220V to 12V 2Amp power supply modification

[Nauman Muhammad]

Hi!

Dear All,

I want to modify my 220V AC to 12V DC 2Amp power supply to give me 14V to charge 12v lead acid batteries.

Basically my needs are to charge 12v 6Amp battery and run a 0.6A router simultaneously.

Please see attached pictures for possible modifications and also guide me how to discharge this circuit for e.g capacitors etc so that while soldering new thing there remains no risk of electric shock or burn.

 

[duke37]

The 10μF capacitor should be discharged with a resistor, say 10kΩ.

The board is split with high voltage on the left and low voltage on the right linked by an opto isolator.
To get a higher voltage, the feedback needs to be reduced or the reference voltage increased. What is IC2?

 

[davenn]

What is IC2?

most likely a KA431 or TL431 or similar

here's a pretty common layout for a SMPS ......
As can be seen, this one has 2 outputs 12V and 5V

I'm more interested in what is the 3 leg device up in the top left corner to the right of that inductor
The OP's SMPS obviously has a more basic primary side that lacks the usual controller IC


DAve

 

[Nauman Muhammad]

The 10μF capacitor should be discharged with a resistor, say 10kΩ.

The board is split with high voltage on the left and low voltage on the right linked by an opto isolator.
To get a higher voltage, the feedback needs to be reduced or the reference voltage increased. What is IC2?

I will tell you about IC2 shortly.. I will tell you its number!

 

[Nauman Muhammad]

most likely a KA431 or TL431 or similar

here's a pretty common layout for a SMPS ......
As can be seen, this one has 2 outputs 12V and 5V

I'm more interested in what is the 3 leg device up in the top left corner to the right of that inductor
The OP's SMPS obviously has a more basic primary side that lacks the usual controller IC


DAve

Actually I am very much begineer and my understanding is poor. However I will check the number on 3 legged IC and will tell you..

 

[73's de Edd]

Sir Nauman Muhammad . . . . . . .

Until I saw this separate posting. . . . I was afraid that you just had a mere sealed wall
wart . . . . this is a unit that we should be able to slightly modify for your needs..

Initially filling in on suppositions . . . .

This circuitry design is just a power oscillator with a feed back loop control.

The questioned component between the line inductor RF filter and the DW1 diode should just be a common
transistor (Pass its part number to us) involved with interfacing the photocell aspect of the Optical Isolator IC1
into the power oscillator transistor that is mounted to the heat sink.

IC2 will be an adjustable shunt regulator .

What I can't make out is the color banding on R7 so pass all of its sequential color bands back to us .
Confirm if I am reading R6 correct as being color coded as ORANGE*BLACK*RED for 4K ohm.
R5 I am reading as RED*RED*BLACK for 22 ohms.
What I also can not see, since the marking is covered up, is the marking on the PCB for DW1 and D6.
You are familiar with the standard schematic diode symbol which uses a black triangle with a black bar placed
at one tip of the triangle.
That would designate a standard diode which D6 will probably be . . .but is hidden from view.
The one of interest will be the assigned "DW1" if its black bar has some angled tiplets, that would be designating a Zener type of diode, can you confirm that ? . . . . . .by its PCB marked symbol.
( Up above those diodes I have placed both types of symbols I suspect D6 to be as shown, but am uncertain of the "DW1" marked unit.)

This might just be the right time to use your NEW DMM by using it to measure those three resistors.
For our further acquaintance with it, can you give us a full frontal CLOSEUP photo of it.

You already have been informed of R6 and R5 values . . . .to confirm . . .so you now have R7 to figure out . . . . but I suspect that it might be similar to the value of R6.

We can even start a feasibility test now, if you will sort through your old electronics devices components
sourcings to find a small resistor anywhere from 4700 YellowVioletRed through 10,000 BrownBlackOrange ohms in value . . . . . hoping to find close to 4700.

