are slightly different capacities and will be at various
stages of discharge. What I want to do is have some way of limiting the
current going to each battery to about 2 amps. What is the easiest way
to do this?

Note, I want to get a minimum of 14V to the batteries, so whatever
circuit I use needs to drop no more than 1 volt, preferably less.

I have searched Mouser and Digikey for current regulators but haven't
found anything that looks like it will do the job. Obviously I don't
know exactly what to look for.




--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more at http://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.
http://hrrdb.com

 

[John Fields]

On Thu, 20 Dec 2007 13:40:07 -0600, Chris W <1qazse4@cox.net> wrote:

I have a 0 - 15V, 30 amp power supply that I want to use to charge
several small SLA batteries at the same time. The problem is the
batteries are slightly different capacities and will be at various
stages of discharge. What I want to do is have some way of limiting the
current going to each battery to about 2 amps. What is the easiest way
to do this?

---
Assuming, worst case, that you've allowed any of the batteries to be
completely depleted, then you're faced with: (View in Courier)


E 15V
R = --- = ----- = 7.5 ohms
I 2A

and the circuit will look like this: (View in Courier)

+15V-+-[7R5]-[+BT1]-+
| |
+-[7R5]-[+BT2]-+
. .
+-[7R5]-[+BTn]-+
|
GND>----------------+


and the peak power dissipated by the resistors, for each completely
dissipated battery, will be:

E˛ 15V˛
P = --- = ------ = 30 watts
R 7R5
---

Note, I want to get a minimum of 14V to the batteries, so whatever
circuit I use needs to drop no more than 1 volt, preferably less.

---
With all due respect, since this is seb, you don't know what you're
talking about. If you're interested in finding out why, post back
and ask for an explanation.


--
JF

 

[Chris W]

John Fields wrote:

Assuming, worst case, that you've allowed any of the batteries to be
completely depleted, then you're faced with: (View in Courier)

If the zero load voltage of a lead acid battery is 0V then you are
wasting your time trying to recharge it. A much more realistic value
would be discharged voltage of 10V.

E 15V
R = --- = ----- = 7.5 ohms
I 2A

E 5V
R = --- = ----- = 2.5 ohms
I 2A

If the battery is at 12V and we use the 2.5 ohm resistor...

E 3V
I = --- = ------ = 1.2A
R 2.5ohms


by the time the battery reaches 13V



E 2V
I = --- = ------ = .8A
R 2.5ohms

At this rate it is going to take a very long time to recharge the batteries.

and the circuit will look like this: (View in Courier)

+15V-+-[7R5]-[+BT1]-+
| |
+-[7R5]-[+BT2]-+
. .
+-[7R5]-[+BTn]-+
|
GND>----------------+


and the peak power dissipated by the resistors, for each completely
dissipated battery, will be:

E˛ 15V˛
P = --- = ------ = 30 watts
R 7R5
---

Note, I want to get a minimum of 14V to the batteries, so whatever
circuit I use needs to drop no more than 1 volt, preferably less.

---
With all due respect, since this is seb, you don't know what you're
talking about. If you're interested in finding out why, post back
and ask for an explanation.

Please do explain? I guess I should have said this in my first post, I
was hoping for some kind of current regulator similar to what is used
to drive LEDs at a constant current. That way the battery would charge
at a constant (or nearly constant) 2 amps till the battery voltage
reached in the area 14V to 14.7V, at which point the current would drop
as the charge topped off.



--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more at http://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.
http://hrrdb.com

 

[Marra]

On Dec 20, 7:40 pm, Chris W <1qaz...@cox.net> wrote:

I have a 0 - 15V, 30 amp power supply that I want to use to charge
several small SLA batteries at the same time.  The problem is the
batteries are slightly different capacities and will be at various
stages of discharge.  

Problems problems problems !!!

Forget it and go watch TV !

 

[ehsjr]

Chris W wrote:

John Fields wrote:

Assuming, worst case, that you've allowed any of the batteries to be
completely depleted, then you're faced with: (View in Courier)


If the zero load voltage of a lead acid battery is 0V then you are
wasting your time trying to recharge it. A much more realistic value
would be discharged voltage of 10V.

You've missed the point. I'll address that later. Here,
I'll address your 10 volt figure: a 12 volt lead acid
battery that is discharged to 10 volts is definitely
damaged. 10.5 volts is regarded as the absolute minimum,
and a battery discharged that low is probably damaged.
It would be better to set up your circuits to discharge
to no lower than say 11 volts and then automatically
disconnect.

As far as 10 volts being a much more realistic figure:
you either did not read, or do not understand what
John said. I'll emphasise: "Assuming, *worst case* ..."

E 15V
R = --- = ----- = 7.5 ohms
I 2A


E 5V
R = --- = ----- = 2.5 ohms
I 2A

What you don't realize is that you are missing the
point John made. You *must* use the 7.5 ohm
resistor to guarantee that the current cannot go above
2 amps. Incidently, you would do very well to read
carefully whatever John has to say and any time that
you disagree with it, assume that you don't understand
it.

If the battery is at 12V and we use the 2.5 ohm resistor...

E 3V
I = --- = ------ = 1.2A
R 2.5ohms


by the time the battery reaches 13V



E 2V
I = --- = ------ = .8A
R 2.5ohms

At this rate it is going to take a very long time to recharge the
batteries.

