Lead Acid Battery Charging

[Jessica Shaw]

Hi,

Battery: http://www.digikey.com/product-search/en?lang=en&site=us&KeyWords=522-1032-nd&x=12&y=27

Charger:
http://www.digikey.com/scripts/DkSearch/dksus.dll?WT.z_header=search_go&lang=en&keywords=BA500120500003NK&x=12&y=11&cur=USD

Circuit: http://img841.imageshack.us/img841/1848/batterycharger.jpg

I am trying to power up my circuit using the above mentioned battery
and the circuit. My load works fine. It requires 300mA at +5V. The
problem comes when I tried to charge up the battery with the charger.
The charger tries to charge the battery and than gives the error after
half an hour of charging.

I replace the Load with the charger. So, in the schematic the point A
is connected to positive of the charger and Ground is connected the
charger's ground.

Any suggestions!

Thanks
Jess

 

[Fred Abse]

On Thu, 19 Jul 2012 11:03:30 -0700, Jessica Shaw wrote:

Hi,

Battery:
http://www.digikey.com/product-search/en?lang=en&site=us&KeyWords=522-1032-nd&x=12&y=27

Charger:
http://www.digikey.com/scripts/DkSearch/dksus.dll?WT.z_header=search_go&lang=en&keywords=BA500120500003NK&x=12&y=11&cur=USD

Circuit: http://img841.imageshack.us/img841/1848/batterycharger.jpg

I am trying to power up my circuit using the above mentioned battery and
the circuit. My load works fine. It requires 300mA at +5V. The problem
comes when I tried to charge up the battery with the charger. The charger
tries to charge the battery and than gives the error after half an hour of
charging.

I replace the Load with the charger. So, in the schematic the point A is
connected to positive of the charger and Ground is connected the charger's
ground.

What on EARTH are you trying to do?

Why the cockamamy string of regulators?

Any suggestions!

(1) Use a single 5V regulator, eg. LM7805,, well thermally sinked. A 7805
will stand up to 35V on its input. Total dissipation will be less, since
each separate regulator draws about 8 milliamps, internally, just to stay
in business, so you save 16 milliamps. You'll have (300+8) milliamps times
(12-5) volts = a tad over 2 watts in the 7805, to a first approximation,
which you need to get out with an adequate heatsink.

(2) Connect the charger direct to the battery, not the load end. Lose the diodes.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

Hi,

My concerns are

1. Will the charger not affect the 7805 in any way, if I connect it to the output of the 7805? The diode D1 is protecting the output of the 7805 from charger. Am I right?

2. If I do not use the D2 than the output of the 7805 will be connected to +12 V out put of the battery. And the load will get 12 V instead of 5 volts.

I can only use one ( same) connector to provide power to the load and for getting the battery charged via battery charger.

Thanks
jess

 

[Jamie]

jsscshaw88@gmail.com wrote:

Hi,

My concerns are

1. Will the charger not affect the 7805 in any way, if I connect it to the output of the 7805? The diode D1 is protecting the output of the 7805 from charger. Am I right?

2. If I do not use the D2 than the output of the 7805 will be connected to +12 V out put of the battery. And the load will get 12 V instead of 5 volts.

I can only use one ( same) connector to provide power to the load and for getting the battery charged via battery charger.

Thanks
jess

How about this?


o-----------------o
/-\ GND
( ) charging unit
\+/
o
|
|
o
|
D1 V
- ____ ____ ____
| | | | | | |
o----o------o|7810|-o----o|7808|o----o-|7805|o-----o-+
| |____| |____| |____| +
o | | | |
SLA Battery - GND .-.
--- GND GND | |
| Load | |
GND '-'
o
|
|
|
o
GND

Don't forget about the minimum required for these 78xx type regulators.
it looks like you are on the minimum threshold.

Jamie

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 19 Jul 2012 11:03:30 -0700 (PDT), Jessica Shaw <jsscshaw88@gmail.com>
wrote:

Hi,

Battery: http://www.digikey.com/product-search/en?lang=en&site=us&KeyWords=522-1032-nd&x=12&y=27

Charger:
http://www.digikey.com/scripts/DkSearch/dksus.dll?WT.z_header=search_go&lang=en&keywords=BA500120500003NK&x=12&y=11&cur=USD

Circuit: http://img841.imageshack.us/img841/1848/batterycharger.jpg

I am trying to power up my circuit using the above mentioned battery
and the circuit. My load works fine. It requires 300mA at +5V. The
problem comes when I tried to charge up the battery with the charger.
The charger tries to charge the battery and than gives the error after
half an hour of charging.

This looks like a fairly smart charger. It's probably sensing that the charge
current is not dropping off as it should )because you have a parallel current
path). Often these chargers will charge with a constant current up until some
threshold, then switch to constant voltage to finish off the charge. Drawing
significant current in parallel with the battery can cause problems.

