12v battery protector circuit

[andy]

I'm looking for a circuit that will protect a lead acid battery against
excessive discharge - i.e. cut off the supply to the main circuit when the
voltage from the battery goes below about 11V. Ideally, once the circuit
has switched off, the protection circuit shouldn't use any current itself,
either, which is the tricky bit. The best I've come up with so far is:


Vin o--------------------o------------------------------o
|
|<
+------|
| |\
R1 .-. | Vg
100k | | o---------+
| | | |
'-' .-. |
| | | R2 |
| | |100k |
- '-' |
zener, 9.5v ^ | |
+--------o |
| |
.-. |
| | R3 |
| | 3.3k |
'-' ===
| |^|
Gns o-------------------o---------+|+-----------------o

I built it without the mosfet, and it does work - the circuit switches
right off at Vin=~10V, and Vg goes high at Vin=~10.5V, which would switch
the mosfet on. But there's still the 0.5V gap between the two, and it's
not very adjustable - is there a better way of doing this?


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[andy]

On Thu, 26 Aug 2004 04:02:12 +0100, andy wrote:

I'm looking for a circuit that will protect a lead acid battery against
excessive discharge - i.e. cut off the supply to the main circuit when the
voltage from the battery goes below about 11V. Ideally, once the circuit
has switched off, the protection circuit shouldn't use any current itself,
either, which is the tricky bit. The best I've come up with so far is:


Vin o--------------------o------------------------------o
|
|
+------|
| |\
R1 .-. | Vg
100k | | o---------+
| | | |
'-' .-. |
| | | R2 |
| | |100k |
- '-' |
zener, 9.5v ^ | |
+--------o |
| |
.-. |
| | R3 |
| | 3.3k |
'-' ===
| |^|
Gns o-------------------o---------+|+-----------------o

I built it without the mosfet, and it does work - the circuit switches
right off at Vin=~10V, and Vg goes high at Vin=~10.5V, which would switch
the mosfet on. But there's still the 0.5V gap between the two, and it's
not very adjustable - is there a better way of doing this?

P.S. R1 is just to stop the base current going too high when the
transistor saturates - it doesn't have much effect on the switching (with
Vg=~10V, the voltage across R1 was about 0.4 V)

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[Gene]

What about a low-power comparator (always connected to the battery) with a
suitable hysteresis whose output drives a NPN transistor which de/energizes
a relay?
Gene

 

[Rich Grise]

andy wrote:

I'm looking for a circuit that will protect a lead acid battery against
excessive discharge - i.e. cut off the supply to the main circuit when the
voltage from the battery goes below about 11V. Ideally, once the circuit
has switched off, the protection circuit shouldn't use any current itself,
either, which is the tricky bit. The best I've come up with so far is:


Vin o--------------------o------------------------------o
|
|
+------|
| |\
R1 .-. | Vg
100k | | o---------+
| | | |
'-' .-. |
| | | R2 |
| | |100k |
- '-' |
zener, 9.5v ^ | |
+--------o |
| |
.-. |
| | R3 |
| | 3.3k |
'-' ===
| |^|
Gns o-------------------o---------+|+-----------------o

I built it without the mosfet, and it does work - the circuit switches
right off at Vin=~10V, and Vg goes high at Vin=~10.5V, which would switch
the mosfet on. But there's still the 0.5V gap between the two, and it's
not very adjustable - is there a better way of doing this?


What's wrong with 0.5V hysteresis, especially in a low voltage cutout? Or

do you mean that you need more?

Thanks,
Rich

 

[andy]

On Thu, 26 Aug 2004 05:11:18 +0000, Rich Grise wrote:

andy wrote:

I'm looking for a circuit that will protect a lead acid battery against
excessive discharge - i.e. cut off the supply to the main circuit when the
voltage from the battery goes below about 11V. Ideally, once the circuit
has switched off, the protection circuit shouldn't use any current itself,
either, which is the tricky bit. The best I've come up with so far is:


Vin o--------------------o------------------------------o
|
|
+------|
| |\
R1 .-. | Vg
100k | | o---------+
| | | |
'-' .-. |
| | | R2 |
| | |100k |
- '-' |
zener, 9.5v ^ | |
+--------o |
| |
.-. |
| | R3 |
| | 3.3k |
'-' ===
| |^|
Gns o-------------------o---------+|+-----------------o

I built it without the mosfet, and it does work - the circuit switches
right off at Vin=~10V, and Vg goes high at Vin=~10.5V, which would switch
the mosfet on. But there's still the 0.5V gap between the two, and it's
not very adjustable - is there a better way of doing this?