So what? If you use the resistors John specified,
it will limit the current to no more than 2 amps,
which is what you specified. You did not specify
anything about how long it takes to charge them.

and the circuit will look like this: (View in Courier)

+15V-+-[7R5]-[+BT1]-+
| |
+-[7R5]-[+BT2]-+
. .
+-[7R5]-[+BTn]-+
|
GND>----------------+


and the peak power dissipated by the resistors, for each completely
dissipated battery, will be:

E˛ 15V˛
P = --- = ------ = 30 watts
R 7R5
---

Note, I want to get a minimum of 14V to the batteries, so whatever
circuit I use needs to drop no more than 1 volt, preferably less.


---
With all due respect, since this is seb, you don't know what you're
talking about. If you're interested in finding out why, post back
and ask for an explanation.



Please do explain? I guess I should have said this in my first post, I
was hoping for some kind of current regulator similar to what is used to
drive LEDs at a constant current. That way the battery would charge at
a constant (or nearly constant) 2 amps till the battery voltage reached
in the area 14V to 14.7V, at which point the current would drop as the
charge topped off.

If you wanted a constant current 2 amps, that is what you
should have said. You can't have what you want with the
equipment you have (the 0-15V 30 amp supply), because any
regulator you use to provide constant current has an
"overhead", meaning it will drop the available voltage
to some lower value.

Also, 14.7 is too high. I'd add a 1N5408 diode in series
with each resistor in John's design for two reasons: to
drop the voltage from the supply down to ~14.3 maximum,
and to prevent batteries from discharging into each
other if the power fails. Even at 14.3, you'll damage
the batteries if you leave them charging too long.

If you want faster charging and/or constant current,
you need a different source - the 0-15 volt supply
won't support that. And if you want to protect your
batteries, you need a proper charging circuit. You
either buy it or build it. So, if you are going to
build, you may as well forget that 0-15V supply
as the heart of it.

Ed

 

[gearhead]

On Dec 20, 11:40 am, Chris W <1qaz...@cox.net> wrote:

I have a 0 - 15V, 30 amp power supply that I want to use to charge
several small SLA batteries at the same time.  The problem is the
batteries are slightly different capacities and will be at various
stages of discharge.  What I want to do is have some way of limiting the
current going to each battery to about 2 amps.  What is the easiest way
to do this?

Note, I want to get a minimum of 14V to the batteries, so whatever
circuit I use needs to drop no more than 1 volt, preferably less.

I have searched Mouser and Digikey for current regulators but haven't
found anything that looks like it will do the job.  Obviously I don't
know exactly what to look for.

--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more athttp://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.http://hrrdb.com

I'm not sure why everybody is being so hard on you.
You want a constant current source that has compliance down to .5
volt, and connect one such circuit to each battery.
I have no experience with the constant-current circuits using FET's,
but perhaps they would have a compliance range that goes as low. I
know Win Hill mentioned recently in another thread that the LND150
depletion-mode fet, when connected as a constant current source
(requires only one resistor), has a compliance down to .5 volts.
Dunno if it would carry 2 amps. Just an idea to get you started, do a
little research. I'll get back with a link to Win's post.

 

[gearhead]

On Dec 21, 8:37 am, gearhead <nos...@billburg.com> wrote:

On Dec 20, 11:40 am, Chris W <1qaz...@cox.net> wrote:





I have a 0 - 15V, 30 amp power supply that I want to use to charge
several small SLA batteries at the same time.  The problem is the
batteries are slightly different capacities and will be at various
stages of discharge.  What I want to do is have some way of limiting the
current going to each battery to about 2 amps.  What is the easiest way
to do this?

Note, I want to get a minimum of 14V to the batteries, so whatever
circuit I use needs to drop no more than 1 volt, preferably less.

I have searched Mouser and Digikey for current regulators but haven't
found anything that looks like it will do the job.  Obviously I don't
know exactly what to look for.

--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more athttp://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.http://hrrdb.com

I'm not sure why everybody is being so hard on you.
You want a constant current source that has compliance down to .5
volt, and connect one such circuit to each battery.
I have no experience with the constant-current circuits using FET's,
but perhaps they would have a compliance range that goes as low.  I
know Win Hill mentioned recently in another thread that the LND150
depletion-mode fet, when connected as a constant current source
(requires only one resistor), has a compliance down to .5 volts.
Dunno if it would carry 2 amps.  Just an idea to get you started, do a
little research.  I'll get back with a link to Win's post.- Hide quoted text -

- Show quoted text -

Well, the jfet idea won't work. That's typically a few milliamps.

If you can deal with losing close to a volt, try this:

15v
|
,---------+
| |
1K batt
| |
| |
| _|
+-------|_ NTD110N02R
| |
| |
'------\c |
npn |-+
/e |
| |
| 0.3
| |
| |
'---+
|
gnd

The mosfet has on-resistance of 4 milliohms or something, so at 2 amps
the mosfet does not add significantly to the "dropout." You have one
base-emitter drop subtracted from your 15 volts, leaving you with 14.3
or 14.4 volts... not bad, actually.

"ground" in the circuit diagram is obviously not battery negative,
it's the negative terminal of the power supply, which is not common to
the battery. You can get common ground if you want it, though, by
flipping the whole circuit upside down and using p-channel mosfet and
pnp transistor.

Fiddle with the sense resistor to get the current you want. Resistor
power is I squared R and about doubled for safety margin, you want a
two or three watt resistor.

The circuit will show some temperature dependence, not enough to
notice unless the circuit is exposed to extreme winter/summer outdoor
temperatures.

