Modifying old low output charger to make it usable

Thread Starter

Roxx.R

Joined Mar 14, 2023
47
Hello friends, here is another small project i am starting with.

From my electronics collection i took old Samsung charger which was useless until now. The old charger has low output and recently i learned about step up booster and came to know that we can increase the output so i bought a step up booster, but its a bit different than the images i saw online and on Amazon.

Charger's output is 4.76V 0.55A this output is not enough for our mobile devices. Therefore, i bought a step up booster and it seems an easy job to assemble.

For this i am thinking to make a wooden case and i will try to make it look attractive.
Moreover, i do have few questions about changing the potentiometer and other queries. All those questions and doubts will be coming in the next post.

To be continued...
Until then please check the pictures.

Thank you.
 

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LowQCab

Joined Nov 6, 2012
4,026
~2-Watts is the maximum amount of Power that You can expect by adding a Boost-Converter.

A Boost-Converter only increases the available Voltage,
but it will only provide approximately ~80% of the available Power.
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Thread Starter

Roxx.R

Joined Mar 14, 2023
47
~2-Watts is the maximum amount of Power that You can expect by adding a Boost-Converter.

A Boost-Converter only increases the available Voltage,
but it will only provide approximately ~80% of the available Power.
Thanks.. i am new to the this but i am learning about it from YouTube and Google.. i mostly dont understand the symbols in a circuit diagram, the formulas and different components used in circuits.

The charger i have is useless but i think i can make it useful by modifying it.. i am still learning and searching information about potentiometer and an output display .. i want to make a charger where we can change the output.
 

LowQCab

Joined Nov 6, 2012
4,026
One of the most important things that You will ever learn
about Electricity/Electronics is "Ohms-Law".
You can't do anything without it.
Below are some illustrations that, hopefully,
will make remembering all the formulas easier for You.

Electricity/Electronics is ~90% Math-Formulas,
and ~10% odd anomalies that occur in all "real-world" devices.
Nothing is 100% perfect.
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Ohms Law FLAT .png
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Power-Watage Formulas FLAT  .png
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Ohms-Law-Pie-Chart .png
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Thread Starter

Roxx.R

Joined Mar 14, 2023
47
Thanks @LowQCab .. this is helpful... This was taught in school but dont remember it now.. i had left studies incomplete. Anyways, its interesting so i use my time to learn and make new things. Thanks again.
 

LowQCab

Joined Nov 6, 2012
4,026
Instead of putzing-around with what you've got,
go to your local "Thrift-Store" ( Goodwill ),
and get a Laptop-Computer-Power-Supply, for probably ~$3.oo,
they produce a ton of Power for their size.

Be careful, they put out around ~20-Volts,
and are easily capable of smoking other devices that are rated for ~5-Volts.
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Thread Starter

Roxx.R

Joined Mar 14, 2023
47
Yes, thats a good option, but i want to do something practically, make things myself. and i have got plenty electronic stuff.. i mean broken or damaged, unknown circuits, computer parts and so on... I don't want to throw away all that, i want to use it and make something out of it.
 

Alec_t

Joined Sep 17, 2013
14,280
The charger i have is useless but i think i can make it useful by modifying it.
Why is it 'useless'? 4.76V is within the USB spec tolerance so it should be fine for charging things at a modest rate. If you need a higher output current then you need a different charger, as no booster or simple modification can boost the present charger's current while maintaining that voltage.
 

Thread Starter

Roxx.R

Joined Mar 14, 2023
47
I say it useless as it had a different connector, and didn't had a usb output. currently most devices we have, they all have micro usb and usb type C connectors. The main problem is the output Volt and Ampere is too low and most devices we have need charger to be 5V/9V12V and 2A or 2.5A but this old charger has low ampere of 0.55A .. so i think it will take a long time to charge devices.

For modification, i want to add a display to check the output and an extended potentiometer so i can adjust the output using a knob, then i can use it for a variety of purposes.

I checked it gives out 39.6V maximum. i think the ampere also increases which I haven't checked. Please take a look at the images. Sorry for the low quality images.

About the potentiometer, i want to use the rotary type with 3 pins but i don't know which pin to connect to the square shaped blue potentiometer on the circuit board. Please guide me here. They both have 3pins.