What's wrong with 0.5V hysteresis, especially in a low voltage cutout? Or
do you mean that you need more?

Thanks,
Rich

It's not hysteresis - the mosfet having switched on or off doesn't make
any different to the switching point of the transistor circuit. A bit of
hysteresis would be good, but that's not what's happening - it's that the
mosfet and hence the main circuit will switch off at Vin=~10.5V, but the
transistor circuit won't switch right off until Vin drops to ~10V. It
would probably do the job OK (in practice, switching off the main circuit
would probably be enough, because the battery would then get charged back
up from the solar panel on the next day's sunshine), but I'm wondering if
there's a better way, where the whole thing switches off at a single
cutoff voltage. (Ideally with a bit of hysteresis to stop it dithering or
browning out).

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[andy]

On Thu, 26 Aug 2004 03:17:40 +0000, Gene wrote:

What about a low-power comparator (always connected to the battery) with a
suitable hysteresis whose output drives a NPN transistor which de/energizes
a relay?
Gene

Could do, but I'm still looking for a way where there's no current draw at
all apart from leakage once the circuit has turned off.

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[Gene]

I dare to say it is extremely difficult, because once the circuit is turned
off, one needs some form of energy to turn it on. This "energy" has to come
from somewhere, either external to the system (external battery, manual
reset, etc.) or internal. In any case, micro-power comparators or OpAmps
draw ~50 microA, this would be the only device, together with the voltage
sensing and reference circuit, that uses power from the battery in the
off-status. Probably you could achieve confortably less than 100microA
"leakage". (BTW, is cost an issue?)
Gene

 

[John Fields]

On Thu, 26 Aug 2004 04:02:12 +0100, andy
<news4@earthsong.free-online.co.uk> wrote:

I'm looking for a circuit that will protect a lead acid battery against
excessive discharge - i.e. cut off the supply to the main circuit when the
voltage from the battery goes below about 11V. Ideally, once the circuit
has switched off, the protection circuit shouldn't use any current itself,
either, which is the tricky bit. The best I've come up with so far is:


Vin o--------------------o------------------------------o
|
|
+------|
| |\
R1 .-. | Vg
100k | | o---------+
| | | |
'-' .-. |
| | | R2 |
| | |100k |
- '-' |
zener, 9.5v ^ | |
+--------o |
| |
.-. |
| | R3 |
| | 3.3k |
'-' ===
| |^|
Gns o-------------------o---------+|+-----------------o

I built it without the mosfet, and it does work - the circuit switches
right off at Vin=~10V, and Vg goes high at Vin=~10.5V, which would switch
the mosfet on. But there's still the 0.5V gap between the two, and it's
not very adjustable - is there a better way of doing this?

---
Yes, check "Low voltage cutoff circuit" on
alt.binaries.schematics.electronic. Jim Thompson and I have both
posted circuits which should work.

On mine you'll have to change R4 to 12.1k because of your 11V cutoff
requirement and on Jim's, you'll need to change R3 to 2.21k.

--
John Fields

 

[Jonathan Kirwan]

On Thu, 26 Aug 2004 17:06:12 +0100, andy <news4@earthsong.free-online.co.uk>
wrote:

Could do, but I'm still looking for a way where there's no current draw at
all apart from leakage once the circuit has turned off.

Andy, your current circuit draws current because the BJT applies some gain to
the small leakage current through the zener (perhaps 100X or so) and this is
then allowed to flow through the collector. The zener leakage is probably
something like a microamp or so, so this suggests something like 100uA of
consumption for your circuit.

Frankly, I'm not even sure what your circuit is supposed to be doing. Those
100k resistors seem very big. For one thing, zeners are spec'd to operate at
much higher currents than you'll get with your 100k resistor. For another, your
zener will leak current to the base of the BJT and the BJT will apply gain to it
to yield collector currents that are some 100X higher.