 

[Chris W]

ehsjr wrote:

You've missed the point. I'll address that later. Here,
I'll address your 10 volt figure: a 12 volt lead acid
battery that is discharged to 10 volts is definitely
damaged. 10.5 volts is regarded as the absolute minimum,
and a battery discharged that low is probably damaged.

What part of "more realistic" don't you understand? I never said that
10V was the lowest voltage you would see in lead acid battery that was
still worth recharging, just that is was real more realistic value than 0V.

It would be better to set up your circuits to discharge
to no lower than say 11 volts and then automatically
disconnect.

When did discharging become part of this discussion.

As far as 10 volts being a much more realistic figure:
you either did not read, or do not understand what
John said. I'll emphasise: "Assuming, *worst case* ..."

Well let's think about that for a minute. Why might I have asked to
limit the current to 2A. A reasonable person could conclude that it was
to prevent damaging the battery by charging it too fast. Since .3C is
the highest rate most SLA batteries are rated to charge at, you can also
reasonably conclude that the batteries in question are at least 7Ah. So
"worst case" is the battery is dead to never returned to a charged
state. Why would I care if such a battery were to see more than 2A from
my charger? I wouldn't, it's dead and no matter what I do to it, it
will still be dead. Remember the underlying goal here is to charge
batteries.

If I were using the simple resistor to limit the current, I would turn
the supply voltage down to around 14.5V. Using your number of 10.5 for
fully discharged but still usable, we get the following.


E 4V
R = --- = ----- = 2 ohms
I 2A


If the battery is defective and has an internal short we would have this...

E 14.5V
I = --- = ------- = 7.25A
R 2ohms

A 7Ah battery can easily absorb that much current, but it would require
a ridiculous 7.25A * 14.5V = 105.125W resistor.

A 2A fuse would be a much cheaper way to go.

What you don't realize is that you are missing the
point John made. You *must* use the 7.5 ohm
resistor to guarantee that the current cannot go above
2 amps. Incidently, you would do very well to read
carefully whatever John has to say and any time that
you disagree with it, assume that you don't understand
it.

I am well aware that John is very knowledgeable when it comes to
electronics, obviously far more than I. John has also been very helpful
in that past, which is why I was a little surprised he didn't post some
kind of constant current circuit that could be used instead.

If you wanted a constant current 2 amps, that is what you
should have said.

On this point I will agree, instead of saying "limiting the current" in
my OP I should have said, "regulate the current". However when I added,

"I want to get a minimum of 14V to the batteries, so whatever circuit I
use needs to drop no more than 1 volt, preferably less."

I thought that would have be a good clue that I was looking for some
kind of current regulator. If not when I said, "I have searched Mouser
and Digikey for current regulators..." It should have been even more
clear what I was after. I will admit that I could have made it much
more clear in the original question better.

You can't have what you want with the
equipment you have (the 0-15V 30 amp supply), because any
regulator you use to provide constant current has an
"overhead", meaning it will drop the available voltage
to some lower value.

I am aware of that, as is obvious if you read my whole post. But as I
mentioned I have 15V and said I was looking to see at least 14V at the
battery. That may be an unrealistic goal, but I don't think so. I
could live with 13.5V if I had to. I know there are linear voltage
regulators that drop less than 1V.

Also, 14.7 is too high. I'd add a 1N5408 diode in series
with each resistor in John's design for two reasons: to
drop the voltage from the supply down to ~14.3 maximum,
and to prevent batteries from discharging into each
other if the power fails. Even at 14.3, you'll damage
the batteries if you leave them charging too long.

I don't remember ever asking for a set it and forget it charger. I
personally don't believe in charging batteries unmonitored. Especially
when the goal is to charge them quickly with out damaging them. Which I
probably should have also stated in my OP. Since I will not be leaving
them on the charger once they are charged, 14.7V is not a problem. Some
chargers go higher than that. However 14V is plenty.

I have two chargers right now, they both limit the current to whatever
value you set till they reach their voltage set point then drop the
current to maintain that voltage till the current drops below a certain
value, at which point they stop charging. On one charger the voltage
set point is 14V and the other it is 14.7. I charged a 50Ah battery on
the charger with the 14V set point till the current dropped below 1A. A
few hours after it had finished, I hooked it up to the other charger and
in less than 10 min it had dropped it's current to less than 1A, so
clearly there isn't much point in going over 14V.

If I were going to leave the batteries connected to a charger for an
extend "float" charge I would turn the voltage to no more than 13.5. To
be on the safe side 13.2 would be better. However I don't see the point
in leaving a lead acid battery continuously hooked to a float charger.
They have such a slow self discharge rate, that charging them every 2
months is plenty to keep the ready for use.



--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more at http://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.
http://hrrdb.com

 

[John Fields]

On Thu, 20 Dec 2007 20:55:50 -0600, Chris W <1qazse4@cox.net> wrote:

John Fields wrote:

Assuming, worst case, that you've allowed any of the batteries to be
completely depleted, then you're faced with: (View in Courier)


If the zero load voltage of a lead acid battery is 0V then you are
wasting your time trying to recharge it. A much more realistic value
would be discharged voltage of 10V.

---
Worst case, if you've allowed a battery to become totally depleted,
will be zero volts and, if you want to try to recharge it and limit
the initial current to 2 amps, you'll need a 7.5 ohm resistor.
---

E 15V
R = --- = ----- = 7.5 ohms
I 2A


E 5V
R = --- = ----- = 2.5 ohms
I 2A

If the battery is at 12V and we use the 2.5 ohm resistor...

