Change output voltage of a laptop power supply brick.

[rnorman3]

Hi all, I have a fairly new modern Dell power brick which puts out 19V and 7A. I want to drop the output to about 15V. Do you think I would be able to do that somehow? I figure I can cut it open. I can then post pictures if you think it would help. I'm sure I'd have to somehow change the reference voltage. It would be great if it was adjustable but somehow I doubt it.
Thanks,
Richard

 

[KeepItSimpleStupid]

See if you can locate switching IC and then look at the datasheet or application note. Because the supplies are isolated, it's harder to find the feedback point. You may have to "reverse engineer" the schematic.

 

[tronitech]

It may even have a pot you can adjust if it's used on other products?

 

[ronsimpson]

It is typical that there is a opto-isolator and a TL431 that is the error amplifier. If you could attach a picture of the PCB it will help.

 

[rnorman3]

Thanks guys, I've attached pictures.

OK, I tried to get a good picture for you. I put some notes on the smaller picture. I believe the isolator is probably the 8 pin chip on the left. I can't get a picture of the other side of the board as yet, if you need it let me know. If we find the opto-isolator, where then is the voltage reference and voltage divider for the feedback I guess is the question. I'm hoping to just be able to pop off one of these resistors and change the value. let me know what you can see. Also the numbers on the chips are almost impossible to pick out, seems they're covered with something. Any ideas for getting that coating off? Perhaps some acetone?

Thanks,
Richard

 

[ronsimpson]

The opto isolator is across the primary to secondary isolation. There is a 0.3 to 0.4" gap between the parts on each side of the board. I think the isolator is the 6 pin part at the bottom of the board. Through hole on the other side of the board.

There are too may parts to easy reverse engineer the power supply. There is probably a voltage divider to get the output voltage down to 2.5 volts. Then a 8 pin IC which might be a TL431 or a op-amp then off to the isolator.

 

[rnorman3]

OK, got more info:

Hi Ron, check out the attached picture "half board with markup". I found the optocouplers where you said, there are two. I believe the chip (which I can't identify) is special purpose, it contains the 2.5vref, as well as the comparator and possibly some other functions. I highlighted in blue what I think is the voltage divider resistors. I'm going to try and put a fairly large ~100K resistor across one of these and see if the output changes and is still regulated. BTW this is a Dell Mode PA-1131-02D2, which I like to put in the post here because then it'll show up on google if someone else ever looks. Hopefully we can nail this down!
Richard

 

[rnorman3]

Found the parallel network!

Picture attached.
The network consists of:
+V through a:
50C = 32400
Then though two resistors in parallel:
94B = 9310
95B = 9530
Then Ground

So the voltage in the center is:

1/(1/9310 + 1/9530) = 4709

4709/(4709+32400) = 4709/37109 = 0.1269

19.6 x 0.1269 = 2.487V

So if I put a 100Kohm resistor in parallel with the 50C(32400ohm) I get a factor of:

1/(1/100000+1/32400)=24471

4709/(4709+24471) = 0.1614

2.487/0.1614=15.4V which is exactly what I get when I measure it! And the regulation seems to be working fine. This should work great. It'll be difficult to get the 100kOHM resistor in place but I'm off to try that now. I might try to order a 0 to 100k pot that I can put in series with a 50k resistor and in parallel with the 50C SMT resistor. This would allow me to adjust the voltage for other projects. It's a good brick so it should provide lots of power!
Thanks all!! I'll let you know if anything goes awry.
Richard
****************************************************
OK, I'm updating this same post...final modified picture of Dell power Supply attached. Works great, output voltage is 15.36 on this cheapie meter.
Thanks again!

 

[tmad]

Picture attached.
The network consists of:
+V through a:
50C = 32400
Then though two resistors in parallel:
94B = 9310
95B = 9530
Then Ground

So the voltage in the center is:

1/(1/9310 + 1/9530) = 4709

4709/(4709+32400) = 4709/37109 = 0.1269

19.6 x 0.1269 = 2.487V

So if I put a 100Kohm resistor in parallel with the 50C(32400ohm) I get a factor of:

1/(1/100000+1/32400)=24471

4709/(4709+24471) = 0.1614

2.487/0.1614=15.4V which is exactly what I get when I measure it! And the regulation seems to be working fine. This should work great. It'll be difficult to get the 100kOHM resistor in place but I'm off to try that now. I might try to order a 0 to 100k pot that I can put in series with a 50k resistor and in parallel with the 50C SMT resistor. This would allow me to adjust the voltage for other projects. It's a good brick so it should provide lots of power!
Thanks all!! I'll let you know if anything goes awry.
Richard
****************************************************
OK, I'm updating this same post...final modified picture of Dell power Supply attached. Works great, output voltage is 15.36 on this cheapie meter.
Thanks again!

Hi all.
I am having the same case here, but my charger is 18.5V and I want to drop the output to 15V.
I cant see the attached photos, could you please reattach them if you still have them.

Thanx.

 

[rnorman3]

Sorry man, I lost the pictures too...if you try to find the opto couplers and post pictures someone might help you identify them...