In the case where the zener hasn't yet reached its zener voltage, it will just
leak at say, 1uA. Thus, your Vg will be:

Izener ~= 1uA (rough guess)
V(R1) = Izener * R1 ~= 0.1V (negligible)
so, Ib(BJT) ~= Izener, Ic(BJT) = beta * Ib(BJT) ~= 100uA (estimate, for now)

With an Ic of 100uA and an (R2+R3) of slightly over 100k, you'd expect to see
about 10V across them. Since we already know as a rough guess that the voltage
supply is below the zener voltage and thus below 10V, we can be pretty sure that
V(CE) is small and that the transistor is probably saturated (reducing the beta
so that the actual Ic matches up with (R2+R3) to yield about the supply voltage.

So, in effect Vg would basically just track your supply voltage before the zener
voltage kicks in, placing the supply voltage directly on your MOSFET gate.
Mostly because of that huge R2 value, though.

But what happens after the zener has kicked in? Well, Ib will be even higher of
course. And V(CE) will just be that much less (not much less, because it was
already saturated to begin with.) So Vg would be about the same then, too.

So you might as well just hook up the supply to Vg and be done with it.

This all changes some if you make R2 a smaller value. I'm not sure if you are
telling us correctly about the 100k value on R2, or not. But if R2 is a lot
smaller than 100k, then the case before breakdown and the case after breakdown
look different, which is probably what you are after. Then Vg is held down to a
low value by the much smaller R2 and R3 against the relatively smaller looking
Ic=beta*(Ib=leakage) until the zener breakdown when the Ib rises (although
again, I suspect your R1 should be something less than 100k as even these tiny
Ib currents present a real voltage drop across something that big and your zener
really needs more substantial currents after breakdown, anyway) and then more
substantial Ic values can begin to work against R2+R3 and cause Vg to rise up
towards the supply voltage.

But any reasonable way you work it, your Ic is going to be at least beta times
the zener leakage and that's going to set the total leakage of your circuit,
when off.

I also think Gene's point about having to have some kind of comparison circuit
operating, even when the MOSFET is disabled, is true enough. You'll need
something running. But I believe you can get this down to about a microamp or
two with predictable performance, with some careful design. (I'm not a good or
professional designer, but I do have an idea how this might be achieved.)

Jon

 

[chuck h]

"andy" <news4@earthsong.free-online.co.uk> wrote in message
newsan.2004.08.26.03.02.09.376509@earthsong.free-online.co.uk...

I'm looking for a circuit that will protect a lead acid battery against
excessive discharge - i.e. cut off the supply to the main circuit when the
voltage from the battery goes below about 11V. Ideally, once the circuit
has switched off, the protection circuit shouldn't use any current itself,
either, which is the tricky bit. The best I've come up with so far is:

www.homepower.com/files/lvdhp60.pdf

 

[Jonathan Kirwan]

On Thu, 26 Aug 2004 13:24:58 -0500, John Fields <jfields@austininstruments.com>
wrote:

Yes, check "Low voltage cutoff circuit" on
alt.binaries.schematics.electronic. Jim Thompson and I have both
posted circuits which should work.

Yes, at:

abqpi0dp3dcoacl0hp4ab74ij1r7dn909u@4ax.com

and,

b92qi095nboqk2ei2h7ub3fgshkkiog251@4ax.com

I just looked.

On mine you'll have to change R4 to 12.1k because of your 11V cutoff
requirement and on Jim's, you'll need to change R3 to 2.21k.

Both of these appear to require a switch to turn them back on, though, since
power to the rest of the circuit is blocked by both of these once the source to
gate resistor takes over and holds the MOSFET off. I'm guessing, but Andy is
looking for something that does this automatically when the voltage rises *AND*
where it has very low Iq when the power is disconnected from the load.

Jon

 

[andy]

On Thu, 26 Aug 2004 19:16:16 +0000, Jonathan Kirwan wrote:

On Thu, 26 Aug 2004 17:06:12 +0100, andy <news4@earthsong.free-online.co.uk
wrote:

Could do, but I'm still looking for a way where there's no current draw at
all apart from leakage once the circuit has turned off.

Andy, your current circuit draws current because the BJT applies some gain to
the small leakage current through the zener (perhaps 100X or so) and this is
then allowed to flow through the collector. The zener leakage is probably
something like a microamp or so, so this suggests something like 100uA of
consumption for your circuit.

I hadn't thought of that, I admit.

Frankly, I'm not even sure what your circuit is supposed to be doing. Those
100k resistors seem very big.

