Converting mains battery charger for 12v operation

  • Thread starter Daniel Kelly (AKA Jack)
  • Start date
On Tue, 27 Jul 2004 20:43:14 +0000 (UTC), Peter A Forbes <diesel@easynet.co.uk>
wrote:

Just a few thoughts, I have been watching the posts but haven't had time to
absorb all of the detail.

The Li-Ion battery will have its own regulating circuitry in the battery, and
most batteries take a raw DC feed from whatever is charging them. They monitor
overvoltage and undervoltage and charge current. We have done very large
series/parallel Li-Ion battery chargers at 110V 50A X 4 outputs for a subsea
application and also a more conventional multi-way version for a
series/parallelled cells, but generally the battery defines what it needs.
Click to expand...
I can assure you that IMOE that is NOT the topology of most small (aka
hand-held) appliances that employ Li-Ion cells. They employ (at most) a pack
protection module which monitors the parameters you describe but this does NOT
provide the charge control function, merely providing the safety element. In
appliances such as typical cellphones which employ a single cell (i.e. operate
at 3v6 nominal) there is frequently NO electronics whatsoever in the pack.

Re the high DC voltage on the charger, it sounds like a conventional switchmode
step-down and isolating circuit, probably to keep the volume and weight down.
It's cheaper to buy a volume switcher in from thr far east that make a
conventional transformer/rectifier capacitor DC supply over here, and it has the
advantage of being universal input, 90 - 260V at almost any frequency.

My 2c worth.

Peter
Click to expand...
 
Hi,

The measured battery voltage (no load) is 7.65v. The battery claims to be a
7.2v 3Ah LiIon pack. I guess it is a 2S Li-Ion pack.

Thanks,
Jack


"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:jtadnXe2gJmXKpvcRVn-ig@comcast.com...
Jack,
What is the nominal battery voltage? I will take a look later, but the
320VDC sounds bad. The 8.45V sounds about right for a charging voltage
for
a nominal 7.5V or so battery. Unregulated 8.45V would be on the low side
for a charger input that charges a 6V or higher battery.

The charger for my Motorola GSM phone appears to use a switching regulator
type of charger. The blob that plugs into the wall is not big or heavy
enough to contain a 50/60 Hertz transformer.

Tam
"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:ce64eq$2ldg$1@uns-a.ucl.ac.uk...
Hmmm... having looked again at the PCB, I'm not so sure!

Take a look:

http://www.ucl.ac.uk/~ucgadak/charger_components.jpg
http://www.ucl.ac.uk/~ucgadak/charger_merged.jpg
http://www.ucl.ac.uk/~ucgadak/charger_tracks.jpg

I want to put 8.45v onto C22 (it's marked on the last 2 JPGs). There
are
a
total of 3 transformers. 2 of which have 240v on both sides (i.e. their
coils are symetric). And there's definitely circuitry to produce 320v
DC
(D1 is a high voltage rectifier).

Urg. I dunno anymore. Any thoughts?

Thanks,
Jack


"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:ce647t$1qe8$1@uns-a.ucl.ac.uk...
Hiya,

I'm 99.999% sure my charger works in way "A".

All the control circuitry for the LiIon charging is on a little
daughter
board, which is definitely downstream of the 8.45v I measured across
the
smoothing cap.

Thanks,
Jack




"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:zb-dnTGSQPl56JvcRVn-ug@comcast.com...
Jack,

I want to point out that there are basically two ways the charger
can
work:

A. The AC line is stepped down to a low voltage, rectified, and fed
to
the
charger. The 8.45V is the input to the charger. This is what I am
assuming.

B. The AC is rectified to give 160 -340 VDC, which then goes to a
switching
regulator. The 8.45V is what goes to the battery. If this is the
case,
forget it.

Tam
"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:0Y-dne4Ly4wn-5vcRVn-qg@comcast.com...

"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote
in
message
news:ce5d6i$1oeu$1@uns-a.ucl.ac.uk...
Hi Tam,

Yes, I've taken the lid off the battery charger. It turns out
the
voltage
across the smoothing capacitor (downstream of the step-down
transformer
and
diode rectifier) is 8.45v.