E 3V
I = --- = ------ = 1.2A
R 2.5ohms


by the time the battery reaches 13V



E 2V
I = --- = ------ = .8A
R 2.5ohms

At this rate it is going to take a very long time to recharge the batteries.

---
As Ed noted, you didn't specify a particular recharge time.
---

and the circuit will look like this: (View in Courier)

+15V-+-[7R5]-[+BT1]-+
| |
+-[7R5]-[+BT2]-+
. .
+-[7R5]-[+BTn]-+
|
GND>----------------+


and the peak power dissipated by the resistors, for each completely
dissipated battery, will be:

E˛ 15V˛
P = --- = ------ = 30 watts
R 7R5
---

Note, I want to get a minimum of 14V to the batteries, so whatever
circuit I use needs to drop no more than 1 volt, preferably less.

---
With all due respect, since this is seb, you don't know what you're
talking about. If you're interested in finding out why, post back
and ask for an explanation.



Please do explain? I guess I should have said this in my first post, I
was hoping for some kind of current regulator similar to what is used
to drive LEDs at a constant current.

---
Yes, that would have helped.
---

That way the battery would charge
at a constant (or nearly constant) 2 amps till the battery voltage
reached in the area 14V to 14.7V, at which point the current would drop
as the charge topped off.

---
Assuming a 15V supply, a battery discharged to 10V, and a desired
charging current of 2 amperes would require, as you stated, a series
resistance of 2.5 ohms between the supply and the battery. At the
high end, with a 15V supply, the battery charged to 14.7V and a
charging current of 2 amperes would require a resistance of 0.15
ohms.

That means that as the battery voltage increased, the resistance
would have to automatically decrease in order to keep a constant 2
ampere current into the battery.

Also, at the high end, things start to get a little tricky with only
a differential of 0.3V between the battery and the supply.

But... this ought to work: (View in Courier)


..+15V>-+------------+--------------+---+------------------------+
.. | | | | |
.. | | E | |
.. [4K7] | 2N3906 B-|--[10K]--------------+ |
.. | | C | | |
.. | | | | | |
.. | +--+ LT1782 | | | |
.. | | | / | | | |
.. | +-|E\ | S | O |
.. +-[34K]-+---|- >-+--[10K]---+-G IRF7204 | O |
.. | | +-|+/ | | D | |R
.. | | | | +-[ZENER>]-+ | | |E
.. | | | | 1N4744A | | |S
.. | | | | | LT1716 | |E
.. | +-|--|------------------|----------|-\ | |T
.. | | | | | 1% | >---A | |
.. | | | | +-[150K]-+-|+/ 4001 Y-+ |
.. | | | | |+ | +-B | |
.. |K | | | [BAT] [2100] | | |
.. [LM4040-10] | | | | |1% | A-+ |
.. | | +--|------------------+ | +-Y 4001 |
.. | | | | | B----+
.. | [698] | [0.1R] | |
.. | |1% | | | [10K]
.. | | | | | |
..GND>--+-------+----+------------------+--------+---------------+

According to this, anyway:

Version 4
SHEET 1 880 936
WIRE -304 208 -768 208
WIRE -32 208 -304 208
WIRE 32 208 -32 208
WIRE 176 208 32 208
WIRE 448 208 176 208
WIRE 32 240 32 208
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WIRE 688 288 304 288
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WIRE -256 464 -768 464
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WIRE 688 560 688 288
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WIRE -464 624 -512 624
WIRE 544 640 688 592
WIRE 688 656 544 608
WIRE 544 672 544 640
WIRE 576 672 544 672
WIRE -768 688 -768 464
WIRE -640 688 -640 624
WIRE -464 688 -464 624
WIRE -384 704 -384 400
WIRE 176 704 176 656
WIRE 176 704 -384 704
WIRE 688 720 688 656
WIRE 688 720 640 720
WIRE 176 736 176 704
WIRE 352 736 352 528
WIRE 576 736 544 736
WIRE 544 752 544 736
WIRE -768 848 -768 768
WIRE -640 848 -640 768
WIRE -640 848 -768 848
WIRE -464 848 -464 768
WIRE -464 848 -640 848
WIRE -304 848 -304 448
WIRE -304 848 -464 848
WIRE 176 848 176 816
WIRE 176 848 -304 848
WIRE 352 848 352 816
WIRE 352 848 176 848
WIRE 448 848 448 544
WIRE 448 848 352 848
WIRE 544 848 544 832
WIRE 544 848 448 848
WIRE -768 896 -768 848
FLAG -768 896 0
SYMBOL pmos 128 448 M180
SYMATTR InstName M1
SYMATTR Value IRF7204
SYMBOL voltage 176 560 R0
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SYMATTR Value 698
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SYMATTR SpiceLine trise 10e-9 tfall 10e-9 vhigh 15V
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SYMBOL res 320 512 R90
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SYMATTR Value PULSE(0 15 1e-3 1e-6 1e-6 1e-3)
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SYMBOL zener -16 320 R180
WINDOW 0 45 30 Left 0
WINDOW 3 24 0 Left 0
SYMATTR InstName D1
SYMATTR Value BZX84C15L
TEXT -754 872 Left 0 !.tran 2
TEXT -760 784 Left 0 ;SUPPLY
TEXT -632 784 Left 0 ;REFERENCE
TEXT 568 832 Left 0 ;RESET
TEXT 56 672 Left 0 ;BATTERY


Note that an IRF7204 won't work in the real world because of its
power dissipation limits.

Be aware that with a battery discharged to 10V the MOSFET will have
5 volts from drain to source and 2 amps through it, so it'll be
dissipating 10 watts.