I replace the Load with the charger. So, in the schematic the point A
is connected to positive of the charger and Ground is connected the
charger's ground.

One at a time or is the charger connected to the battery at the same time as
the load is?

 

[Phil Allison]

"Jessica Shaw"

Circuit: http://img841.imageshack.us/img841/1848/batterycharger.jpg

I am trying to power up my circuit using the above mentioned battery
and the circuit. My load works fine. It requires 300mA at +5V.

** Your schem gives 4.3 volts out.

Why are there no caps for stability ?



..... Phil

 

[Jamie]

jsscshaw88@gmail.com wrote:

The thing is that the there is only only connector for the output and battery charging. The same connector goes to batter charger's output and into the load.
jess

Is this home work or a trick question? I suppose putting the charger
in series with the load may do something, that is, as long as the math
comes out.


Jamie

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 19 Jul 2012 16:52:44 -0700 (PDT), jsscshaw88@gmail.com wrote:

Remove the load completely and than connect the charger. charge the battery . Remove the battery and than connect the load. i am not conncting the charger and the battery at the same time in parallel.

Ok, you can safely eliminate your load as the problem. Have you measured the
terminal voltage on the battery? If your battery has a shorted cell or is
seriously discharged (don't do that), the charger may refuse to charge it.

If this is the case, you may need to give it a jump-start with a power supply.
Set a power supply to an open-circuit voltage of ~14V, current limited to .25A
or maybe .5A.

 

The thing is that the there is only only connector for the output and battery charging. The same connector goes to batter charger's output and into the load.
jess

 

Either the load will be atttached to the circuit (at the ouput of the 7805) OR battery charger.

 

Its just a rough schematic. There are capacitors on board.

Lets see, if the load is connected than the charger is not connected. And if the charger is connected than the load cannot be connected.

I did not understand about parallel path. Where is parallel path?

jess

 

no its not a home work. how will give a home work like this

 

Remove the load completely and than connect the charger. charge the battery . Remove the battery and than connect the load. i am not conncting the charger and the battery at the same time in parallel.

 

[John Fields]

On Thu, 19 Jul 2012 19:09:37 -0400, Jamie
<jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:

jsscshaw88@gmail.com wrote:

Hi,

My concerns are

1. Will the charger not affect the 7805 in any way, if I connect it to the output of the 7805? The diode D1 is protecting the output of the 7805 from charger. Am I right?

2. If I do not use the D2 than the output of the 7805 will be connected to +12 V out put of the battery. And the load will get 12 V instead of 5 volts.

I can only use one ( same) connector to provide power to the load and for getting the battery charged via battery charger.

Thanks
jess

How about this?


o-----------------o
/-\ GND
( ) charging unit
\+/
o
|
|
o
|
D1 V
- ____ ____ ____
| | | | | | |
o----o------o|7810|-o----o|7808|o----o-|7805|o-----o-+
| |____| |____| |____| +
o | | | |
SLA Battery - GND .-.
--- GND GND | |
| Load | |
GND '-'
o
|
|
|
o
GND

Don't forget about the minimum required for these 78xx type regulators.
it looks like you are on the minimum threshold.

---
It looks like you missed the part about that she has to use the same
physical connector for the load _and_ the charger.
--
JF

 

[Ian Field]

"Fred Abse" wrote in message
newsan.2012.07.19.19.15.28.328903@invalid.invalid...

On Thu, 19 Jul 2012 11:03:30 -0700, Jessica Shaw wrote:

Hi,

Battery:
http://www.digikey.com/product-search/en?lang=en&site=us&KeyWords=522-1032-nd&x=12&y=27

Charger:
http://www.digikey.com/scripts/DkSearch/dksus.dll?WT.z_header=search_go&lang=en&keywords=BA500120500003NK&x=12&y=11&cur=USD

Circuit: http://img841.imageshack.us/img841/1848/batterycharger.jpg

I am trying to power up my circuit using the above mentioned battery and
the circuit. My load works fine. It requires 300mA at +5V. The problem
comes when I tried to charge up the battery with the charger. The charger
tries to charge the battery and than gives the error after half an hour of
charging.

I replace the Load with the charger. So, in the schematic the point A is
connected to positive of the charger and Ground is connected the charger's
ground.

What on EARTH are you trying to do?

Why the cockamamy string of regulators?

Any suggestions!

(1) Use a single 5V regulator, eg. LM7805,, well thermally sinked. A 7805
will stand up to 35V on its input. Total dissipation will be less, since
each separate regulator draws about 8 milliamps, internally, just to stay
in business, so you save 16 milliamps. You'll have (300+8) milliamps times
(12-5) volts = a tad over 2 watts in the 7805, to a first approximation,
which you need to get out with an adequate heatsink.

