Design 5V-1A AC-DC converter with LNK626DG

Posted by: Roni Daffan on 09/25/2016

Hello,

My name is Roni.

I which to design a very samll size AC-DC converter, for 5V 1A, in a close encapsulation (Not much heat dissipation).

For some reason, both the LNK626DG and the transformer gets very host for current consumption of 600mA.

For a current consumption above 600mA, the output is turned off for ~1sec, and then on for 250ms, and re-cycle.

The LNK626DG should support up to 13W of output power, but I'm not able to get even 5W.

Another funny thing. When I tested the recommended values of 6.3K and 3.9K for the resistor divider (R352,R353 in the drawing) , the output voltage was 9V. So I changed R352 to 1.8K to get 5V output.

BTW, the output zener diode Z350 of 5.6Vis not installed, so there is no problem with that.

See the attached circuit and the transformer I was using - 2 page document

Could you please point an error in the design or in the selection of the transformer ?

What do you suggest to change ?

Thank you so much in advance !!!

Roni

Files

AC-DC converter circuit with LNK626 (258.69 KB)

Comments

Hello Roni,

Thanks a lot for considering using Power Integrations product.

For quick debug and fault trouble shooting, we suggest you use our PIexpert tool. you can find this online, below is the link.

What you need to do is input your requirement and then you can check your design automatically. it is very convenient. below is the link.

https://piexpertonline.power.com/site/login

When I run your design in the PIexpert tool. I notice your reflected voltage is too high which means your transformer turns ratio is too big.

this will cause your secondary diode condution time to be short which will afftect the FB pin smapling.

you will find a warning in the diode condution time section in the PIexpert, DCON. This is the calculated diode conduction time of the output diode. This value should always be greater than 3.1 ms, the point at which the voltage on the FEEDBACK pin of LinkSwitch-CV is sampled. for your design, it is only 1.96us, which is too low. You should decrease your turn ratio to at least 2.4.

Please try with a low turn ratio transformer to see if it helps or not. you can also try the one PIexpert suggests.

Best Regards

Thank you very much for your reply.

However, I want to understant, you wrote that the reflected voltage is too high becasue the transformer turning ratio is too high. Should not it be the opposite ? I mean, increasing the transformer ratio will decrease the reflected voltage (V1/V2 = N2/N1). Should not I increase the transformer ratio in order to decrease the reflected voltage ?

Thank you again,

Roni

Hello Roni,

the equation you write should be Vprimary/Vsecondary=Nprimary/Nsecondary. the the reflect voltage will be low if you decrease the turn ratio as your output voltage keeps the same.

Best Regards