# Rechargeable Torch Light repair

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#### indigo

##### New Member
Hello There ....
I'm new to electronics and interesting into it.

Attached pic is my rechargeable torch light circuit.
One day while charging it came out smoke and after open it I saw the capicitor was blown.
I tried to read the value and it says 50V 100uf.
so I replaced with new one and charge again.
It blown out again.
I measure all the resistors, diodes ... all are look good.
so I measure the output DC and it is around 80V.
even I increase the resistor value of near the Film Capicitor to double and output voltage is the same.
How can I reduce the voltage?

All your suggestions are very much appreciated.
TQ.

#### JimB

##### Super Moderator
Welcome to ETO.

Your charger appears to be a transformerless type, as an electronics beginner you need to be aware that these do not provide isolation from the mains supply and are a shock hazard when connected to the mains.

A picture of the other side of the circuit board would be useful so that we can work out the circuit of this charger.

I increase the resistor value of near the Film Capicitor to double and output voltage is the same.
At a quick guess the purpose of that resistor is to ensure that there is not a high voltage stored on the red capacitor when the charger is disconnected from the mains.
I suggest that you return the resistor to its original value, changing it will not make a difference in this case.

There are six wire connections to that circuit board,
two to the battery,
two to the mains supply,
what do the other two connect to?

I notice that R3 has been replaced, why?

JimB

#### indigo

##### New Member
Welcome to ETO.

Your charger appears to be a transformerless type, as an electronics beginner you need to be aware that these do not provide isolation from the mains supply and are a shock hazard when connected to the mains.

A picture of the other side of the circuit board would be useful so that we can work out the circuit of this charger.

At a quick guess the purpose of that resistor is to ensure that there is not a high voltage stored on the red capacitor when the charger is disconnected from the mains.
I suggest that you return the resistor to its original value, changing it will not make a difference in this case.

There are six wire connections to that circuit board,
two to the battery,
two to the mains supply,
what do the other two connect to?

I notice that R3 has been replaced, why?

JimB

Thanks for prompt reply and help JimB.
And for the shock hazard warning.
Will put it back to original value.
Now I attached back side of the circuit together with R2,3,4.
The other two are connected to switch.
The R3,4 are for the light bulb if I trace correctly. Those R are not respond to test meter and I've taken out.

R2 also burnt and can't read the value but that R is connect to charging indicator LED.
That R2 is replaced with 100 Ohm.
Now when I connect to main and within 30sec the capacitor is bursted.

Can I replace the capacitor with 150v 100uf?
Does the diodes has the resistance?
I've only left to replace 4diodes.

Thanks again.

#### alec_t

##### Well-Known Member
My guess is that one or more of the diodes is dead, so AC is getting to the capacitor.

#### Les Jones

##### Well-Known Member
Hi Alec,
I suspect the fault is that there is not enough load on the board to limit the voltage. (Normaly I would expect there to be a zener diode to limit the voltage across the electrolytic capaacitor.)
indigo,
can you post a schematic of how the board is wired to the battery and bulb. Also what is the battery voltage and type ? What is the rating of the bulb ? (Voltage and current oe wattage.) If there is no load on this type of power supply the voltage can go up to almost 1.414 times the RMS value of the AC supply.

Les.

#### JimB

##### Super Moderator
I have just sketched the circuit for this.

If the battery is not connected, or is open circuit, the capacitor will fail and explode. The battery is the only load if the torch is not switched on.

JimB

#### Nigel Goodwin

##### Super Moderator
If the battery is not connected, or is open circuit, the capacitor will fail and explode.
I suspected that from the top view, there's nothing there to prevent it - bit of a 'nasty' design.

#### indigo

##### New Member
My guess is that one or more of the diodes is dead, so AC is getting to the capacitor.
The D4 is faulty on board but normal at off the board.
I replaced that D4 with old one from other circuit, after connect to the main the charge indicator was blown.
So I put it back with original one and now it D4 is normal on board again.
But still suspect that. As I mentioned above my post I should replace all Ds.

Hi Alec,
I suspect the fault is that there is not enough load on the board to limit the voltage. (Normaly I would expect there to be a zener diode to limit the voltage across the electrolytic capaacitor.)
indigo,
can you post a schematic of how the board is wired to the battery and bulb. Also what is the battery voltage and type ? What is the rating of the bulb ? (Voltage and current oe wattage.) If there is no load on this type of power supply the voltage can go up to almost 1.414 times the RMS value of the AC supply.