I was trying to keep the current down as much as possible - the main
circuit only uses 0.8 mA, so I'd like this part to use maybe up to 0.2 or
0.3mA when it's on, and under 10 uA when it's off.

The idea of it is -

- when Vin is below Vzener plus 0.7 volts, there's no current except for
leakage, because both devices are off (but I forgot about this being
amplified, as you said).
- when Vin goes above Vz+0.7, then current will start to flow in the
zener. This will create a collector current, which will start to increase
the voltage across R2 and R3. R3 is meant to give a negative feedback - as
the collector current increases, this will push the base voltage on the
transistor closer to Vin again, tending to switch it off.
- so Vg will rise until the transistor saturates. The maths I worked out
has this happening when Vin=(Vz+0.7)*(R2+R3)/R2
- then the voltage across R1 starts to rise, which stops the base current
going too high after the transistor has saturated.

For one thing, zeners are spec'd to operate at
much higher currents than you'll get with your 100k resistor. For another, your
zener will leak current to the base of the BJT and the BJT will apply gain to it
to yield collector currents that are some 100X higher.

In the case where the zener hasn't yet reached its zener voltage, it will just
leak at say, 1uA. Thus, your Vg will be:

Izener ~= 1uA (rough guess)
V(R1) = Izener * R1 ~= 0.1V (negligible)
so, Ib(BJT) ~= Izener, Ic(BJT) = beta * Ib(BJT) ~= 100uA (estimate, for now)

With an Ic of 100uA and an (R2+R3) of slightly over 100k, you'd expect to see
about 10V across them. Since we already know as a rough guess that the voltage
supply is below the zener voltage and thus below 10V, we can be pretty sure that
V(CE) is small and that the transistor is probably saturated (reducing the beta
so that the actual Ic matches up with (R2+R3) to yield about the supply voltage.

So, in effect Vg would basically just track your supply voltage before the zener
voltage kicks in, placing the supply voltage directly on your MOSFET gate.
Mostly because of that huge R2 value, though.

But what happens after the zener has kicked in? Well, Ib will be even higher of
course. And V(CE) will just be that much less (not much less, because it was
already saturated to begin with.) So Vg would be about the same then, too.

So you might as well just hook up the supply to Vg and be done with it.

It didn't do that, honest. I'll try building again to be sure.

This all changes some if you make R2 a smaller value. I'm not sure if you are
telling us correctly about the 100k value on R2, or not. But if R2 is a lot
smaller than 100k, then the case before breakdown and the case after breakdown
look different, which is probably what you are after. Then Vg is held down to a
low value by the much smaller R2 and R3 against the relatively smaller looking
Ic=beta*(Ib=leakage) until the zener breakdown when the Ib rises (although
again, I suspect your R1 should be something less than 100k as even these tiny
Ib currents present a real voltage drop across something that big and your zener
really needs more substantial currents after breakdown, anyway) and then more
substantial Ic values can begin to work against R2+R3 and cause Vg to rise up
towards the supply voltage.

But any reasonable way you work it, your Ic is going to be at least beta times
the zener leakage and that's going to set the total leakage of your circuit,
when off.

I also think Gene's point about having to have some kind of comparison circuit
operating, even when the MOSFET is disabled, is true enough. You'll need
something running.

The idea of that circuit was that you don't, because it's relying on the
transistor and zener shutting off when Vin goes below their combined
switch off voltage. But I hadn't thought about the transistor amplifying
the leakage, as you've said.

But I believe you can get this down to about a microamp or
two with predictable performance, with some careful design. (I'm not a good or
professional designer, but I do have an idea how this might be achieved.)

Jon

I did build it as in the diagram, and it worked as I said with those
resistor values - Vg stuck at 0V for Vin = 0 to ~10V, and swung up to near
Vin as Vin went from ~10 to ~10.5V. I probably am using the devices
outside the specs though, so maybe I was just lucky.

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[andy]

On Thu, 26 Aug 2004 20:12:06 +0000, Jonathan Kirwan wrote:

On Thu, 26 Aug 2004 13:24:58 -0500, John Fields <jfields@austininstruments.com
wrote:

Yes, check "Low voltage cutoff circuit" on
alt.binaries.schematics.electronic. Jim Thompson and I have both
posted circuits which should work.

