Maximum output power for universal input

Posted by: ysba on 12/08/2011

On the LinkSwitch LNK409 design, why the Continuous Output Power must be below 18W for an universal voltage input? Image and spreadsheet are attached. The image shows the spreadsheet of my design, and the device datasheet. According to datasheet, the maximum power is 50W, over a voltage range from 85 to 300V. Thanks.

Files

FNT01 P40_AP1 100-240V.pxlph (37.87 KB)

Comments

Hello,

thanks for your post. I agree this is incorrect. I spoke to software team leader and confirmed that this is a bug. The spreadsheet is pulling the minimum power numbers not the maximum power numbers from the datasheet power table. This doesn't impact the design results - it is just a display issue. It will be corrected in an update next month.

One thing to note. The datasheet power table assumes Flyback topology. The buck-boost topology (true when you use the flyback spreadsheet with a VOR value equal to the output voltage (1:1 turns ratio) requires higher peak currents for a given output voltage and power (no turns ratio transformation as with the Flyback). This will result in lower power capability as a function of the the output voltage (lower voltage = lower power).

Cheers PI-Chekov

So, I can use the Flyback spreadsheet for a Buck-Boost topology, with a 1:1 turns ratio. But, about the NS parameter (number of secundary turns), how should I treat it?

Correct.

With the turns ratio set 1:1 (VOR=VOUT+VD, where VD is the VF of the freewheeling/ouptut diode) the number of secondary turns entered also set the number of primary turns in the spreadsheet ie NP=NS.

Use the warnings in the spreadsheet to guide you, set NS to give you acceptable values for peak flux density (BP) and operating flux density (BM).

If you post a design file I'd be happy to review it for you.

Cheers,

Chekov

I already did an prototype, and everything seems normal. The main difference from the spreadsheet is the inductor. I have only PQ3220 cores with 0.85mm air gap. The spreadsheet recommends 67 turns, for a 0.43mm air gap. To deal with this, I used the relation N=sqtr(L/AL), so I applied my core's AL and obtained a higher number of turns. I used a 24 AWG copper wire, 105 turns on 7 layers. The resulting inductance is about 2400uH.
My design file is attached. Can you review it? Thanks.

The attached is different from what you are describing. For example it uses EI35 and the primary inductance is 1958 uH.
In general you should not worry too much abut the air gap. Try to get the correct inductance. as you say the AL value has a bigger impact on gap size. Just make sure that its not too small (smaller than 0.1 mm)