An IRF4905 would be a good choice:

http://www.irf.com/product-info/datasheets/data/irf4905.pdf

but I don't have it in my library yet, so there ya go...


--
JF

 

[ehsjr]

Chris W wrote:

ehsjr wrote:

You've missed the point. I'll address that later. Here,
I'll address your 10 volt figure: a 12 volt lead acid
battery that is discharged to 10 volts is definitely
damaged. 10.5 volts is regarded as the absolute minimum,
and a battery discharged that low is probably damaged.


What part of "more realistic" don't you understand?

Look, check your attitude at the door. You responded
to this:

Assuming, worst case, that you've allowed any of the batteries to be
completely depleted, then you're faced with: (View in Courier)

If the zero load voltage of a lead acid battery is 0V then you are
wasting your time trying to recharge it. A much more realistic value
would be discharged voltage of 10V.

*** end quote***

There is only one realistic value for *worst case* discharged
voltage and that is a 0 volts and a short circuit. 0 ohms.
You *must* consider that in the design.

Until you understand that, you cannot design a safe charger.
And unless you abandon your argumentative attitude, you can't
be helped.

Ed

 

[Ross Herbert]

On Thu, 20 Dec 2007 13:40:07 -0600, Chris W <1qazse4@cox.net> wrote:

I have a 0 - 15V, 30 amp power supply that I want to use to charge
several small SLA batteries at the same time. The problem is the
batteries are slightly different capacities and will be at various
stages of discharge. What I want to do is have some way of limiting
the
current going to each battery to about 2 amps. What is the easiest
way
to do this?

Note, I want to get a minimum of 14V to the batteries, so whatever
circuit I use needs to drop no more than 1 volt, preferably less.

I have searched Mouser and Digikey for current regulators but haven't
found anything that looks like it will do the job. Obviously I don't
know exactly what to look for.

Why not use a simple current limited charger employing a common linear
regulator or an ST L200 device - one for each battery you wish to
charge. Several battery charger circuits have been published on the
web;

eg. http://www.vt52.com/diy/myprojects/other/charger/charger.htm

Digikey L200CV 497-1382-5-ND Adj. Voltage/Current Regulator
5pin-Pentawatt 1.92(per 1) 1.60(per 10) 1.35(per 100)

Data sheet http://www.st.com/stonline/books/pdf/docs/1318.pdf

 

[John Fields]

On Fri, 21 Dec 2007 15:10:53 -0600, John Fields
<jfields@austininstruments.com> wrote:

Snip...

Hey, Chris, did you read my last post in this thread?


--
JF

 

[Chris W]

John Fields wrote:

On Fri, 21 Dec 2007 15:10:53 -0600, John Fields
jfields@austininstruments.com> wrote:

Snip...

Hey, Chris, did you read my last post in this thread?

I read it and have been trying to figure out what v2 and v4 are for.
I'm also not entirely sure what A1 and A2 are, they look like some kind
of or gate with a standard and inverse output.

I also tried to simulate the circuit posted by gearhead. I couldn't
find a NTD110N02R in LTspice, NTLMS4504N was the closest I could find.
With the .3 ohm resister the current was 2.7A so I tried changing it and
the simulation kept giving me errors. Also the voltage seemed to be all
over the map but the current was very flat.

I also couldn't make much sense of your circuit using the simulator.

--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more at http://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.
http://hrrdb.com

 

[Chris W]

John Fields wrote:

On Fri, 21 Dec 2007 15:10:53 -0600, John Fields
jfields@austininstruments.com> wrote:

Snip...

Hey, Chris, did you read my last post in this thread?

I was also wondering if something as simple as the modification you
posted to the circuit from the Ramsey Electronics BL1 kit I mentioned in
my post back on Nov-28, would work.

For reference here it is..

Version 4
SHEET 1 880 708
WIRE -928 -144 -1088 -144
WIRE -640 -144 -928 -144
WIRE -496 -144 -640 -144
WIRE -160 -144 -496 -144
WIRE -16 -144 -160 -144
WIRE -640 -96 -640 -144
WIRE -496 -96 -496 -144
WIRE -160 -96 -160 -144
WIRE -16 -96 -16 -144
WIRE -640 32 -640 -32
WIRE -368 32 -640 32
WIRE -16 32 -16 -32
WIRE -16 32 -288 32
WIRE -928 80 -928 -144
WIRE -640 80 -640 32
WIRE -16 80 -16 32
WIRE -496 128 -496 -16
WIRE -496 128 -576 128
WIRE -464 128 -496 128
WIRE -368 128 -288 32
WIRE -368 128 -400 128
WIRE -288 128 -368 32
WIRE -256 128 -288 128
WIRE -160 128 -160 -16
WIRE -160 128 -192 128
WIRE -80 128 -160 128
WIRE -640 256 -640 176
WIRE -640 256 -832 256
WIRE -16 256 -16 176
WIRE -16 256 -640 256
WIRE -928 304 -928 160
WIRE -896 304 -928 304
WIRE -800 352 -832 352
WIRE -1088 400 -1088 -144
WIRE -928 400 -928 304
WIRE -896 400 -928 400
WIRE -800 448 -800 352
WIRE -800 448 -832 448
WIRE -896 496 -928 496
WIRE -800 496 -800 448
WIRE -1088 624 -1088 480
WIRE -928 624 -928 496
WIRE -928 624 -1088 624
WIRE -800 624 -800 576
WIRE -800 624 -928 624
WIRE -1088 688 -1088 624
FLAG -1088 688 0
SYMBOL npn -80 80 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL npn -576 80 M0
SYMATTR InstName Q2
SYMATTR Value 2N3904
SYMBOL res -512 -112 R0
SYMATTR InstName R3
SYMATTR Value 100k
SYMBOL res -176 -112 R0
SYMATTR InstName R4
SYMATTR Value 100k
SYMBOL cap -400 112 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 10e-6
SYMBOL cap -192 112 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C2
SYMATTR Value 10e-6
SYMBOL npn -896 256 R0
SYMATTR InstName Q3
SYMATTR Value 2N3904
SYMBOL npn -832 400 M0
SYMATTR InstName Q4
SYMATTR Value 2N3904
SYMBOL res -944 64 R0
WINDOW 0 -42 40 Left 0
WINDOW 3 -42 70 Left 0
SYMATTR InstName R5
SYMATTR Value 10k
SYMBOL res -816 480 R0
SYMATTR InstName R6
SYMATTR Value 30
SYMBOL voltage -1088 384 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value 6
SYMATTR InstName V2
SYMBOL LED -656 -96 R0
WINDOW 0 -29 -1 Left 0
WINDOW 3 -135 70 Left 0
SYMATTR InstName D1
SYMATTR Value NSPW500BS
SYMBOL LED -32 -96 R0
SYMATTR InstName D2
SYMATTR Value NSPW500BS
TEXT -1072 656 Left 0 !.tran 0 3 2 uic