Also, does the rotary potentiometer has to be of some value like 10K or 100K? ..or just any rotary potentiometer will work like we have for volume? I haven't bought the potentiometer yet.

Now i will search and select an output display for volt and Ampere if its available at a local shop.
 

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LowQCab

Joined Nov 6, 2012
4,026
"" I checked it gives out 39.6V maximum.
i think the ampere also increases which I haven't checked. ""


Are You taking about using your "useless" Wall-Wort-Charger to Power a Boost-Converter ????,
and the Boost-Converter is then producing 39.6-Volts ?

"39.6-Volts" is what "may" be possible with Zero-Load,
as soon as You apply any significant Load to the Boost-Converter,
the Voltage will drop drastically.

"" ........... and most devices we have need charger to be 5V/9V/12V and 2A or 2.5A ""

This is exactly the performance that You can expect from a Used Laptop-Computer-Power-Supply.
With a Laptop-Computer-Power-Supply,
You can add a Voltage-Regulator after it to adjust it's Output to
any Voltage You need, and at a respectable amount of Current.

Using a "Knob" to adjust the Voltage is a bad idea.
Knobs are too easy to bump, or forget where they are set.
You need a pre-set Switch to select an exact Voltage,
and even a Switch is easy to make an expensive mistake with.

For Charging, the Voltage needs to be fixed to an exact value that the device needs for Charging.
Applying too much Voltage to a device can destroy it instantly.
There may be no "second chances" if You get the Voltage wrong even for a split-second.

You CAN NOT increase the Charging-Rate of a device by increasing the Voltage.
Some cheapo USB-Chargers deliver substantially less than the maximum-USB-Specifications,
which may slow-down Charging on certain items
that are designed to utilize the full USB-Specified-Output-Voltage and Current.

Don't say I didn't warn You.

If You exceed ~5.5-Volts Input into a USB device, You will probably smoke it instantly.
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BobTPH

Joined Jun 5, 2013
8,813
A boost converter can increase voltage while decreasing current. It cannot boost power, because that would violate the laws of physics, it would be free energy.
 

Thread Starter

Roxx.R

Joined Mar 14, 2023
47
@LowQCab i am a bit confused now.. i am sorry as i lack required knowledge here and also some terms are still not understandable to me.

the Voltage will drop drastically
I agree to this, but lets see once i connect potentiometer and digital volt display to it and then check.

Laptop-Computer-Power-Supply
you mean the laptop charger? Like the photo i have attached, its my laptop's charger and its output is 19.5V 3.33A and i cannot increase or decrease that output. If you mean something else, i request you to please share a link or photo which would be very helpful.

Using a "Knob" to adjust the Voltage is a bad idea.
i mean using rotary potentiometer instead of the blue coloured potentiometer to increase or decrease the output volt. Like externally be able to use the potentiometer.. check the attached image.

If You exceed ~5.5-Volts Input into a USB device, You will probably smoke it instantly
I agree.. example: lets say i have 3bulbs of 5V, 9V and 18V so then i can adjust the output and check the bulbs. I don't want to give more volts to a device than specified.

To sum up, what i want is to make the output adjustable. So that i can use it for different purposes. The volt can be adjusted with the help of potentiometer and i want to use a digital display to see the output volt. I have attached an image of the volt meter. Here is the link to the volt meter.

Please let me know if this can be bulit or not, what are your thoughts? Is there something else i need to add? ..i will try to make a diagram of the circuit and share. Then surely you guys can find my mistakes if any..

Thanks.
 

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BobTPH

Joined Jun 5, 2013
8,813
You can use the phone charger and boost converter to make an adjustable supply providing about 2W from 5 to 10V.

It will not help you to charge you phone faster.
 

Thread Starter

Roxx.R

Joined Mar 14, 2023
47
yes, i dont want to charge mobile phone faster. The phone's original charger would work faster instead, the phone's charger has output of 5V/9V 2A.

Now last question i have is about the potentiometer. Does it have to be any specific potentiometer of some value? Because i see there are different values on it. Like in these images each one has a value, B100k, B10k, B22k... Which one should i use?

Then i need to figure out the 3 pins to connect to potentiometer and the other potentiometer.
 

