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Battery charger part

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LSMFT

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Sep 10, 2010, 9:53:36 AM9/10/10
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Can somebody tell me what this part is?
It's a plate with 2 contacts. 2 12vac outputs from the transformer are
connected to the contacts. The output is the positive charging lead.
The other output of the transformer goes through what looks like a 12v
12a diode though I'm not sure if it is a diode. Then goes through the
meter to the negative charging lead.
It's an old charger, puts out 13.5vdc on the charger, tapers down to
zero when the battery is charged. Tough charger, dropped it a couple
times on the concrete floor. Unknown maker.

http://tinypic.com/m/bimp1g/2 Diode
http://tinypic.com/m/bimp6p/2 Plate
http://tinypic.com/m/bimp83/2 Transformer
http://tinypic.com/m/bimp9h/2 Plate contact
http://tinypic.com/m/bimpah/2 Meter connection
http://tinypic.com/m/bimpeb/2 Charger

--
LSMFT

I look outside this morning and everything was in 3D!

nesesu

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Sep 10, 2010, 10:08:40 AM9/10/10
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On Sep 10, 6:53 am, LSMFT <bole...@aol.com> wrote:
> Can somebody tell me what this part is?
> It's a plate with 2 contacts. 2 12vac outputs from the transformer are
> connected to the contacts. The output is the positive charging lead.
> The other output of the transformer goes through what looks like a 12v
> 12a diode though I'm not sure if it is a diode. Then goes through the
> meter to the negative charging lead.
> It's an old charger, puts out 13.5vdc on the charger, tapers down to
> zero when the battery is charged. Tough charger, dropped it a couple
> times on the concrete floor. Unknown maker.
>
> http://tinypic.com/m/bimp1g/2      Diodehttp://tinypic.com/m/bimp6p/2      Platehttp://tinypic.com/m/bimp83/2      Transformerhttp://tinypic.com/m/bimp9h/2      Plate contacthttp://tinypic.com/m/bimpah/2      Meter connectionhttp://tinypic.com/m/bimpeb/2      Charger

>
> --
> LSMFT
>
> I look outside this morning and everything was in 3D!

THe 'plate' is the rectifier. The two brown plasitc 'pucks' are the
actual diodes. The 'diode' in your photo is the circuit breaker [12A].
Very typical simple full wave rectifier battery charger.

Neil S.

Mr...@totally.invalid

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Sep 10, 2010, 10:11:37 AM9/10/10
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On Fri, 10 Sep 2010 09:53:36 -0400, LSMFT <bol...@aol.com> wrote:

>Can somebody tell me what this part is?
>It's a plate with 2 contacts. 2 12vac outputs from the transformer are
>connected to the contacts. The output is the positive charging lead.
>The other output of the transformer goes through what looks like a 12v
>12a diode though I'm not sure if it is a diode. Then goes through the
>meter to the negative charging lead.
>It's an old charger, puts out 13.5vdc on the charger, tapers down to
>zero when the battery is charged. Tough charger, dropped it a couple
>times on the concrete floor. Unknown maker.
>

>http://tinypic.com/m/bimp1g/2 Diode - This seems to be a 12v- 12amp thermal self-reset circuit breaker
>http://tinypic.com/m/bimp6p/2 Plate - Probably the rectifiers (2) - may be selenium

Mr.E

LSMFT

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Sep 10, 2010, 10:20:31 AM9/10/10
to

So I should be able to replace the 'plate' with a solid state full wave
bridge rectifier?
What causes the tapering down effect as the battery charges?

Jeffrey Angus

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Sep 10, 2010, 10:30:33 AM9/10/10
to
On 9/10/2010 9:20 AM, LSMFT wrote:
> So I should be able to replace the 'plate' with a solid state
> full wave bridge rectifier?

You have solid state full wave rectifier. A bridge will give
you twice the output voltage.

> What causes the tapering down effect as the battery charges?

The output voltage of the charger is fixed, as the battery
voltage comes us (charges) the current goes down.

Jeff

LSMFT

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Sep 10, 2010, 10:33:01 AM9/10/10
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Jeffrey Angus wrote:
> On 9/10/2010 9:20 AM, LSMFT wrote:
>> So I should be able to replace the 'plate' with a solid state
>> full wave bridge rectifier?
>
> You have solid state full wave rectifier. A bridge will give
> you twice the output voltage.

Are you sure?

>
>> What causes the tapering down effect as the battery charges?
>
> The output voltage of the charger is fixed, as the battery
> voltage comes us (charges) the current goes down.
>
> Jeff
>

Jeffrey Angus

unread,
Sep 10, 2010, 11:02:58 AM9/10/10
to
On 9/10/2010 9:33 AM, LSMFT wrote:
> Jeffrey Angus wrote:
>> On 9/10/2010 9:20 AM, LSMFT wrote:
>>> So I should be able to replace the 'plate' with a solid state
>>> full wave bridge rectifier?
>>
>> You have solid state full wave rectifier. A bridge will give
>> you twice the output voltage.
>
> Are you sure?

Yes, the key here is the word bridge.

There are two diodes in a simple full wave rectifier.
AND a center tap on the transformer. Only one half of the
winding is conducted through a diode at a time. But since
they alternate, it's a full wave rectifier.

A bridge rectifier has four diodes, and goes across the
entire output winding. So you'll get twice the voltage.

Jeff

Allodoxaphobia

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Sep 10, 2010, 11:43:20 AM9/10/10
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Not if you hook it up simply as a full wave rectifier -- leaving
the other 'side' (probably marked with a minus ("-") sign) floating.