--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more at http://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.
http://hrrdb.com

 

[John Fields]

On Sat, 22 Dec 2007 23:23:59 -0600, Chris W <1qazse4@cox.net> wrote:

John Fields wrote:
On Fri, 21 Dec 2007 15:10:53 -0600, John Fields
jfields@austininstruments.com> wrote:

Snip...

Hey, Chris, did you read my last post in this thread?

I read it and have been trying to figure out what v2 and v4 are for.

---
V2 is an LM4040-10, a voltage reference used to set the charging
current to 2 amps and to disconnect the battery when its voltage
rises to 14.7V.

V4 is used to reset the A1 A2 RS flip-flop.

In the real world it's the reset switch which is used to start the
charger after it's disconnected the battery once it's charged up to
14.7V.
---

I'm also not entirely sure what A1 and A2 are, they look like some kind
of or gate with a standard and inverse output.

---
Those are CD4001 NOR gates used to make an RS flip-flop. LTSPICE
allows their ORs to be used as either ORs or NORs depending on which
output you use.
---

I also tried to simulate the circuit posted by gearhead. I couldn't
find a NTD110N02R in LTspice, NTLMS4504N was the closest I could find.
With the .3 ohm resister the current was 2.7A so I tried changing it and
the simulation kept giving me errors. Also the voltage seemed to be all
over the map but the current was very flat.

---
I don't know what you mean by: "the voltage seemed to be all over
the map."

I tried it and it worked OK with any N MOSFET and a resistance of
0.37 ohms except that there's no end of charge and the supply will
keep on pumping nearly 2 amps into the battery no matter what its
voltage rises to.

Here:

Version 4
SHEET 1 892 680
WIRE -16 -80 -192 -80
WIRE 208 -80 -16 -80
WIRE -16 -16 -16 -80
WIRE 208 -16 208 -80
WIRE 208 96 208 64
WIRE -16 176 -16 64
WIRE 160 176 -16 176
WIRE -16 240 -16 176
WIRE 208 288 208 192
WIRE 208 288 48 288
WIRE 208 336 208 288
WIRE -192 384 -192 -80
WIRE -192 528 -192 464
WIRE -16 528 -16 336
WIRE -16 528 -192 528
WIRE 208 528 208 416
WIRE 208 528 -16 528
WIRE -192 576 -192 528
FLAG -192 576 0
SYMBOL voltage -192 368 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 15
SYMBOL voltage 208 -32 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value PULSE(0 14.28 0 1)
SYMBOL npn 48 240 M0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res -32 -32 R0
SYMATTR InstName R1
SYMATTR Value 1000
SYMBOL res 192 320 R0
SYMATTR InstName R2
SYMATTR Value .37
SYMBOL nmos 160 96 R0
SYMATTR InstName M2
SYMATTR Value FDS6612A
TEXT -170 552 Left 0 !.tran 2

I also couldn't make much sense of your circuit using the simulator.

---
Oh, well...

It's basically an accurate 2 amp constant current source with a
battery disconnect when it charges up to 14.7V with only 300
millivolts of headroom to the 15V supply.


--
JF

 

[John Fields]

On Sat, 22 Dec 2007 23:39:01 -0600, Chris W <1qazse4@cox.net> wrote:

John Fields wrote:
On Fri, 21 Dec 2007 15:10:53 -0600, John Fields
jfields@austininstruments.com> wrote:

Snip...

Hey, Chris, did you read my last post in this thread?


I was also wondering if something as simple as the modification you
posted to the circuit from the Ramsey Electronics BL1 kit I mentioned in
my post back on Nov-28, would work.

---
It would, except it would suffer from the same problems as
gearhead's circuit.


--
JF

 

[gearhead]

On Dec 23, 4:00 am, John Fields <jfie...@austininstruments.com> wrote:

On Sat, 22 Dec 2007 23:23:59 -0600, Chris W <1qaz...@cox.net> wrote:

I also tried to simulate the circuit posted by gearhead.  I couldn't
find a NTD110N02R in LTspice, NTLMS4504N was the closest I could find.
With the .3 ohm resister the current was 2.7A so I tried changing it and
the simulation kept giving me errors.  Also the voltage seemed to be all
over the map but the current was very flat.