Yes, at:

abqpi0dp3dcoacl0hp4ab74ij1r7dn909u@4ax.com

and,

b92qi095nboqk2ei2h7ub3fgshkkiog251@4ax.com

I just looked.

On mine you'll have to change R4 to 12.1k because of your 11V cutoff
requirement and on Jim's, you'll need to change R3 to 2.21k.

Both of these appear to require a switch to turn them back on, though, since
power to the rest of the circuit is blocked by both of these once the source to
gate resistor takes over and holds the MOSFET off. I'm guessing, but Andy is
looking for something that does this automatically when the voltage rises *AND*
where it has very low Iq when the power is disconnected from the load.

Jon

That is what I'm looking for. I'll have a proper look at the circuits and
the other comments when my head's a bit clearer, and see if I can come up
with something that does what I want. The TL431 looks like a
possibly-useful component apart from the 1mA minimum current - The whole
circuit only uses 0.8 mA, so this is a bit high for what I want.

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[andy]

On Thu, 26 Aug 2004 13:24:58 -0500, John Fields wrote:

On Thu, 26 Aug 2004 04:02:12 +0100, andy
news4@earthsong.free-online.co.uk> wrote:

I'm looking for a circuit that will protect a lead acid battery against
excessive discharge - i.e. cut off the supply to the main circuit when the
voltage from the battery goes below about 11V. Ideally, once the circuit
has switched off, the protection circuit shouldn't use any current itself,
either, which is the tricky bit. The best I've come up with so far is:


Vin o--------------------o------------------------------o
|
|
+------|
| |\
R1 .-. | Vg
100k | | o---------+
| | | |
'-' .-. |
| | | R2 |
| | |100k |
- '-' |
zener, 9.5v ^ | |
+--------o |
| |
.-. |
| | R3 |
| | 3.3k |
'-' ===
| |^|
Gns o-------------------o---------+|+-----------------o

I built it without the mosfet, and it does work - the circuit switches
right off at Vin=~10V, and Vg goes high at Vin=~10.5V, which would switch
the mosfet on. But there's still the 0.5V gap between the two, and it's
not very adjustable - is there a better way of doing this?

---
Yes, check "Low voltage cutoff circuit" on
alt.binaries.schematics.electronic. Jim Thompson and I have both
posted circuits which should work.

On mine you'll have to change R4 to 12.1k because of your 11V cutoff
requirement and on Jim's, you'll need to change R3 to 2.21k.

Thanks - I finally found a free news server that carries that group
(news.tehnicom.net). What I want is a circuit that doesn't need resetting,
so I'll have a proper look at both those designs when my head's a bit
clearer, and see if I can come up with something.

R3 is meant to be a pot, like you said in your reply on abse - I just
tested with a static resistor because it was easier.

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[Robert C Monsen]

"andy" <news4@earthsong.free-online.co.uk> wrote in message
newsan.2004.08.27.01.28.59.370737@earthsong.free-online.co.uk...

How about using something a bit more straightforward:

P-MOSFET Q3
-----------o---------o------------o-------+^+-----------
VIN | | | ||| VOUT
.-. | .-. ===
| |47k | | |220k |
| |R1 | | |R4 |
'-' | '-' |
| ___ |< | |
o-|___|-| PNP o-------'
| 100k |\ Q1 |
| R2 | ___ |/
| o---|___|--| NPN
10V - | 47k |> Q2
Zener ^ .-. R5 |
D1 | | | |
| | |220k |
| '-'R3 |
| | |
| | |
GND | | |
-----------o---------o------------o----------------------

D1 = 1N4740A (10V Zener)
Q1 = 2N3906 (PNP)
Q2 = 2N3904 (NPN)
Q3 = IRF9Z34S for example

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

When VCC dips below Vzener + 0.7, Q1 turns off, making Q2 turn off and
Q3 turn off!

This circuit also has a fairly sharp knee, meaning it won't drag VOUT
through an extended period of low voltage, it'll drop to zero fairly
quickly. This is affected adversely by high impedance of the voltage
source.

It also uses very little current; at VIN = 15V, it uses around 600uA.
At VIN = 10, it uses almost none.

The zener should get more current to be accurate, but it doesn't
really matter; it'll probably be within 5%. If you are worried about
accuracy, you should instead use a voltage reference and a comparator.