Thanks,
Jack

This is the unregulated voltage, right? I can't come up with a
reason
why
it
should not work if you run 8 - 9 V from a 3 terminal regulator to
this
point. Your rectifier diodes will prevent the transformer from
shorting
out
the DC. I assume the actual battery voltage is 6V nominal. I guess
this
is
what you proposed originally. The thing to watch for is if the
thing
uses
positive ground. Could cause fireworks if any grounded metal on
the
camcorder touched grounded metal on the car. Of course, the
camcorder
probably has no exposed metal. At any rate I would be sure to
include
a
fuse. Measure the DC current when running off AC.

I think some of us are confused as to what the topology is. Is the
actual
charger in the camera, in the brick, or do you remove the battery
from
the
camera and connect it to the charger? Any power jack on the camera
should
be
labeled as to what the voltage range is. Either on the camera, or
in
the
instruction book. My Ricoh Hi8, for instance, uses a 6V battery.
The
camera
has a label that states 6 - 7.5VDC. You remove the battery for
charging.

Let us know how things work out

Tam
Click to expand...
 
Hi Peter,

Thanks loads for your reply.

Hmm... I don't think my battery pack does have any protection circuitry in
it. I may be wrong. I'm only guessing... but there definitely is a pretty
complex looking daugter board inside the charger - it's got 2 SMD ICs and 7
SMD FETs (at least, I assume they're FETs).

Jack

"Peter A Forbes" <diesel@easynet.co.uk> wrote in message
news:mbfdg0df9vvainio232flb5o3lkb1odk01@4ax.com...
On Tue, 27 Jul 2004 10:29:25 -0400, "Tam/WB2TT" <t-tammaru@c0mca$t.net
wrote:


"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:ce5d6i$1oeu$1@uns-a.ucl.ac.uk...
Hi Tam,

Yes, I've taken the lid off the battery charger. It turns out the
voltage
across the smoothing capacitor (downstream of the step-down transformer
and
diode rectifier) is 8.45v.

Thanks,
Jack

This is the unregulated voltage, right? I can't come up with a reason why
it
should not work if you run 8 - 9 V from a 3 terminal regulator to this
point. Your rectifier diodes will prevent the transformer from shorting
out
the DC. I assume the actual battery voltage is 6V nominal. I guess this
is
what you proposed originally. The thing to watch for is if the thing uses
positive ground. Could cause fireworks if any grounded metal on the
camcorder touched grounded metal on the car. Of course, the camcorder
probably has no exposed metal. At any rate I would be sure to include a
fuse. Measure the DC current when running off AC.

I think some of us are confused as to what the topology is. Is the actual
charger in the camera, in the brick, or do you remove the battery from
the
camera and connect it to the charger? Any power jack on the camera should
be
labeled as to what the voltage range is. Either on the camera, or in the
instruction book. My Ricoh Hi8, for instance, uses a 6V battery. The
camera
has a label that states 6 - 7.5VDC. You remove the battery for charging.

Let us know how things work out

Tam


Just a few thoughts, I have been watching the posts but haven't had time
to
absorb all of the detail.

The Li-Ion battery will have its own regulating circuitry in the battery,
and
most batteries take a raw DC feed from whatever is charging them. They
monitor
overvoltage and undervoltage and charge current. We have done very large
series/parallel Li-Ion battery chargers at 110V 50A X 4 outputs for a
subsea
application and also a more conventional multi-way version for a
series/parallelled cells, but generally the battery defines what it needs.

Re the high DC voltage on the charger, it sounds like a conventional
switchmode
step-down and isolating circuit, probably to keep the volume and weight
down.
It's cheaper to buy a volume switcher in from thr far east that make a
conventional transformer/rectifier capacitor DC supply over here, and it
has the
advantage of being universal input, 90 - 260V at almost any frequency.

My 2c worth.

Peter
--
Peter & Rita Forbes
diesel@easynet.co.uk
Engine pages for preservation info:
http://www.oldengine.org/members/diesel
Click to expand...
 
Jack,

I took a look at your pictures. Hard to tell what is there, but a lot of
stuff. In line with what some others have been saying, the 8.45V may be the
actual battery charging voltage. Can you measure the voltage coming off the
charger at the two spring loaded(?) contacts that connect to the battery to
see if it is 8.45V. I would still look at the Canon web site if they make
some kind of car adapter.