You will initially power up and measure the DC output from the supply and write it down.
Unplug the main AC power to the unit.
Then take your soldering irons power plugs blades and shunt them across the 12VDC power supply connections at my RED DOTS.
Hold that connection to bleed down the power supply output for a count to 15, aloud . . . . . . .in Urdu . . . . just so that your father / wife will have to ask " WHAT are you doing ? ? ? ? ".
Then, use just one blade of that connector to finally totally discharge the supply capacitor.
Take your ~4.7 K resistor and solder tack it across R7 on the solder / foil side of the board.
Apply AC power to the board.
Measure DC output of the supply now and write down.
Unplug AC power and use the same procedure in bleeding your DC output from the supply.
Lift the test resistor from across R7 and move it down to solder tack across R6 and shunt it.
Power up the supply and test the DC voltage output and write it down.
Our three test voltages will now clue us as how we need to alter two resistors values to get
~ 20% more voltage output from this supply.
If this is beyond their range we will just move over to the other side of the board and accomplish
it with a slight modification of the feedback loop to the oscillator circuitry.

Thats . . .it . . . . unless you need any clarifications . . . . plus I need all of those bits of requested
information mentioned earlier..


TECHNO DISPLAY / REFERENCING . . . .of howtodoittoit:

73's de Edd

 

[duke37]

If the IC is a kA431, then the sense terminal will be at 5V.
To get 29% more voltage across the 'top' resistor, the 'bottom' resistor from the sense terminal to negative common needs to pass 29% more current . This can be done by tacking a resistor of about three times its value across it.

 

[73's de Edd]

My schematic at the bottom is just a " closer approximation " of actual circuitry to be found . .e.g. feedback controlled power oscillator.
Compared to one using a UA 3824 series of S.M. controller I.C.


73's de Edd

 

[Nauman Muhammad]

Sir Nauman Muhammad . . . . . . .

Until I saw this separate posting. . . . I was afraid that you just had a mere sealed wall
wart . . . . this is a unit that we should be able to slightly modify for your needs..

Initially filling in on suppositions . . . .

This circuitry design is just a power oscillator with a feed back loop control.

The questioned component between the line inductor RF filter and the DW1 diode should just be a common
transistor (Pass its part number to us) involved with interfacing the photocell aspect of the Optical Isolator IC1
into the power oscillator transistor that is mounted to the heat sink.

IC2 will be an adjustable shunt regulator .

What I can't make out is the color banding on R7 so pass all of its sequential color bands back to us .
Confirm if I am reading R6 correct as being color coded as ORANGE*BLACK*RED for 4K ohm.
R5 I am reading as RED*RED*BLACK for 22 ohms.
What I also can not see, since the marking is covered up, is the marking on the PCB for DW1 and D6.
You are familiar with the standard schematic diode symbol which uses a black triangle with a black bar placed
at one tip of the triangle.
That would designate a standard diode which D6 will probably be . . .but is hidden from view.
The one of interest will be the assigned "DW1" if its black bar has some angled tiplets, that would be designating a Zener type of diode, can you confirm that ? . . . . . .by its PCB marked symbol.
( Up above those diodes I have placed both types of symbols I suspect D6 to be as shown, but am uncertain of the "DW1" marked unit.)

This might just be the right time to use your NEW DMM by using it to measure those three resistors.
For our further acquaintance with it, can you give us a full frontal CLOSEUP photo of it.

You already have been informed of R6 and R5 values . . . .to confirm . . .so you now have R7 to figure out . . . . but I suspect that it might be similar to the value of R6.

We can even start a feasability test now, if you will sort throught your old electronics devices components
sourcings to find a small resistor anywhere from 4700 YellowVioletRed through 10,000 BrownBlackOrange ohms in value . . . . . hoping to find close to 4700.