Jonesy
--
Marvin L Jones | jonz | W3DHJ | linux
38.24N 104.55W | @ config.com | Jonesy | OS/2
* Killfiling google & XXXXbanter.com: jonz.net/ng.htm

LSMFT

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Sep 10, 2010, 1:06:24 PM9/10/10
to

The wave pulse is doubled but the voltage is not I believe. The peaks
remain at the same level in my experience.

Plain...@yawho.com

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Sep 10, 2010, 1:33:52 PM9/10/10
to

No. You could replace the 'plate' with two stud mount rectifiers in a
suitable heat sink. However, the voltage drop of of the silicon
rectifiers is much less than the voltage drop of the original selenium
rectifier. This will result in overcharging, and release of hydrogen
and oxygen gas as the water in the electrolyte is disassociated.

>What causes the tapering down effect as the battery charges?

Good grief, don't you even understand lead-acid batteries? As the
battery charges the voltage across it increases (although the voltage
is not linear with percent of charge). The output of the battery
charger is the 'upper' portion of the sinusoidal AC waveform. Current
will flow into the battery only when the instantaeous voltage out of
the rectifier is greater than the charging voltage of the battery.
As the battery voltage increases, the fraction of the waveform
actually charging the battery decreases.

The transformer was designed taking into consideration the
characteristics of the original rectifiers. In other words, as the
battery became fully charged, only a small 'trickle' charge would
flow.

PlainBill

PeterD

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Sep 10, 2010, 7:51:40 PM9/10/10
to
On Fri, 10 Sep 2010 10:33:01 -0400, LSMFT <bol...@aol.com> wrote:

>Jeffrey Angus wrote:
>> On 9/10/2010 9:20 AM, LSMFT wrote:
>>> So I should be able to replace the 'plate' with a solid state
>>> full wave bridge rectifier?
>>
>> You have solid state full wave rectifier. A bridge will give
>> you twice the output voltage.
>
>Are you sure?

Yes, he's sure. So are most of the rest of us. Two diodes in a full
wave rectifier.

nesesu

unread,
Sep 11, 2010, 10:16:58 AM9/11/10
to
On Sep 10, 10:06 am, LSMFT <bole...@aol.com> wrote:
> Jeffrey Angus wrote:
> > On 9/10/2010 9:33 AM, LSMFT wrote:
> >> Jeffrey Angus wrote:
> >>> On 9/10/2010 9:20 AM, LSMFT wrote:
> >>>> So I should be able to replace the 'plate' with a solid state
> >>>> full wave bridge rectifier?
>
> >>> You have solid state full wave rectifier. A bridge will give
> >>> you twice the output voltage.
>
> >> Are you sure?
>
> > Yes, the key here is the word bridge.
>
> > There are two diodes in a simple full wave rectifier.
> > AND a center tap on the transformer. Only one half of the
> > winding is conducted through a diode at a time. But since
> > they alternate, it's a full wave rectifier.
>
> > A bridge rectifier has four diodes, and goes across the
> > entire output winding. So you'll get twice the voltage.
>
> > Jeff
>
> The wave pulse is doubled but the voltage is not I believe. The peaks
> remain at the same level in my experience.
>
> --
> LSMFT
> - Show quoted text -

I am not sure what you are saying here. Jeff is quite correct. With
the existing rectifier you get one positive peak of DC output every
half cycle, therefore it is 'Full Wave'.
There is little point in replacing the rectifiers since thay appear to
be [or similar to] Motorola MR type 50A silicon diodes attached to the
heat sink. anything you replace them with is not likely to be any
better and could be much worse.

Neil S.

Paul Nelson

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Sep 11, 2010, 11:41:01 AM9/11/10
to
The part in the first pic ("diode") is a circuit breaker. The two
ceramic buttons on the "plate" are the diode rectifiers, that'd be a
half-wave bridge configuration.

Hope that helps!


--
Paul Nelson W5GNF "When I go, I want to go quietly, in my
Ames, Iowa sleep, like my grandfather- not
Senior Engineer (Retired) screaming, like his passengers."
Sauer-Danfoss Company
(DrH...@qwest.net)
"More hay, Trigger?"
ex-Cessna 140 N77149 (sigh) "No thanks, Roy, I'm stuffed."

Paul Nelson

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Sep 11, 2010, 1:28:15 PM9/11/10
to
oops. full-wave, but not bridge.

Grant

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Sep 11, 2010, 10:25:49 PM9/11/10
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On Sat, 11 Sep 2010 12:28:15 -0500, Paul Nelson <drh...@qwest.net> wrote:

>oops. full-wave, but not bridge.

But you could replace the diodes with a bridge, just ignore either the
plus or minus terminal on bridge and use centre tap of transformer as
per existing circuit. Those high current large bridges are probably
easier to buy and mount on the heatsink than a pair of power diodes.

The unused pair of diodes in the bridge wont do any harm.

Grant.

Wild_Bill

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Sep 13, 2010, 8:08:24 AM9/13/10
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As mentioned, the 12V 12A part is a circuit breaker, a self-resetting type
which opens when heat from high current causes it to open, then closes again
when it cools. The opening action will repeat if the current draw/load is
excessive.

The plate assembly has 3 connections, not 2. A (black?) wire lead is
connected directly to the plate.
The plate is a heatsink for the 2 red diodes. The red diodes are silicon
types with terminals/contacts on each side/end. Two diode terminals contact
the plate, and the other terminals contact the metal holders that the other
2 wire leads are connected to (green & green?).

The diodes need to be in their original specific orientations for the
correct output of the charger.
Mark the diodes before removal, or make sure of their orientation before
removal, as the anode and cathode may not be marked. The part number or
other printing around the body may be the only indication of the diode
polarity. With the printing viewed right-side-up, the top contacts should be
the same terminals.. A or C.
Or, there may be a diode symbol printed to indicate A and C.
Or use an ohmeter to determine polarity of the diodes.

--
Cheers,
WB
.............


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