(2) Connect the charger direct to the battery, not the load end. Lose the
diodes.

***I'd delete the forward conducting diode (D1), if there was any good
reason to use all 3 regulators; I'd consider the effect of any
reservoir/smoothing/decoupling caps on the output of one to the input of the
next, and decide whether each regulator needed its own individual reverse
protection diode - the regulators do of course need decoupling caps as close
to the pins as possible to prevent oscillation. Although I'd probably do as
suggested and let a single 5V reg carry the load with the help of a "tub
thumping" great heatsink.

 

[David Eather]

On Fri, 20 Jul 2012 04:03:30 +1000, Jessica Shaw <jsscshaw88@gmail.com>
wrote:

Hi,

Battery:
http://www.digikey.com/product-search/en?lang=en&site=us&KeyWords=522-1032-nd&x=12&y=27

Charger:
http://www.digikey.com/scripts/DkSearch/dksus.dll?WT.z_header=search_go&lang=en&keywords=BA500120500003NK&x=12&y=11&cur=USD

Circuit: http://img841.imageshack.us/img841/1848/batterycharger.jpg

I am trying to power up my circuit using the above mentioned battery
and the circuit. My load works fine. It requires 300mA at +5V. The
problem comes when I tried to charge up the battery with the charger.
The charger tries to charge the battery and than gives the error after
half an hour of charging.

I replace the Load with the charger. So, in the schematic the point A
is connected to positive of the charger and Ground is connected the
charger's ground.

Any suggestions!

Thanks
Jess

OK the 'fault' is easy.

D2 interferes with the chargers ability to monitor the battery.

The picture of the battery charger shows a 5.5mm power plug. The sockets
for this plug normally have a switch built in. Use that switch for
isolation rather than D2 and the problems should go away.

Also (and as PA said) the output voltage of the regulator will be low. Put
a 1n400x in series with the Gnd pin to jack up the output voltage by the
'missing' .7 volt. You can dump all the extra regulators and just use a
7805 with a small heatsink of about 15 degC/watt - use thermal paste. Also
remember the caps for stability.


--
Using Opera's revolutionary email client: http://www.opera.com/mail/

 

Hi,

The Load and the battery has following female connector on it

http://www.digikey.com/product-search/en?lang=en&site=us&KeyWords=CP-040DH

The battery has the following male connector on it

http://www.digikey.com/scripts/dksearch/dksus.dll?vendor=0&keywords=CP-2211-ND

This connector has two different connector on each end. I cut the connector on one end and solder it to the output of the 7805 ( positive terminal) and the ground of the battery.

I do not know whether this connector has an internal switch or not ...

jess

 

On Saturday, 21 July 2012 01:39:14 UTC+10, (unknown) wrote:

Hi,

The Load and the battery has following female connector on it

http://www.digikey.com/product-search/en?lang=en&site=us&KeyWords=CP-040DH

The battery has the following male connector on it

http://www.digikey.com/scripts/dksearch/dksus.dll?vendor=0&keywords=CP-2211-ND

This connector has two different connector on each end. I cut the connector on one end and solder it to the output of the 7805 ( positive terminal) and the ground of the battery.

I do not know whether this connector has an internal switch or not ...

jess

It doesn't matter - on further reflection a plugable switch won't solve the problem.

 

[Fred Abse]

On Fri, 20 Jul 2012 14:48:41 +0100, Ian Field wrote:

***I'd delete the forward conducting diode (D1)

Many (most) integrated regulators don't like being backdriven.

if there was any good
reason to use all 3 regulators; I'd consider the effect of any
reservoir/smoothing/decoupling caps on the output of one to the input of
the next, and decide whether each regulator needed its own individual
reverse protection diode - the regulators do of course need decoupling
caps as close to the pins as possible to prevent oscillation. Although I'd
probably do as suggested and let a single 5V reg carry the load with the
help of a "tub thumping" great heatsink.

Heatsinking requirements would be the same with one regulator, as with
three, just the distribution would be different. Apart from having the
thermal resistance of a single TO-220 tab, instead of distributed
over three. The same amount of power has to be dissipated.

Dissipation should be slightly less, by losing the quiescent of two
regulators.


--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[Fred Abse]

On Thu, 19 Jul 2012 12:43:43 -0700, jsscshaw88 wrote:

My concerns are

1. Will the charger not affect the 7805 in any way, if I connect it to the
output of the 7805? The diode D1 is protecting the output of the 7805 from
charger. Am I right?

2. If I do not use the D2 than the output of the 7805 will be connected to
+12 V out put of the battery. And the load will get 12 V instead of 5
volts.

I can only use one ( same) connector to provide power to the load and for
getting the battery charged via battery charger.

You still haven't answered the burning question:

Why the string of three regulators, when one will do the job?

Phil Allison's comment is correct. You won't get 5 volts out.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)