Tam
"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in message
news:ce89k6$fbe$1@uns-a.ucl.ac.uk...
Hi,

The measured battery voltage (no load) is 7.65v. The battery claims to be
a
7.2v 3Ah LiIon pack. I guess it is a 2S Li-Ion pack.

Thanks,
Jack


"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:jtadnXe2gJmXKpvcRVn-ig@comcast.com...
Jack,
What is the nominal battery voltage? I will take a look later, but the
320VDC sounds bad. The 8.45V sounds about right for a charging voltage
for
a nominal 7.5V or so battery. Unregulated 8.45V would be on the low
side
for a charger input that charges a 6V or higher battery.

The charger for my Motorola GSM phone appears to use a switching
regulator
type of charger. The blob that plugs into the wall is not big or heavy
enough to contain a 50/60 Hertz transformer.

Tam
"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:ce64eq$2ldg$1@uns-a.ucl.ac.uk...
Hmmm... having looked again at the PCB, I'm not so sure!

Take a look:

http://www.ucl.ac.uk/~ucgadak/charger_components.jpg
http://www.ucl.ac.uk/~ucgadak/charger_merged.jpg
http://www.ucl.ac.uk/~ucgadak/charger_tracks.jpg

I want to put 8.45v onto C22 (it's marked on the last 2 JPGs). There
are
a
total of 3 transformers. 2 of which have 240v on both sides (i.e.
their
coils are symetric). And there's definitely circuitry to produce 320v
DC
(D1 is a high voltage rectifier).

Urg. I dunno anymore. Any thoughts?

Thanks,
Jack


"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:ce647t$1qe8$1@uns-a.ucl.ac.uk...
Hiya,

I'm 99.999% sure my charger works in way "A".

All the control circuitry for the LiIon charging is on a little
daughter
board, which is definitely downstream of the 8.45v I measured across
the
smoothing cap.

Thanks,
Jack




"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:zb-dnTGSQPl56JvcRVn-ug@comcast.com...
Jack,

I want to point out that there are basically two ways the charger
can
work:

A. The AC line is stepped down to a low voltage, rectified, and
fed
to
the
charger. The 8.45V is the input to the charger. This is what I am
assuming.

B. The AC is rectified to give 160 -340 VDC, which then goes to a
switching
regulator. The 8.45V is what goes to the battery. If this is the
case,
forget it.

Tam
"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:0Y-dne4Ly4wn-5vcRVn-qg@comcast.com...

"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote
in
message
news:ce5d6i$1oeu$1@uns-a.ucl.ac.uk...
Hi Tam,

Yes, I've taken the lid off the battery charger. It turns out
the
voltage
across the smoothing capacitor (downstream of the step-down
transformer
and
diode rectifier) is 8.45v.

Thanks,
Jack

This is the unregulated voltage, right? I can't come up with a
reason
why
it
should not work if you run 8 - 9 V from a 3 terminal regulator
to
this
point. Your rectifier diodes will prevent the transformer from
shorting
out
the DC. I assume the actual battery voltage is 6V nominal. I
guess
this
is
what you proposed originally. The thing to watch for is if the
thing
uses
positive ground. Could cause fireworks if any grounded metal on
the
camcorder touched grounded metal on the car. Of course, the
camcorder
probably has no exposed metal. At any rate I would be sure to
include
a
fuse. Measure the DC current when running off AC.

I think some of us are confused as to what the topology is. Is
the
actual
charger in the camera, in the brick, or do you remove the
battery
from
the
camera and connect it to the charger? Any power jack on the
camera
should
be
labeled as to what the voltage range is. Either on the camera,
or
in
the
instruction book. My Ricoh Hi8, for instance, uses a 6V battery.
The
camera
has a label that states 6 - 7.5VDC. You remove the battery for
charging.

Let us know how things work out

Tam
Click to expand...
 
Hi,

OK - I've actually bought a 12v to 240v inverter - it's pretty good.