You will initially power up and measure the DC output from the supply and write it down.
Unplug the main AC power to the unit.
Then take your soldering irons power plugs blades and shunt them across the 12VDC power supply connections at my RED DOTS.
Hold that connection to bleed down the power supply output for a count to 15, aloud . . . . . . .in Urdu . . . . just so that your father / wife will have to ask " WHAT are you doing ? ? ? ? ".
Then, use just one blade of that connector to finally totally discharge the supply capacitor.
Take your ~4.7 K resistor and solder tack it across R7 on the solder / foil side of the board.
Apply AC power to the board.
Measure DC output of the supply now and write down.
Unplug AC power and use the same procedure in bleeding your DC output from the supply.
Lift the test resistor from across R7 and move it down to solder tack across R6 and shunt it.
Power up the supply and test the DC voltage output and write it down.
Our three test voltages will now clue us as how we need to alter two resistors values to get
~ 20% more voltage output from this supply.
If this is beyond their range we will just move over to the other side of the board and accomplish
it with a slight modification of the feedback loop to the oscillator circuitry.

Thats . . .it . . . . unless you need any clarifications . . . . plus I need all of those bits of requested
information mentioned earlier..


TECHNO DISPLAY / REFERENCING . . . .of howtodoittoit:

73's de Edd

Got it! Before proceeding I would like to share the details of ICs. The three leg IC with heat sink is E13005F.

Rest are shown in pics attached. Please have a look.

 

[Nauman Muhammad]

If the IC is a kA431, then the sense terminal will be at 5V.
To get 29% more voltage across the 'top' resistor, the 'bottom' resistor from the sense terminal to negative common needs to pass 29% more current . This can be done by tacking a resistor of about three times its value across it.

It looks like 431A

 

[Nauman Muhammad]

Got it! Before proceeding I would like to share the details of ICs. The three leg IC with heat sink is E13005F.

Rest are shown in pics attached. Please have a look.View attachment 27703View attachment 27704View attachment 27705

 

[Nauman Muhammad]

And yeah whenever open my box which contains all my equipments my Dad and wife says what are you doing and you will end up doing nothing haha... But I am happy with my hobby of electronics and aviation as accounting and auditing is a boring subject

 

[duke37]

What superb pictures, how do you do it?

You are only concerned with the low voltage side of the board.
In the picture showing the TL431A, the sense terminal is on the left. From here there will be one resistor connected to the +output and one connected to the -output. What is the value of these resistors, give us the colours if you are unsure of the code.

 

[73's de Edd]

Sir Nauman Muhammad . . . . . . .

With your info, looks like the better picture actually revealed R5 to be 220 ohms vice 22 ohms.

The 3 terminal "IC" is actually being a MJE1300x family of power transistor, which Motorola introduced in the 80-90's and has been a mainstay since, in these types of circuits.

Looks like you are ready to go to post # 6 and do the steps that are NOW marked in BLUE.


73's de Edd

 

[Nauman Muhammad]

What superb pictures, how do you do it?

You are only concerned with the low voltage side of the board.
In the picture showing the TL431A, the sense terminal is on the left. From here there will be one resistor connected to the +output and one connected to the -output. What is the value of these resistors, give us the colours if you are unsure of the code.

I use my phone with flash on to take these pics

Okay, I will try to de code the code first to learn

 

[Nauman Muhammad]

Sir Nauman Muhammad . . . . . . .

With your info, looks like the better picture actually revealed R5 to be 220 ohms vice 22 ohms.

The 3 terminal "IC" is actually being a MJE1300x family of power transistor, which Motorola introduced in the 80-90's and has been a mainstay since, in these types of circuits.

Looks like you are ready to go to post # 6 and do the steps that are NOW marked in BLUE.


73's de Edd

That's great info..

I am still in search of 4700 ohm resistor... Can higher one work?

Sorry for my poor understanding but I couldn't understand these two steps:

1. Take your ~4.7 K resistor and solder tack it across R7 on the solder / foil side of the board.

2. Lift the test resistor from across R7 and move it down to solder tack across R6 and shunt it.

Can you please provide me a diagram plan for easy understanding?

