Modifying old laptop charger for ham radio use

[oldmannewstuff]

Hello everyone!

I need a power supply that can deliver 13.8Vdc (15% tolerance) at 3A for my radio.

I have an old (but working) HP laptop power brick, that supplies 18.5Vdc at 6.5A. The PSU is heavily shielded against rf emissions, so I think it might serve ok in it's future role.

The idea is to change the power supply to deliver 13.8Vdc by manipulating the feedback circuitry as described in https://hackaday.io/project/3469-modifying-a-notebook-power-supply. The process involves feeding the secondary side with lab power supply.

My PSU uses TSM103/A comparator. I think I've managed to find the resistors that make up the voltage divider, but would like your opinion if I have the right ones. I'm trying to avoid unnecessary desoldering of super tiny smd components 

If I understood correctly, TSM103/A pin 2 is the pin which I have to follow to find the desired voltage divider.

The divider I found consists of 49C (31,6kOhm) and two resistors 51D (332kOhm) and 556 (55MOhm) in parallel. Do these values sound reasonable? To me they seem very high, but that is how I interpreted the markings.

If the values are correct, that would make resistors 51D and 556 be seen as one 313.1kOhm.

What might be the idea behind using that 55MOhm resistor?
I gather it does not matter which part of the voltage divider I change?

The PSU output has three leads. V+, GND and ID.  V+ and ID are connected together through 1,5kOhm resistor. What purpose might that ID lead have? Can I just ignore it, or does it have a vital role in the operation of the PSU?


Thanks for your time 

[ealex]

the ID pin is used by the notebook to identify the charger as "it's own" - so you can leave it alone.

the ref voltage seems to be 2.5V: http://www.st.com/content/ccc/resource/technical/document/datasheet/98/e6/12/29/34/54/4f/d1/CD00001691.pdf/files/CD00001691.pdf/jcr:content/translations/en.CD00001691.pdf

your divider values seem off: 18.5V*31.6k/(313.1k+31.6k)= 1.6959V on the divider output.
it would need over 27V to match the reference voltage: 27V*31.6k/(313.1k+31.6k) = 2.47V

are you sure you're looking at the correct divider ? you can check by adding a large value in parallel with the 332k one - to shift it lower by 10% - and check if the output voltage drops by the same amount.

i'm also trying to use a notebook psu for my SDR, but I've observed that just by connecting the ground / the notebook ground to the SDR the noise floor jumps up enough to kill all reception => i've ended up using the SDR on a linear PSU

[oldmannewstuff]

Thanks for your quick reply! 

Those were the only resistors with continuity from IC pin 2.

I've fed the secondary side with about 19.3V measured at the optocoupler. Increasing the input voltage will drop the voltage at the optocoupler - as it is supposed to do.

Voltage divider output to pin 2 (when using 19.3V input) is 2.48V. So I must have read the markings on the resistors incorrectly?
Pin 3 input voltage is exactly 2.5V, and pin 1 measures 18.03V.

are you sure you're looking at the correct divider ? you can check by adding a large value in parallel with the 332k one - to shift it lower by 10% - and check if the output voltage drops by the same amount.

Do you mean the voltage divider output? Or voltage at IC pin 1?

[ealex]

by what you've measured, it looks like that's the feedback divider.

can you check the resistors with a meter ? just to confirm the decoded values.
measure resistance from pin 2 to GND and to power supply output - then check if the divider ratio fits what you need:  2.5V/18.5V ~= 0.135

[oldmannewstuff]

The resistor with marking "49C" measures 4.2kOhm. But the ohms start slowly creeping higher, if I keep on measuring. After about 30secs it was still sloooowly rising, so I don't know how trustworthy the measurement is.. 

Resistor pair "51D" and "556" measure 4,09kOhm.

Pin 2 to GND = 4.08kOhm
Pin 2 to + = 4.1kOhm

Not sure how to check the divider ratio 

But if the "49C" is actually 31.6kOhm and the resistor pair is 4.09kOhm, that would give me 2.21V output with 19.3V input. Not exactly 2.48V but close.

Do I need to drop the voltage coming from the voltage divider to IC pin 2 to decrease PSU output voltage, or how does this work?

Or do I first have to select the desired output voltage of 13.8V and then the resistor values for the voltage divider so that it produces 2.5V to pin 2? I've used the online calculator at https://www.allaboutcircuits.com/tools/voltage-divider-calculator/.


Highly confusing   

[ealex]

You should adjust the 2 resistors so the divider outputs 2.5V when the divider input is 13.8V

For example try adding a 10-47k resistor in parallel with "Resistor pair "51D" and "556""  - the output voltage should drop somewhat.

Another thing I forgot : some PSU's will not like being pushed to much out of their original design values. I've played with a 12V notebook psu : pushing the output to 15V was  ok, but trying to make it output 9V made it unstable => you need to test it before you power some expensive gear with it.

[oldmannewstuff]

Well, I put a 47kOhm resistor in parallel to the mystery "49C" resistor, and managed to get the voltage to 12.5V. That is well withing the 15% tolerance. 

Another thing I forgot : some PSU's will not like being pushed to much out of their original design values. I've played with a 12V notebook psu : pushing the output to 15V was  ok, but trying to make it output 9V made it unstable => you need to test it before you power some expensive gear with it.

What type of unstability do you mean? Did the output voltage spike, or had the psu trouble producing 9V?

I thought I'd load test my psu with a car light bulb. 55W bulb should draw 4.4A of 12.5V - If I'm not mistaken 

[oldmannewstuff]

Now I know what type of instability you meant.
I tweaked the psu a little bit more, and got to 13.8V, but on powering it from mains, the output voltage fluctuates between 12.23V - 13.79V.

Anyway, thanks for your help! Even if I didn't get to repurpose my laptop brick, I learned a lot 

[Kalvin]

If adjusting the feedback resistor value will make the power supply unstable, you could possibly buy a cheap dc-dc buck module to convert from 18V to 12V. Just add a fuse and a 12V or 13V TVS-diode (which will start conducting around 13.5 - 14 Volts) as simple overvoltage protection, if the cheap dc-dc converter should fail so your radio won't fry. Depending of the quality of the dc-dc converter you may or may not need to add some noise filtering. Just my 2c.

[oldmannewstuff]

Thanks for your 2c 

Thought about that too, but I think I'll try to modify an old pc psu next. The psu has plenty of oompfh, and I don't feel like buying a separate converter...