Les.
First time after replaced the C with new one , that time the battery is connected.
When I connect to the main within few seconds the C is emittimg smoke and burst.
So I remove the battery and check the voltage and right here to this forum.
Below is the similar diagram I found on net. According to that diagram the output voltage is just 12V and used 20V C .
Howcome my one is using 50v? I don't know what is the actual output voltage of mine.

I try to draw and post it later.
The battery is 3.5in long , 2.3in width, 0.8in thick and there is no printing on it. 4v when measure.

I have just sketched the circuit for this.
If the battery is not connected, or is open circuit, the capacitor will fail and explode. The battery is the only load if the torch is not switched on.
JimB
So in your opinion what is the voltage should have my charger.

I suspected that from the top view, there's nothing there to prevent it - bit of a 'nasty' design.
Well, that's why we called it "China Made"

Thank you Guys.
Appreciated.

#### Les Jones

##### Well-Known Member
This is the schematic that I traced for the board.

R3 may be 0.22R rather than 2.2R as I could not decide if the decimal multiplyer was gold or silver. From the way you describe the battery it is probably a single cell lithium polymer type. You still have not told us how the three wires are connected to the battery, bulb and probably a switch. If we assume that the mylar capacitor is 470 nF it would have a reactance of about 6.8K ohms at 50 Hz (Slightly less at 60 Hz) . So the power supply would behave like a 30 mA constant current source. (For relativly low output voltages.) To keep the output below 50 volts (The electrolytic capacitor rating.) the load would need to be greater than 25 mA.
I have no practical experience of lithium polymer cells so I don't know if they tend to fail with high internal resistance. If the cell had developed a high internal resistance then this would explain the electrolytic capacitor.
Edit. I have just noticed that I have probably drawn the LED polarity the wrong way round.

Les.

#### JimB

##### Super Moderator
What Les Jones just said.

I came back to do a quick analysis of the circuit and Les has beaten me to it.

Without a load, ie a battery to charge, the voltage across the capacitor will try to rise to 1.4 x 220 = 310 volts.

If the battery has failed open circuit, the capacitor will explode. There is no point in putting a higher rated capacitor.

JimB

#### indigo

##### New Member
R3 may be 0.22R rather than 2.2R as I could not decide if the decimal multiplyer was gold or silver. From the way you describe the battery it is probably a single cell lithium polymer type. You still have not told us how the three wires are connected to the battery, bulb and probably a switch. If we assume that the mylar capacitor is 470 nF it would have a reactance of about 6.8K ohms at 50 Hz (Slightly less at 60 Hz) . So the power supply would behave like a 30 mA constant current source. (For relativly low output voltages.) To keep the output below 50 volts (The electrolytic capacitor rating.) the load would need to be greater than 25 mA.
I have no practical experience of lithium polymer cells so I don't know if they tend to fail with high internal resistance. If the cell had developed a high internal resistance then this would explain the electrolytic capacitor.
Edit. I have just noticed that I have probably drawn the LED polarity the wrong way round.
Les.
I'm trying draw that could electronics techinician understand.
Below is how the board connected to battery, switch and bulb.

The switch has 2steps, first is normal brightness and second is more.
I put all the info about components on that.
Could u please calculate what is the output voltage should be according to that circuit.
What voltage should be used to charge this kind of battery?

What Les Jones just said.

I came back to do a quick analysis of the circuit and Les has beaten me to it.

Without a load, ie a battery to charge, the voltage across the capacitor will try to rise to 1.4 x 220 = 310 volts.

If the battery has failed open circuit, the capacitor will explode. There is no point in putting a higher rated capacitor.

JimB
Buying the new one is the easiest. It cost around just $5.00 only. Why I'm doing this is, I wanted to learn. How about if I replace all the diodes, capacitor? I'm trying to replace and see the outcome. #### JimB ##### Super Moderator Most Helpful Member Buying the new one is the easiest. It cost around just$5.00 only.
OK

Why I'm doing this is, I wanted to learn.
A very good reason for doing things the hard and expensive way.

How about if I replace all the diodes, capacitor?
I'm trying to replace and see the outcome.
You are failing to grasp what we are trying to tell you.

This is not a constant voltage supply.
It is a very simply constant current supply. The current is determined by the value of capacitor C1.
Replacing all the other components will not help.

The main problem is either:
The battery has failed
or
Capacitor C1 has failed

The battery is what provides a load to the constant current circuit, with no load the voltage will rise to 300v or so.
On the way to 300v, capacitor C2 will break down and explode.

If the battery is not in good condition, there is no point in doing anything with the rest of the circuit.

Could u please calculate what is the output voltage should be according to that circuit.
Without the load it will be 300 volts.
With the battery it will be the battery charging voltage.