Regards,
Bob Monsen

 

[John Fields]

On Fri, 27 Aug 2004 02:37:01 +0100, andy
<news4@earthsong.free-online.co.uk> wrote:

On Thu, 26 Aug 2004 20:12:06 +0000, Jonathan Kirwan wrote:

On Thu, 26 Aug 2004 13:24:58 -0500, John Fields <jfields@austininstruments.com
wrote:

Yes, check "Low voltage cutoff circuit" on
alt.binaries.schematics.electronic. Jim Thompson and I have both
posted circuits which should work.

Yes, at:

abqpi0dp3dcoacl0hp4ab74ij1r7dn909u@4ax.com

and,

b92qi095nboqk2ei2h7ub3fgshkkiog251@4ax.com

I just looked.

On mine you'll have to change R4 to 12.1k because of your 11V cutoff
requirement and on Jim's, you'll need to change R3 to 2.21k.

Both of these appear to require a switch to turn them back on, though, since
power to the rest of the circuit is blocked by both of these once the source to
gate resistor takes over and holds the MOSFET off. I'm guessing, but Andy is
looking for something that does this automatically when the voltage rises *AND*
where it has very low Iq when the power is disconnected from the load.

Jon

That is what I'm looking for. I'll have a proper look at the circuits and
the other comments when my head's a bit clearer, and see if I can come up
with something that does what I want. The TL431 looks like a
possibly-useful component apart from the 1mA minimum current - The whole
circuit only uses 0.8 mA, so this is a bit high for what I want.

---
Try this:

PCH
BAT+>----+-------+-----+-----------------S D------+
| | | G |
| [845K] | | |
[360K] | | | |
| +----|-\ | |
| | | >--------------+ | |
+-------|----|+/ | | |
| | U1A 4001 A--+ | |
| | LMC6762 +--Y U2A | |
| [22.6K] | B-----+ [LOAD]
| | | | |
| | U1B +--B | |
+-------|----|-\ U2B Y-----+ |
| | | >-----+--A | |
| +----|+/ | | |
[LM385-2.5] | | +--A | |
| [226K] | | U2C Y-----+ |
| | | +--B | |
| | | | | |
| | | +--A | |
| | | | U2D Y-----+ |
| | | +--B |
| | | |
BAT------+-------+-----+----------------------------+

All of the resistors are +/- 1% except for the 360k, which is +/- 5%.

The resistor string values were selected for a VBAT cutoff of <=11V
and a turn-on of >=12V. You'll have to recalculate them if you want
something else.

Operating currents at 11V are: for the LM385, 24ľA
for the resistor string, 10ľA
for the comparator, 25ľA
for the 4001, 1ľA
for the MOSFET, 0
-----
For a total of ....................................... 60ľA

I haven't added any hysteresis because the latch makes it unnecessary,
but if the circuit chatters and you find that objectionable, connect
0.1ľF ceramic caps across the inputs of each comparator.

--
John Fields

 

[CFoley1064]

Subject: Re: 12v battery protector circuit
From: andy news4@earthsong.free-online.co.uk
Date: 8/26/2004 8:45 PM Central Daylight Time
Message-id: <pan.2004.08.27.01.45.18.484766@earthsong.free-online.co.uk

snip
What I want is a circuit that doesn't need resetting,
so I'll have a proper look at both those designs when my head's a bit
clearer, and see if I can come up with something.

Hi, Andy.
You got very lucky. There have been an amazing number of really well
thought-out answers to your original post, especially the two circuits posted
by Mr. Thompson and Mr. Fields in binaries. If you start out with an
inadequate problem description and keep adding to it as you go along, you're
kind of wasting the time of the people who respond, as well as an opportunity
to learn something. But...