---
I don't know what you mean by: "the voltage seemed to be all over
the map."

I tried it and it worked OK with any N MOSFET and a resistance of
0.37 ohms except that there's no end of charge and the supply will
keep on pumping nearly 2 amps into the battery no matter what its
voltage rises to.  

All right guys, staring at all that spice has gone to your heads.
Let's look at a 7 amp hour SLA. The printed information on the side
says:

Cycle use: 14.5 - 15.0 V (25 C)
(Initial current: less than 2.10 A)
Standby use: 13.6 - 13.8 V (25 C)

This means that you can supply 15 volts directly to the battery,
as long as you keep the initial current limited to about 2 amps.
Read: ***initial current***
The battery will not draw 2 amps forever! As the battery takes on
more
of a charge, it starts to draw less and less current.
I am not talking through my hat. This is very well known
lead-acid battery behavior. I also know it from experience.

So no, the supply will not "keep on pumping nearly 2 amps into
the battery no matter what..." After a point, the battery itself
cannot
draw that much current. You would have to measure the current it
draws
in milliamps, eventually -- and this is with the battery connected
directly
to a low impedance 15 volt source (such as Chris's 30 amp supply).
You can charge an SLA directly
from a 15 volt power supply without harming it at all --
you do need to limit current, but only during the initial,
"bulk" phase of the charging.
The OP's power supply is regulated to 15 volts: use
a 2 amp current-limiting circuit, no harm
can come to the battery.
Unleess he leaves it on there for a year, that's another matter;
then he needs lower voltage regulation (13.7 volts).

 

[Chris W]

John Fields wrote:

I also tried to simulate the circuit posted by gearhead. I couldn't
find a NTD110N02R in LTspice, NTLMS4504N was the closest I could find.
With the .3 ohm resister the current was 2.7A so I tried changing it and
the simulation kept giving me errors. Also the voltage seemed to be all
over the map but the current was very flat.

---
I don't know what you mean by: "the voltage seemed to be all over
the map."

Here is what I mean by that.
http://hp15c.org/ChargerGraph.gif
That is a graph showing the voltage and current into and through the
battery as it varies from 10 to 15V.

Version 4
SHEET 1 996 808
WIRE -16 0 -208 0
WIRE 368 0 -16 0
WIRE -16 48 -16 0
WIRE -208 112 -208 0
WIRE -16 176 -16 128
WIRE 224 176 -16 176
WIRE -16 192 -16 176
WIRE 224 224 224 176
WIRE 368 240 368 0
WIRE -208 272 -208 192
WIRE -64 272 -208 272
WIRE -208 336 -208 272
WIRE -80 336 -208 336
WIRE -16 336 -16 288
WIRE 224 336 224 288
WIRE 224 336 -16 336
WIRE -16 384 -16 336
WIRE -16 448 -16 384
WIRE -80 576 -80 432
WIRE -16 576 -16 528
WIRE -16 576 -80 576
WIRE 368 576 368 320
WIRE 368 576 -16 576
SYMBOL nmos -64 192 R0
WINDOW 3 60 79 Left 0
SYMATTR Value NTLMS4504N
SYMATTR InstName M1
SYMBOL npn -16 336 M0
SYMATTR InstName Q1
SYMBOL voltage 368 224 R0
WINDOW 123 0 0 Left 0
WINDOW 39 24 132 Left 0
SYMATTR InstName V1
SYMATTR Value 15
SYMBOL res -224 96 R0
SYMATTR InstName R1
SYMATTR Value 1k
SYMATTR SpiceLine tol=5 pwr=1
SYMBOL res -32 432 R0
SYMATTR InstName R2
SYMATTR Value .39
SYMATTR SpiceLine tol=1 pwr=5
SYMBOL zener 240 288 R180
WINDOW 0 24 72 Left 0
WINDOW 3 24 0 Left 0
SYMATTR InstName D1
SYMBOL voltage -16 32 R0
WINDOW 123 0 0 Left 0
WINDOW 39 24 132 Left 0
SYMATTR InstName V2
SYMATTR Value 10
SYMATTR SpiceLine Rser=.025
TEXT -136 632 Left 0 !.dc v2 10 15 .1 v1 15 15 1

I added D1 because the data sheet for NTD110N02R showed it as part of
the device.

I tried it and it worked OK with any N MOSFET and a resistance of
0.37 ohms except that there's no end of charge and the supply will
keep on pumping nearly 2 amps into the battery no matter what its
voltage rises to.

Unless it somehow increases the voltage to the battery above 15V I don't
see how it could keep pumping 2 amps into the battery. Let me explain
why I think that. I have used that 0-15V power supply as a kind of
manual charger in the past. I have a DC watt meter that shows current
and voltage. I just plug it into the power supply and adjust the
voltage so it is about a tenth of a volt over what the battery measures
at. Then I plug the battery in. Then I adjust the voltage up till the
current is at the point I want for the battery I am charging. As the
battery charges the current goes down and I keep turning the voltage up
to get the current back up. When I finally get to a little over 14V I
stop and just wait for the current to drop. If it is a big battery,
once the current is below about 1A I stop. If it is a small battery, I
stop when the current goes down to about .1A

If the current the battery pulls from the voltage regulated power supply
reduces as the battery charges why would it not do the same with the
circuit from gear head?

