Lead acid battery trickle charger repair questions. TIA.

R

Rubicon

Guest
Hello,

My uncle bought a little charger from a garage sale and it doesn't
appear to work so passed it on to me to have a look at.

Lead acid battery trickle charger 12/6V 1.2Amp max, 0.5Amp mean.

The mains transformer has three output wires for 12V and 6V. The wire
goes to a, AC~ lead of a bridge rectifier, the second and third to the
outer terminals of a SPDT rocker switch for 12 or 6V. The common
terminal is connected to the second AC~ lead on the bridge rectifier.
A DC lead of the rectifier goes to the + terminal of the car battery.
The batteries negative terminal connects to a thermal switch which
connects to the (the part I don't understand) positive "Charging" LED
lead via its 330 Ohm/1Watt resistor and then goes back to the second
rectifier DC lead. The resistor is so badly burnt I had to measure it
but there's also a broken bare wire on the LEDs negative lead and
another on the thermal switches terminal to the resistor. It's like
fusewire and as if they were joined but if joined they cut the LED and
resistor out of the circuit. I don't know what it's there for. Any
ideas on what its function is?

Perhaps I can't see the woods for the trees here.

Cheers,

Andrew.

Remove the ZZ from E-Mail address to contact me.
Remove the ZZ from E-Mail address to contact me.
 
"Rubicon" <zzrubicon@netaccess.co.nz> wrote in message
news:400cf96d.2540535@news.netaccess.co.nz...
Hello,

My uncle bought a little charger from a garage sale and it doesn't
appear to work so passed it on to me to have a look at.

Lead acid battery trickle charger 12/6V 1.2Amp max, 0.5Amp mean.

The mains transformer has three output wires for 12V and 6V. The wire
goes to a, AC~ lead of a bridge rectifier, the second and third to the
outer terminals of a SPDT rocker switch for 12 or 6V. The common
terminal is connected to the second AC~ lead on the bridge rectifier.
A DC lead of the rectifier goes to the + terminal of the car battery.
The batteries negative terminal connects to a thermal switch which
connects to the (the part I don't understand) positive "Charging" LED
lead via its 330 Ohm/1Watt resistor and then goes back to the second
rectifier DC lead. The resistor is so badly burnt I had to measure it
but there's also a broken bare wire on the LEDs negative lead and
another on the thermal switches terminal to the resistor. It's like
fusewire and as if they were joined but if joined they cut the LED and
resistor out of the circuit. I don't know what it's there for. Any
ideas on what its function is?

Perhaps I can't see the woods for the trees here.

Cheers,

Andrew.

Remove the ZZ from E-Mail address to contact me.
Remove the ZZ from E-Mail address to contact me.
Click to expand...
So remove the LED wiring and check the bridge and thermal and output of the
tranny, if OK fire it it up.
Replace the LED and resistor with values to suit 12v.
If the mysterious "fuse wire" looks original then worry, otherwise you are
cleaning up someones "fiddling" :)
--
Regards ........ Rheilly Phoull
 
in article 400cf96d.2540535@news.netaccess.co.nz, Rubicon at
zzrubicon@netaccess.co.nz wrote on 1/20/04 03:50:

The batteries negative terminal connects to a thermal switch which
connects to the (the part I don't understand) positive "Charging" LED
lead via its 330 Ohm/1Watt resistor and then goes back to the second
rectifier DC lead. The resistor is so badly burnt I had to measure it
but there's also a broken bare wire on the LEDs negative lead and
another on the thermal switches terminal to the resistor. It's like
fusewire and as if they were joined but if joined they cut the LED and
resistor out of the circuit. I don't know what it's there for. Any
ideas on what its function is?
Click to expand...
IF you are saying that as the circuit stands (wire burned), that the
neg. clip connects to thermal cutout, cutout connects to LED, LED connects
to resistor, resistor connects to bridge (or maybe resistor then LED), and
this burned/broken wire appears to connect the thermal switch to the bridge,
bypassing the LED and resistor, THEN it ^may^ be that the wire served as a
shunt, and when current was flowing, the voltage drop across this shunt
would light the indicator.
IF that's the case, and you don't care if the LED lights, just jump it
and forget it. If the lamp needs to work, you need to figure the
size/length/alloy of wire to make it work at whatever minimum current you
choose (also, the LED may no longer be good if the shunt opened and cooked
the resistor).
If this was ^not^ a shunt, the 1.2 amp current sounds like a LOT to pass
through an LED, of any reasonable size.
Hope this helps.
Dave Cole

--
How do you tell a communist? Well,
it's someone who reads Marx and Lenin.
And how do you tell an anti-Communist?
It's someone who understands Marx and Lenin.
Ronald Reagan
 
Thankyou for your replies.

Measuring the charger when set to 12V and attached to a fairly well
charged 12V car battery.

LED and series resistor only.
12.63V 0.02mA
Charging Indicator LED ON

LED, series resistor and wire shunt.
12.87V 0.64A
Charging Indicator LED OFF

Shunt only.
12.87V 065A
Charging Indicator LED OFF

Without the LED it appears to work fine. It would be nice to have it
ON when it should be The 330 ohm resistor is about right for 12V
operation (LED 2.1V 30mA) Damn thing!

Andrew


On Wed, 21 Jan 2004 10:15:11 GMT, Dave Cole <davidwcole@earthlink.net>
wrote:

in article 400cf96d.2540535@news.netaccess.co.nz, Rubicon at
zzrubicon@netaccess.co.nz wrote on 1/20/04 03:50:

The batteries negative terminal connects to a thermal switch which
connects to the (the part I don't understand) positive "Charging" LED
lead via its 330 Ohm/1Watt resistor and then goes back to the second
rectifier DC lead. The resistor is so badly burnt I had to measure it
but there's also a broken bare wire on the LEDs negative lead and
another on the thermal switches terminal to the resistor. It's like
fusewire and as if they were joined but if joined they cut the LED and
resistor out of the circuit. I don't know what it's there for. Any
ideas on what its function is?

IF you are saying that as the circuit stands (wire burned), that the
neg. clip connects to thermal cutout, cutout connects to LED, LED connects
to resistor, resistor connects to bridge (or maybe resistor then LED), and
this burned/broken wire appears to connect the thermal switch to the bridge,
bypassing the LED and resistor, THEN it ^may^ be that the wire served as a
shunt, and when current was flowing, the voltage drop across this shunt
would light the indicator.
IF that's the case, and you don't care if the LED lights, just jump it
and forget it. If the lamp needs to work, you need to figure the
size/length/alloy of wire to make it work at whatever minimum current you
choose (also, the LED may no longer be good if the shunt opened and cooked
the resistor).
If this was ^not^ a shunt, the 1.2 amp current sounds like a LOT to pass
through an LED, of any reasonable size.
Hope this helps.
Dave Cole

--
How do you tell a communist? Well,
it's someone who reads Marx and Lenin.
And how do you tell an anti-Communist?
It's someone who understands Marx and Lenin.
Ronald Reagan
Click to expand...
Remove the ZZ from E-Mail address to contact me.
 
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