If you're going for fleapower both when on and when off, and you can't deal
with a reset switch, look at the available art. There are a number of battery
power management ICs, including some by Maxim. One IC that comes to mind is
the High-Voltage, Low-Current Voltage Monitor in SOT Package made by Maxim,
specifically the MAX6460. Of course, this is a surface mount part, but it
should fill the bill here. (View in fixed font or M$ Notepad):

Fleapower Undervoltage Lockout
Vbat+ Vout
o-----o--------------------o-----------o--+^+--------o
| | |||
| | | ===
.-. | .-. |
2M | | | 47K| | |
| | | | | |
'-' | '-' |
| .----o------. | |
.-. | Vcc | | |
| |<-------------oIN- | | |
25K | | | | | |
'-' | OUTo----o----'
| .-------oREF | |
.-. | | | |
| | .-. | | |
220K | | | | | MAX6460 | |
'-' | |1M | | .-.
| '-' | | | |
| | | | | |10M
| .---o-------oIN+ | '-'
| | | | | |
| | .-. | GND | |
| | | | '-----o-----' |
Vbat- | | | |1M | |
o-----o | '-' | |
| | | | |
=== | === | |
GND | GND | |
| | |
'-----------------o----------'

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

The IC has a micropower voltage reference and a comparator with open drain
output which is optimized for this type of thing. One gotcha is that, since
it's primarily made as an overvoltage switch, the output defaults to low at
less than 4V, but by that time, your battery is cashed anyway, and 4V shouldn't
be enough to turn on the MOSFET anyway, if you choose wisely. This circuit has
current use of a shade more than 10uA when off, and when on will be dependent
primarily on the pullup resistor. You can tweak that resistor if you can live
with a slower turn-off.

Note in the circuit the extra 10M feedback resistor. That will give you about
a half volt of hysteresis, so your turn off should be about 10.5V, and once
off, your turn-on will be at about 11V. That's important because, if you turn
off the power to the load, the battery voltage will go up, which might lead to
power cycling (and might anyway -- if it's a problem, tweak that resistor
value. Speaking of which, the 25K pot will allow you to tweak shutdown point
to just where you want it.

(By the way, there aren't any small lead-acid batteries. You're being way too
conservative on the off-state current requirement of 10uA. A standard circuit
with an LM10 voltage comparator/micropower reference will draw only 270uA
typical, and should be able to run for many hundreds or even thousands of hours
on an 11V nearly discharged lead-acid battery before it goes to deep discharge.
How many hours depends on the size of the battery. If you tweak shutoff
voltage up a little, you can make that time even longer.

By the way, make sure to add a good sized capacitor to Vbat at the IC to reduce
load transients, which may cause false tripping.

If you're going with the Maxim 6460 circuit, take a good look at the data sheet
before you start. Then, if you don't like something, you can tweak it
yourself...

http://www.maxim-ic.com/quick_view2.cfm?qv_pk=3492

Maxim has these in stock. You can order samples with a credit card and get
them in next-day, if you want.

Would you by chance have any more project revisions, sir?

Chris

 

[John Fields]

On Fri, 27 Aug 2004 08:33:24 -0500, John Fields
<jfields@austininstruments.com> wrote:

PCH
BAT+>----+-------+-----+-----------------S D------+
| | | G |
| [845K] | | |
[360K] | | | |
| +----|-\ | |
| | | >--------------+ | |
+-------|----|+/ | | |
| | U1A 4001 A--+ | |
| | LMC6762 +--Y U2A | |
| [22.6K] | B-----+ [LOAD]
| | | | |
| | U1B +--B | |
+-------|----|-\ U2B Y-----+ |
| | | >-----+--A | |
| +----|+/ | | |
[LM385-2.5] | | +--A | |
| [226K] | | U2C Y-----+ |
| | | +--B | |
| | | | | |
| | | +--A | |
| | | | U2D Y-----+ |
| | | +--B |
| | | |
BAT------+-------+-----+----------------------------+

---
Oops... Miswired the NORs

Should be :

PCH
BAT+>----+-------+-----+-----------------S D------+
| | | G |
| [845K] | | |
[360K] | | | |
| +----|-\ | |
| | | >--------------+ | |
+-------|----|+/ | | |
| | U1A 4001 A--+ | |
| | LMC6762 +--Y U2A | |
| [22.6K] | B-----+ [LOAD]
| | | | |
| | U1B +--A | |
+-------|----|-\ | U2B Y-----+ |
| | | >--+--|--B | |
| +----|+/ | | | |
[LM385-2.5] | | | +--A | |
| [226K] | | | U2C Y-----+ |
| | | +--|--B | |
| | | | | | |
| | | | +--A | |
| | | | U2D Y-----+ |
| | | +-----B |
| | | |
BAT------+-------+-----+----------------------------+


--
John Fields

 

[CFoley1064]