Thanks,
Jack



"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:jt6dnbBns6hKM5rcRVn-hQ@comcast.com...
Jack,

I took a look at your pictures. Hard to tell what is there, but a lot of
stuff. In line with what some others have been saying, the 8.45V may be
the
actual battery charging voltage. Can you measure the voltage coming off
the
charger at the two spring loaded(?) contacts that connect to the battery
to
see if it is 8.45V. I would still look at the Canon web site if they make
some kind of car adapter.

Tam
"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:ce89k6$fbe$1@uns-a.ucl.ac.uk...
Hi,

The measured battery voltage (no load) is 7.65v. The battery claims to
be
a
7.2v 3Ah LiIon pack. I guess it is a 2S Li-Ion pack.

Thanks,
Jack


"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:jtadnXe2gJmXKpvcRVn-ig@comcast.com...
Jack,
What is the nominal battery voltage? I will take a look later, but
the
320VDC sounds bad. The 8.45V sounds about right for a charging
voltage
for
a nominal 7.5V or so battery. Unregulated 8.45V would be on the low
side
for a charger input that charges a 6V or higher battery.

The charger for my Motorola GSM phone appears to use a switching
regulator
type of charger. The blob that plugs into the wall is not big or heavy
enough to contain a 50/60 Hertz transformer.

Tam
"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:ce64eq$2ldg$1@uns-a.ucl.ac.uk...
Hmmm... having looked again at the PCB, I'm not so sure!

Take a look:

http://www.ucl.ac.uk/~ucgadak/charger_components.jpg
http://www.ucl.ac.uk/~ucgadak/charger_merged.jpg
http://www.ucl.ac.uk/~ucgadak/charger_tracks.jpg

I want to put 8.45v onto C22 (it's marked on the last 2 JPGs).
There
are
a
total of 3 transformers. 2 of which have 240v on both sides (i.e.
their
coils are symetric). And there's definitely circuitry to produce
320v
DC
(D1 is a high voltage rectifier).

Urg. I dunno anymore. Any thoughts?

Thanks,
Jack


"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:ce647t$1qe8$1@uns-a.ucl.ac.uk...
Hiya,

I'm 99.999% sure my charger works in way "A".

All the control circuitry for the LiIon charging is on a little
daughter
board, which is definitely downstream of the 8.45v I measured
across
the
smoothing cap.

Thanks,
Jack




"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:zb-dnTGSQPl56JvcRVn-ug@comcast.com...
Jack,

I want to point out that there are basically two ways the
charger
can
work:

A. The AC line is stepped down to a low voltage, rectified, and
fed
to
the
charger. The 8.45V is the input to the charger. This is what I
am
assuming.

B. The AC is rectified to give 160 -340 VDC, which then goes to
a
switching
regulator. The 8.45V is what goes to the battery. If this is the
case,
forget it.

Tam
"Tam/WB2TT" <t-tammaru@c0mca$t.net> wrote in message
news:0Y-dne4Ly4wn-5vcRVn-qg@comcast.com...

"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk
wrote
in
message
news:ce5d6i$1oeu$1@uns-a.ucl.ac.uk...
Hi Tam,

Yes, I've taken the lid off the battery charger. It turns
out
the
voltage
across the smoothing capacitor (downstream of the step-down
transformer
and
diode rectifier) is 8.45v.

Thanks,
Jack

This is the unregulated voltage, right? I can't come up with a
reason
why
it
should not work if you run 8 - 9 V from a 3 terminal regulator
to
this
point. Your rectifier diodes will prevent the transformer from
shorting
out
the DC. I assume the actual battery voltage is 6V nominal. I
guess
this
is
what you proposed originally. The thing to watch for is if the
thing
uses
positive ground. Could cause fireworks if any grounded metal
on
the
camcorder touched grounded metal on the car. Of course, the
camcorder
probably has no exposed metal. At any rate I would be sure to
include
a
fuse. Measure the DC current when running off AC.

I think some of us are confused as to what the topology is. Is
the
actual
charger in the camera, in the brick, or do you remove the
battery
from
the
camera and connect it to the charger? Any power jack on the
camera
should
be
labeled as to what the voltage range is. Either on the camera,
or
in
the
instruction book. My Ricoh Hi8, for instance, uses a 6V
battery.
The
camera
has a label that states 6 - 7.5VDC. You remove the battery for
charging.

Let us know how things work out

Tam
Click to expand...
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top