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

On Dec 20, 7:40=A0pm, 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. =A0The problem is the
> batteries are slightly different capacities and will be at various
> stages of discharge. =A0

Problems problems problems !!!

Forget it and go watch TV !

On Dec 20, 11:40=A0am, 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. =A0The problem is the
> batteries are slightly different capacities and will be at various
> stages of discharge. =A0What I want to do is have some way of limiting the=

> current going to each battery to about 2 amps. =A0What 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. =A0Obviously 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.

On Dec 21, 8:37=A0am, gearhead <nos...@billburg.com> wrote:
> On Dec 20, 11:40=A0am, 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. =A0The problem is the
> > batteries are slightly different capacities and will be at various
> > stages of discharge. =A0What I want to do is have some way of limiting t=
he
> > current going to each battery to about 2 amps. =A0What is the easiest wa=
y
> > 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. =A0Obviously 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. =A0I
> 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. =A0Just an idea to get you started, do a
> little research. =A0I'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.

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

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

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/othe ... harger.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

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

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

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
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WIRE -496 -96 -496 -144
WIRE -160 -96 -160 -144
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WIRE -16 32 -16 -32
WIRE -16 32 -288 32
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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
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WIRE -640 256 -832 256
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WIRE -16 256 -640 256
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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

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

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

On Dec 23, 4:00=A0am, 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. =A0I 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. =A0Also the voltage seemed to be al=
l
> >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. =A0
>


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).

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
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WIRE 368 576 -16 576
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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

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

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

On Sun, 23 Dec 2007 09:11:07 -0800 (PST), gearhead
<nospam@billburg.com> wrote:


>All right guys, staring at all that spice has gone to your heads.

---
Yup, good call.


--
JF

On Dec 23, 10:32=A0am, Chris W <1qaz...@cox.net> wrote:
> John Fields wrote:
> >> I also tried to simulate the circuit posted by gearhead. =A0I 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 an=
d
> >> the simulation kept giving me errors. =A0Also the voltage seemed to be =
all
> >> over the map but the current was very flat.

SNIP
>
> > 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. =A0
>
> 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. =A0
> Let me explain why I think that. =A0

Chris, your thinking is correct.

> I have used that 0-15V power supply as a kind of
> manual charger in the past. =A0I have a DC watt meter that shows current
> and voltage. =A0I 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. =A0Then I plug the battery in. =A0Then I adjust the voltage up till th=
e
> current is at the point I want for the battery I am charging. =A0

Yes. Avoid the manual stuff by implementing what you asked for
in the original post: current limiting.

> As the
> battery charges the current goes down and I keep turning the voltage up
> to get the current back up. =A0When I finally get to a little over 14V I
> stop and just wait for the current to drop. =A0If it is a big battery,
> once the current is below about 1A I stop. =A0If it is a small battery, I
> stop when the current goes down to about .1A

The circuit in a smart charger would do exactly that, except
it would probably just reduce the voltage
to about 13.5 instead of turning completely off.

>
> 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. =A0To
> 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/EP=
7-12.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. =A0

If John's circuit continues to apply 14.7 volts, then the battery will
continue charging, and top off; no problem there.

>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.
>

The mosfet has .0047 ohms Rds, according to the data sheet.
At two amps it'll drop only about a hundredth of a volt.

I haven't looked closely at John's circuit, but it seems to me that
in the end the battery will get almost the same treatment as it would
with a mere limit current. His circuit theoretically could charge the
battery
very slightly faster than mine, because at 14.26 volts (with a .37 ohm
resistor), the resistor in my circuit would cause the current to start
to
tapering off a bit, whereas with his it would continue getting 2 amps
until it hits 14.7 --
but ONLY if the battery actually still is drawing 2 amps above 14.26
volts.
Might or might not do so, but won't make much difference anyway.
Pretty much
a wash between the two circuits.

I'm not a big spice user. I'm wondering if it models true battery
behavior.
Perhaps it just uses an ideal voltage source to represent the battery.
If that was the case, then spice would majorly misrepresent the
charging
behaviour of lead-acid batteries or any battery type for that matter.
You would be better off using a breadboard, multimeter and pencil and
paper.

Mike Robinson