Hello Mike,

I would advise you to verify two minor problems with your design. You have to check the specifications for the chip you are using, have a look here:
http://www.powerint.com/sites/default/files/product-docs/pks_family_datasheet.pdf
One problem could be generated by the overly enthusiastic peak power you choose for your design. The specification calls for peak power up to three times of the maximum continuous output power. Your design choice for continuous output power is 3W, and the peak output power is 24.7W. That is more tan eight times higher.
The second test is for the externally supplied current for the bypass pin. As the specification states on page 2, in typical applications this pin must be externally supplied via a bias winding. And on page 10 you will find that all Peak-Switch designs must use a bias winding to feed operating current into BP pin once the supply is operational. What this means for you is that inside the chip is an internal regulator design mainly to provide the start up energy required for any application. Function of many parameters (power levels, efficiency, etc.) you can get away sometimes without extra current for the BP pin but there is no guarantee for this. A quick test you can run is to use a bench power supply connected with a resistor to the BP pin and to validate your original choice of not using an extra bypass winding.

Cheers,
PI_Crusher

Hi Crusher, The peak output power figure is misleading in that the loads applied to each output will not be on at the same time, therefore the max load would only be 15W at any one time. Rather that a bias winding, could I connect to the DC input to the transformer through a 1 meg resistor to satisify the bias requirement? If not, is there a quick fix I could use by switching to another part rather that a total redesign? Thanks, Mike

Welcome Back Mike,

Let’s start with this new stated 15W. It is still too big. Not the actual power, but the ratio between the 3W “Continuous Power” field, line 9 from the spreadsheet and 24.7W “Peak Power” field or the misleading 15w requirement, line 10. You must change this to stay under 3 times ratio (not 5 times as is the case with 15W peak), and be sure that you comply with the constraints of “Peak Power” definition.

It is very probable you can supply some extra current trough an external resistor. Use an extra DC bench supply with a resistor in order to provide additional current into BP. Once you identify the extra current requirement (if any), use a proper resistor to cover your AC input range. This is not giving you the best efficiency, but could do the trick you need without one extra winding.

Cheers,
PI_Crusher

Hi Crusher, Thanks for the suggestion, I'll give it a try and let you know the outcome. Thanks, Mike

Hello Crusher, I tried what you suggested and the input voltage to the Bias pin seems to be between 3.5VDC and 5.8VDC @ about 1ma. Regulation seems to be good across the full load range. At full load (15W) the transformer gets a little warm but I can't vary the duty cycle (10%) the way it will be used. Another problem has shown up. The secondary output that should be 12VDC is at 20VDC as though the secondaries are actually a stacked 6.5VDC + 12VDC. Can you enlighten me on this? Thanks for the help, Mike

Crusher, To correct the previous post, the secondary is at 20VDC at no load and drops to 8VDC @ a .5A load. What do you think? Thanks, Mike

Hello Mike,

The problems you have with cross regulation for your 12V and 5V are quite normal for any design with multiple outputs. I would suggest three techniques for you to try, just select the one working best for you.
1) use resistors between 12V and 5V to balance the outputs as you can see in this example:
http://www.powerint.com/sites/default/files/PDFFiles/der198.pdf
2) use double feedback for the opto-coupler, see this example:
http://www.powerint.com/sites/default/files/PDFFiles/der55.pdf
3) use zenner + diode/resistor from the high voltage output lo low voltage one like here:
http://www.powerint.com/sites/default/files/PDFFiles/der24.pdf

You can not just copy/past what you see in these examples. You have to understand the concept and select one or all solutions for your design with proper values for your outputs. I believe you will get it right.

Cheers,
PI_Crusher

Crusher, Thanks for the input, I'll take a look and get back to you.... Thanks, Mike

Mike,

Check this document page 21 and see how R10 and VR3 is used here, this can help you also.
http://www.powerint.com/sites/default/files/product-docs/tophx_family_datasheet.pdf
Cheers,
PI_Crusher

Hi Crusher, I have modified the schematic and would like for you to give it the once over for glaring errors before I do the prototype. Any help would be grestly appreciated. Thanks, Mike

Hello Mike,

Your schematic now looks good, it is pretty much what I would do to improve cross regulation. I have no data from you, I can only assume you tested this on your breadboard and you're happy with the results. I do not expect significant changes once you move your schematic to a clean PCB.

Cheers,
PI_Crusher

Hey Crusher, Thanks for the input and I'll keep you posted as to the results. Thanks, Mike

Crusher, Would you take a look at the board layout for anything glaring that I need to fix? Thanks, Mike

Regarding PCB implementation.
For the secondary side, I would increase the thickness of any trace with higher currents, especially for the positive rails.
For the primary side you have to be very carefull with proper spacing between high voltage nets. The reason our chip has a missing pin is to provide this minimum space. You have to make sure you comply with regulatory demands (6mm space?) for anything connected directelly to AC before the fuse, and at least 2mm space between any high voltage nets after the fuse.
Also, the current loop formed by C8+C9, T1 primary, and U1 must be minimized as much as possible.
Cheers,
PI_Crusher

Thanks Crusher, I'll look everything over again. I have the 6mm isolation from the AC side to the DC side. I'll check the positive rails to insure they are at least .060". Mike