Subject: Re: 12v battery protector circuit
From: andy news4@earthsong.free-online.co.uk
Date: 8/27/2004 12:51 PM Central Daylight Time
Message-id: <pan.2004.08.27.17.51.52.841504@earthsong.free-online.co.uk


OK, I'll be specific. The main circuit is the watering timer controller I
have been working on the last month (on and off). I'm building it for some
friends who live out in the country, and have a big polytunnel greenhouse
which needs watering once or twice a day from a tank that fills off the
roof of one of the buildings. They don't have mains power - just a
windmill / battery / inverter system which isn't always reliable, so I'm
trying to build it as a standalone solar powered system. It's meant to be
as automatic as possible, so they can mostly just leave it running and
forget about it. E.g. if they want to go away for a few weeks. I'm also
trying to build it so the design can be used by anyone else in a similar
situation, and as cheaply as possible - i.e. use cheap, standard parts and
simple construction rather than exotic components.

The main circuit triggers a 0.5s 12A pulse through an electromagnet when
the light goes from day to night or vice versa. This is all working and
tested on breadboard. The circuit uses about 0.7-0.8 mA when it's waiting
to trigger, and 12A during the pulses, which makes:

24*0.7mA=17 mAh/day for the controller
12*.5*2/3600=3.3 mAh/day for the electromagnet
= 20 mAh/day total.

I've kept the current as low as possible, because the solar panel will be
the most expensive single component - I've found someone who does a 1 W
(rated) panel for about 10 UKP, and these panels only reliably provide 70
mAh/day in uk winters, 250 mAh/day in the summer. The battery will be a
1.2 or 2.1 Ah yuasa NP battery.

The battery protection circuit is probably icing on the cake. There ought
to be enough charge in it to cover a long run of dull days, but I wanted
something like that in there to stop battery damage if something goes
wrong. Also, there are other similar projects I'm thinking about which
would need this more.

The important thing is that when the circuit is switched off, the current
drain should be well under the drain in normal use, so the solar panel can
recharge the battery. Say no more than 2 mAh/day, which would make 80 uA
average current. And during normal use, it shouldn't add to much to the
total current consumption - say 200 uA max. A reset button would be an
annoyance - there's no way to tell if it's working properly without
waiting for the circuit to trigger, or maybe pressing a button to activate
a power test led, but this would mean there would be something else to do
every day to make sure the system was working, which would spoil the point
of making it automatic.

OK. That circuit should do the job. I guess you've got a good idea of where
you're going. A few comments:

* There are few solar power problems that more panels can't help at least a
little.

* If this beast is going to be outside, you need to be careful about things.
In the summer, overheating the battery can cause it to go flat. In winter and
in conditions of fog or high humidity (you never get that on your side of the
pond, right?), the possibility of condensation is going to wreak havoc with
your best-laid plans of microamp current consumption with leakage currents
across the surface of the board and components, unless you either hermetically
seal the circuit from the atmosphere or apply some kind of sealant on the
board. You might have the best luck with a gasketed enclosure which is
shielded from direct sunlight, and is somewhere that convection cooling and any
breeze will help. Along these lines, also make sure any wires enter/leave the
enclosure through gasketed/sealed glands. Resist the temptation to use
all-temp outdoor silicone sealant here - it corrodes wires.

Good luck
Chris

 

[John Fields]

On Fri, 27 Aug 2004 18:55:36 +0100, andy
<news4@earthsong.free-online.co.uk> wrote:

That's looking promising -

---
"That's looking promising. Thank you." would be so much nicer, don't
you think?

the LMC6762 would also do nicely instead of the
393 in the main circuit I reckon, which should cut down the current a
fair bit more.

---
Maybe, maybe not. the LMC6762 has push-pull outputs, so if you need an
open-collector output, you're out of luck. But, you could use an
LMC6772. ISTR that I posted a circuit using it once but you blew it
off for some reason. Oh, well...
---

The only thing I'd like to know is whether this is a fairly
standard part that's easy to get hold of, or is it one that's likely to be
out of stock or discontinued? (Thinking about if I publish the finished
design, and someone wants to build one a year or few from now.)

---
Jesus, Andy, do some of your own leg work for a change. It's easy
enough to do, just point your browser around and find out who's got
what in stock. Since National makes the part that would be the
logical place to start, and here's what they have on it:

http://www.national.com/search/psearch.cgi?keywords=LMC6762


--
John Fields