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Dodgydave

Joined Jun 22, 2012
11,285
Do you just want to increase the output voltage to just over 5V then there's a Zener diode D52 in series with the optocoupler which sets the output voltage. This is probably a 4.5V one you could try a 5V or 5.1V to up the output voltage as a test.

This won't increase the current as it can't be altered due to the transformer.
IMG_20230318_185525.jpg
 

Jon Chandler

Joined Jun 12, 2008
1,029
I think you're still missing the point.

Your charger puts out 4.76V at 0.55A.

4.76V × 0.55A = 2.6 watts.

2.6 watts is the most your charger can supply. Let's say you want 12 volts from the boost converter.

12V × X amps = 2.6 watts -->

2.6 watts / 12V = 0.26A

But it's even worse than that. The boost converter is maybe 80% efficient – it requires energy to work.

0.26A × 0.80 = 0.2A

Higher voltages will produce even less current.
 

Thread Starter

Roxx.R

Joined Mar 14, 2023
47
@Dodgydave you mean to say that i can try replacing the "Zener diode D52" with other diode, Correct? ..well, that is way tiny for me, if i try then i will mess it up. For me those DC outputs are the smallest connections i can solder. I don't have enough precision with soldering iron. Thanks for the the idea.

@Jon Chandler there should be a "SuperLike" for your post. Those formulas will be helpful. I will copy and save it. Now i am getting a clearer picture slowly. watts and current is a fuzzy thing. Give me some time, let me study about current. I think then i can get this straight. Thank you very much.

Another thing, i tried using a computer's fan 12V 0.24A was mentioned on the fan... So fan was working, then while it was turned on, i increased the volt on that blue coloured, square shaped potentiometer and the fan goes faster, .. again increased volt and fan goes even faster... So due to increased volt its rotating faster.. so does ampere has to do anything in this case? Does it matter if ampere was high or low, will it affect the fan's speed?
 

LowQCab

Joined Nov 6, 2012
4,026
Voltage, is the "Pressure" pushing the Electrons through a Wire.

Current, or Amps, is how many Electrons are moving in the Wire.

Resistance, or Ohms, reduces how many Electrons can flow in a Wire.

Each one of these things has a number that can be measured.

If You know any 2 of the above numbers,
then You can find the third, missing number, with a simple Math Formula.

All three of these numbers are necessary to determine what a Circuit will do.
They are all, always, interdependent on each other.
They all follow simple Math-Formulas, all the time, ( Ohms-Law ).
----------------------------------------------------------------------------------------------

A Power-Supply provides the "Pressure" in a Circuit,

The amount of "Current" is determined by the "Load",
and how much "Resistance" it has to the "Pressure" (Voltage), being created by the Power-Supply.

The Power-Supply can only provide a limited "Volume", (Amps), to the Load,
if the "Load" has a very low "Resistance", it will be demanding a lot of "Volume", (Amps),
if the Power-Supply can not provide that much Volume, (Amps),
then the "Pressure", (Voltage), from the Power-Supply will go down.
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Thread Starter

Roxx.R

Joined Mar 14, 2023
47
I read it and understand it but can't remember it... I have to come here again and again to read it when i try to think about it or work on it.. maybe i will learn it by doing and making it practically multiple times.. thanks @LowQCab

For now, something came up and i will continue after 25th march... Most probably i am thinking to finish this by adding a DC voltage meter and a potentiometer.
 

Thread Starter

Roxx.R

Joined Mar 14, 2023
47
Finally, i used a glue gun and sticked both the circuit to each other and kept a 5mm thick wood piece in between... Then covered it with cardboard on the sides and bottom... For now, I have kept the top side open from where i can see the booster and can use the potentiometer.

I tried to add another potentiometer but the soldering points are too small for me on the circuit.. Moreover this potentiometer was just lying around, i don't think this 10K potentiometer would work for this circuit.. that 10k potentiometer may not be in a working condition.

I will use this set up to check cooling fans, led or bulbs if they are working or so.. i will need to use multimeter everytime to check and set the output volt.. please let me know if your have any ideas about what other ways i can use this set up...

Thats the end, i don't think i will make any more changes, except i may add a transparent top cover to it.
Thank you everyone for helping me, guiding me and provided so much to learn.
 

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