 

[73's de Edd]

Sir Nauman Muhammad . . . . . . .


I am still in search of 4700 ohm resistor... Can higher one work?

I think that the use of TWO resistors in series might even make it even EASIER !
Just find two values that will add up to that approximate 4700 ohm value.
In my case I have shown 2200 and 3300, but its what you can find in YOUR salvaged
parts.

Sorry for my poor understanding but I couldn't understand these two steps:

1. Take your ~4.7 K resistor and solder tack it across R7 on the solder / foil side of the board.
(VIOLET lined and circled referencing markings.)
I will use a photo of the foil side of the circuit from an earlier post of yours, but , you can now see that it involves using TWO series arranged resistors, that you won't have to bend the leads on.

2. Lift the test resistor from across R7 and move it down to solder tack across R6 and shunt it.
In this case you will just unsolder the pair and move them on down to R6 position .
(BLUE lined and circled referencing markings.)
With my thinking being that this test voltage, will be the most informative.

Can you please provide me a diagram plan for easy understanding?
Use this referencing below, for teaching yourself resistor color coding . . . .in case its value is not printed upon it.

http://www.electronics-tutorials.ws/resistor/res_2.html


Le Illustration:

73's de Edd

 

[duke37]

There will be 5V across the 2.2k resistor so the current will be 2.27mA
2.27mA through the 3.3k resistor will be 7.5V so the total voltage will be12.5V
If the voltage is to be increased to 14V then the voltage across the 3.3k will be 9V and the current will be 9/3.3k = 2.79mA
There is therefore an excess of 0.59mA to be passed from the sense terminal to ground.
R = 5/0.59m = 8.5k
Try a 8.2k (standard value) tacked across the 2.2k resistor and check output voltage.

There is no need to take anything out of the board.

 

[Nauman Muhammad]

Sir Nauman Muhammad . . . . . . .


I am still in search of 4700 ohm resistor... Can higher one work?

I think that the use of TWO resistors in series might even make it even EASIER !
Just find two values that will add up to that approximate 4700 ohm value.
In my case I have shown 2200 and 3300, but its what you can find in YOUR salvaged
parts.

Sorry for my poor understanding but I couldn't understand these two steps:

1. Take your ~4.7 K resistor and solder tack it across R7 on the solder / foil side of the board.
(VIOLET lined and circled referencing markings.)
I will use a photo of the foil side of the circuit from an earlier post of yours, but , you can now see that it involves using TWO series arranged resistors, that you won't have to bend the leads on.

2. Lift the test resistor from across R7 and move it down to solder tack across R6 and shunt it.
In this case you will just unsolder the pair and move them on down to R6 position .
(BLUE lined and circled referencing markings.)
With my thinking being that this test voltage, will be the most informative.

Can you please provide me a diagram plan for easy understanding?
Use this referencing below, for teaching yourself resistor color coding . . . .in case its value is not printed upon it.

http://www.electronics-tutorials.ws/resistor/res_2.html


Le Illustration:

73's de Edd

Initial voltage without any alteration is 12.54V

I remove the R7 and as I could found the comination of total 6000 Ohm resistor, I installed it in place of R7. The supply then gave me 7.54V

Resistor I removed has 10,000 Ohm.

Now begore placing 6000 Ohm resistor at R6, do I need to put back the original resistor I removed from R7?

 

[Nauman Muhammad]

There will be 5V across the 2.2k resistor so the current will be 2.27mA
2.27mA through the 3.3k resistor will be 7.5V so the total voltage will be12.5V
If the voltage is to be increased to 14V then the voltage across the 3.3k will be 9V and the current will be 9/3.3k = 2.79mA
There is therefore an excess of 0.59mA to be passed from the sense terminal to ground.
R = 5/0.59m = 8.5k
Try a 8.2k (standard value) tacked across the 2.2k resistor and check output voltage.

There is no need to take anything out of the board.

I will also try this one Will 7.6 Ohm resistor work?