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Output Voltage Temperature Stability

Posted by: HDP on

Hello,

we have a design with the LNK564.
Output 2V, 0,9A.
The output voltage changes from
1,9V @ -30°C to 2,4V @ 20°C up to 2,8V @ 80C.
Which is a -20% to +16% change in output voltage.
Far too much.
Ist this typical?
Any hints, what mistakes in layout may have such result?

Thanks,
Kai Tönnies

Comments

Submitted by VCastrellon on 12/08/2008
Is it possible you post your schematic for me to take a look to your design? About the measurement you are posting, are they at no load or full load? Regarding operation at 80C, I do not see much room for the device to opera without going into thermal shutdown.

Here are schematic and layout. I'm quite sure that it is a problem of layout.
The schematic built on breadboard runs fine.
The schematic built on the printed board has the problem.
The measurements are made with 1/10 of full load.
80C ist not the problem, we will have ambient temperature no more than 30C.
The main problem ist the output voltage linear decrease from +20C to -20C.

Thanks a lot,
Kai Tönnies

[PI-Chekov: Files removed at request of forum member]

Submitted by HDP on 12/17/2008
So, I have posted my layout. Is there someone, who knows, has an idea, why this layout is temperature critical?
Submitted by matulla78 on 01/07/2009

In reply to by prasun

Hello, you should try to use a optocoupler feedback circuit. Take a look at AN-40 on page 3. There you can see the behavior of the two feedback circuiits.
Submitted by PI-Chekov on 01/08/2009

Hello Kai,

I suspect the issue is the low output voltage which increases the effect of parasitic elements.

As the temperature changes anything that causes a change in the voltage across C4 vs the output voltage will cause an error. Possibilities are the Vf of the output diode vs D2, the value of C4 (ceramics can have a big tc) or the reverse recovery of D2.

As an experiment connect the resistor devider (R2,4,5) directly to the output. This will bypass the isolation barrier so be sure that you use an AC isolation transformer to power the supply before measuring anything. You will need to adjust the feedback resistor values to match the 2 V output voltage. You should find the voltage doesn't change.

Try replacing D2 with a 1N4005 diode - I wonder if the 1SR154 has a faster recovery characteristic.

Cheers

PI-Chekov