Also as a side note if you stop charging the battery as soon as it
reaches 14.7V at full charge current, it will not be fully charged. To
fully charge it you need to leave it at your stop voltage till the
current drops off, how much it needs to drop off depends on the size of
the battery for a car battery after it is down to around 1A it is done
for a little 7Ah you need to wait till it is more like .1A see the
charge graph in the pdf below.
http://www.bb-battery.com/productpages/EP/EP7-12.pdf

It's basically an accurate 2 amp constant current source with a
battery disconnect when it charges up to 14.7V with only 300
millivolts of headroom to the 15V supply.

Recent experience has shown that charging to 14.7V is doesn't add much
more charge than charging to 14V. I couldn't tell for sure from the
spec sheet for the NTD110N02R suggested by gearhead how much voltage
would be dropped and as you can see from the graph posted above that
wasn't much help either.


--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more at http://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.
http://hrrdb.com

 

[John Fields]

On Sun, 23 Dec 2007 12:32:04 -0600, Chris W <1qazse4@cox.net> wrote:

John Fields wrote:
I also tried to simulate the circuit posted by gearhead. I couldn't
find a NTD110N02R in LTspice, NTLMS4504N was the closest I could find.
With the .3 ohm resister the current was 2.7A so I tried changing it and
the simulation kept giving me errors. Also the voltage seemed to be all
over the map but the current was very flat.

---
I don't know what you mean by: "the voltage seemed to be all over
the map."

Here is what I mean by that.
http://hp15c.org/ChargerGraph.gif
That is a graph showing the voltage and current into and through the
battery as it varies from 10 to 15V.

---
I don't get anything like that, I just get a nice sloped line for
voltage and a flat line for current until it drops off.

<snipped circuit list>

I tried running the list, but it won't run. What is it?
---

I added D1 because the data sheet for NTD110N02R showed it as part of
the device.



I tried it and it worked OK with any N MOSFET and a resistance of
0.37 ohms except that there's no end of charge and the supply will
keep on pumping nearly 2 amps into the battery no matter what its
voltage rises to.


Unless it somehow increases the voltage to the battery above 15V I don't
see how it could keep pumping 2 amps into the battery. Let me explain
why I think that. I have used that 0-15V power supply as a kind of
manual charger in the past. I have a DC watt meter that shows current
and voltage. I just plug it into the power supply and adjust the
voltage so it is about a tenth of a volt over what the battery measures
at. Then I plug the battery in. Then I adjust the voltage up till the
current is at the point I want for the battery I am charging. As the
battery charges the current goes down and I keep turning the voltage up
to get the current back up. When I finally get to a little over 14V I
stop and just wait for the current to drop. If it is a big battery,
once the current is below about 1A I stop. If it is a small battery, I
stop when the current goes down to about .1A

If the current the battery pulls from the voltage regulated power supply
reduces as the battery charges why would it not do the same with the
circuit from gear head?

---
Because gearhead's circuit is a constant _current_, not a constant
_voltage_ source.
---

Also as a side note if you stop charging the battery as soon as it
reaches 14.7V at full charge current, it will not be fully charged. To
fully charge it you need to leave it at your stop voltage till the
current drops off, how much it needs to drop off depends on the size of
the battery for a car battery after it is down to around 1A it is done
for a little 7Ah you need to wait till it is more like .1A see the
charge graph in the pdf below.
http://www.bb-battery.com/productpages/EP/EP7-12.pdf

---
What's happening there is that the current starts to drop off
because the last part of the cycle is charging the battery with a
constant voltage source and as the battery voltage rises the voltage
difference between the battery and the charger gets smaller,
reducing the amount of current into the battery.
---

It's basically an accurate 2 amp constant current source with a
battery disconnect when it charges up to 14.7V with only 300
millivolts of headroom to the 15V supply.

Recent experience has shown that charging to 14.7V is doesn't add much
more charge than charging to 14V. I couldn't tell for sure from the
spec sheet for the NTD110N02R suggested by gearhead how much voltage
would be dropped and as you can see from the graph posted above that
wasn't much help either.

---

Then just do this:

..+15V>-+-----------------+-------------+
.. | | |
.. | | |
.. [4k7] +--+ LT1782 |
.. | | | / |
.. | 1% +-|E\ S
.. +-[34K]---+------|- >--[10K]--G IRF4905
.. | | +-|+/ D
.. | | | | |+
.. |K |1% | | [BAT]
.. [LM4040-10] [698] | | |
.. | | +--|-------------+
.. | | | |
.. | | | [0.1R]
.. | | | |
..GND>--+---------+-------+-------------+

It'll limit the current into the battery to 2 amps until the battery
starts getting close enough to 15V so that the supply won't be able
to pump that much current into the battery and it'll just taper off
after that. Don't worry about 14.7V, the spec says the charging
voltage should be 2.45 volts per cell, which is 15.24V total.


--
JF

 

[John Fields]

On Sun, 23 Dec 2007 14:09:10 -0600, John Fields
<jfields@austininstruments.com> wrote:

Then just do this:

.+15V>-+-----------------+-------------+
. | | |
. | | |
. [4k7] +--+ LT1782 |
. | | | / |
. | 1% +-|E\ S
. +-[34K]---+------|- >--[10K]--G IRF4905
. | | +-|+/ D
. | | | | |+
. |K |1% | | [BAT]
. [LM4040-10] [698] | | |
. | | +--|-------------+
. | | | |
. | | | [0.1R]
. | | | |
.GND>--+---------+-------+-------------+

---
Forget it, that won't work.

More later.


--
JF