What voltage should be used to charge this kind of battery?
Who knows, all we can see is that the battery is a grey oblong thing. Wild guess, less than 10 volts, probably 5 or 6 volts.

JimB

#### indigo

##### New Member
The main problem is either:
The battery has failed
or
Capacitor C1 has failed

The battery is what provides a load to the constant current circuit, with no load the voltage will rise to 300v or so.
On the way to 300v, capacitor C2 will break down and explode.

If the battery is not in good condition, there is no point in doing anything with the rest of the circuit.

Without the load it will be 300 volts.
With the battery it will be the battery charging voltage.

Who knows, all we can see is that the battery is a grey oblong thing. Wild guess, less than 10 volts, probably 5 or 6 volts.

JimB
I did replace the C1 with good one and output voltage is 80v without the load.
I did mention the battery is 4v at post#8
Did u mean the circuit is build to depand on battery? That make nonsense.
From what I did ...
C1 is good.
C2 is good.
Only diodes are need to place.
And see what is the outcome ....
will tell u guys ....

#### Les Jones

##### Well-Known Member
If you don't believe what we are all telling you then when you fit a 450 volt rating capacitor fot C2 you will find it does not explode and you will get about 300 volts across it. It is possible that the diodes will fail then because the original ones would not need to be rated at at such a high voltage when the board was connected to a GOOD battery. Try charging the battery from a 4.2 volt DC supply via a 10 ohm resistor and measure the charging current.

Les.

#### indigo

##### New Member
The main problem is either:
The battery has failed
or
Capacitor C1 has failed

JimB
Yesterday I replaced all diodes and NEW battery.
After connect to main AC , the charging LED blink 1 time and off.
And D3 , D4 are dead.
Now is the time to change C1.
How can I test that C1 with analog meter?

If you don't believe what we are all telling you then when you fit a 450 volt rating capacitor fot C2 you will find it does not explode and you will get about 300 volts across it. It is possible that the diodes will fail then because the original ones would not need to be rated at at such a high voltage when the board was connected to a GOOD battery. Try charging the battery from a 4.2 volt DC supply via a 10 ohm resistor and measure the charging current.

Les.
Not because of I don't believe in all of u. U guys got much more knowledge and experiences.
I'm not satisfy with what I found out.
How could a person build like such circuit?
What I consider to this torch light is like normal mobile phone charger.
Mobile phone chargers or any other DC chargers, whether they are connected to load or not output voltage is rated as they printed and when u measure there is what the voltage is what they printed.
So, Les ....
how can I modify that circuit to use with external 5v charger?

I'm sorry that if u think I've offended u guys.

Last edited:

#### Les Jones

##### Well-Known Member
The reason someone designed it this way is to use the bare minimum of components so as to build it for the minimum cost. To answer the question you asked Jim about testing C1 with an normal multimeter (Either analogue or digital.) it is not possible with just the meter. From what you say in post #15 it does sound like the insulation in C1 is breaking down. As you have managed to source a replacement battery can you say if my guess that it is a lithium polymer battery is right or wrong ? If it is a lithium battery then the design of the charging circuit is not suitable. Also there should be over discharge protection. You can get ready made charger and protection boards but I don't know if exactly 5 volts from a phone charger is quite enough for them to work properly. You would have to find the specification of the board. A phone charger or any other DC power supply will be isolated either by a high frequency transformer in a switch mode power supply or in older versions by a transformer working at the power supply frequency of 50 or 60 Hz. Because there are no connections that you can touch on your torch the can get away with a none isolated design.

Les.

#### indigo

##### New Member
I really don't know the what kind of battery and after googling the net that battery appear to be Lead Acid Rechargeable Battery with the 4v 1000-1500mAh. (My one should be 1200mAh , saw at online selling site)
The circuit is the only 1.5sq.in. and nothing printed on it as u see at post#1. They don't even print the resistor's value.
Ofcourse must need to make isolated.
Cut off the AC circuit line, put the jumper cable to able to charge with DC... , that's what I'm thinking, whether can or not I don't know.

#### Les Jones

##### Well-Known Member
If it is a lead acid battery then you would need about 4.6 volts from the charger to fully charge it. I do not know of any adjustable LDO regulators that will work with an input / output differential of only 0.4 volts. (5.0 - 4.6) It MAY be possible to to do it with a logic level power mosfet and a rail to rail op amp. To do this you would have to learn about electronics. I am not prepared to spend the time researching suitable components and designing a circuit to build an item that you can buy cheaply. I have never seen a circuit board with the component values printed.

Les.

FYI Les ,