UCC28704: Using the calculation sheet with multiple output transformer

LazyHD

Part Number: UCC28704
Other Parts Discussed in Thread: TINA-TI

Hello everyone,

I am trying to design an AC/DC isolated PSU. I want to use an off-the-shelf transformer, I found 750811613 from Wurth that might meet my requirements.

My requirements are to have a universal AC input and 2 outputs 1.6A@5V and 0.6A@12V.

I have few questions :

1- If I use the UCC28704 calculation sheet provided by TI (sluc603b) and make the calculations for one output the 12V for example. For the current consumption I will use the sum of both. In the fields for the secondary winding I will use the ones for the 12V winding, in the primary and auxiliary fields I will use the one provided by the wurth datasheet exactly. The 5V winding should follow the 12V winding on its own ? Will this calculations work ?

2- Is it possible to simulate the above circuit ? Wurth provides an encrypted LTSpice model while TI provides an encrypted PSpice model. Can these two models be made to play together ?
If not can I simulate the Wurth transformer in TINA-TI or Workbench or any other tool ?

3- In general is this transformer suitable for this application ? I read few application notes that "The secondary voltage should be higher than the target voltage, otherwise the target output voltage will be unrealizable. ". When Wurth datasheet states 12v does it mean I can have a target voltage of 12V or my target voltage should be less ?

4- I found other calculators/tools/application notes that require the transformer core and the number of turns. Wurth does not provide this information. TI sheet does not require these entries, only the ratio of turns and inductance. Does this impose any limitations on TI sheet ?

Thanks for any help, sorry if my questions sound idiotic :D .

References:

1- Wurth transformer datasheet

2- TI UCC28704 Calculation sheet:

over 7 years ago

0 Bernard Keogh over 7 years ago

TI__Genius 9130 points

Here is an initial response to your questions.

1) For multiple outputs, you need to add up the total power in W of all the outputs, and then design initially for a single output at this full power level. Then you can choose the other outputs as integer turn-ratios scaled to the main output. E.g. in your case, the total power is (5 * 1.6) + (12 * 0.6) = 15.2 W. If the transformer is implemented with say 12T for the 12-V output, then the 5-V winding would be 6T (to account for rectifying diode Vf drops)

Note that each output cannot be easily individually current or power limited - the primary will only limit the total power, so the full 15 W could be drawn from either 5-V or 12-V rail, if the other is unloaded or lightly loaded. You would need to add a post-regulator to the secondary outputs to implement individual current-limiting.

Note also that since the UCC28704 is a PSR controller, the primary regulator will try to regulate the more heavily-loaded secondary winding, it's not so easy to force preferential good regulation on one output (but is possible in some circumstances). If the loads can exhibit extreme cross-reg conditions (i.e. one output at heavy load with the other at no or very light load), the lightly-loaded output will tend to increase due to the leakage inductance.

2) TI provides a PSpice model for UCC28704 that can be used with a PSpice simulator. TI does not and cannot support LTSpice. For support with the Wurth transformer model, you need to contact Wurth.

3) I will review your Excel design calculator and comment further on the suitability of the chosen Wurth transformer. The design calculator will tell you the required Rcs value (and resulting max Ipeak), magnetising inductance, and turns ratio (primary to main sec and to aux bias). You then need to choose (or request from the transformer vendor) a suitable transformer with this inductance and turns ratio. The other secondary turns ratios will need to scale from the main secondary to get the right voltages. If the Wurth spec sates that the secondary windings are rated for 12 V and 5 V, with a ratio of 2.252 between the secondaries, this should give close to 12 V & 5 V outputs, but it depends on the rectifying diode Vf values.

4) The absolute turns depends on the ferrite core area Ae, and the allowed Bpk value at Ipk - the Bpk is a trade-off between core and copper loss, and is ultimately limited by the saturation capability of the chosen ferrite material. Wurth might be able to give you more details about the actual turns and/or Bpk if you ask them. The TI design calculators do not always include the ability to enter the core Ae and choose the actual number of turns for each winding.

I will reply later with more comments on the design calculator values that you have used.

Thanks,
Bernard

0 LazyHD over 7 years ago in reply to Bernard Keogh

Expert 1060 points

Thank you for this detailed reply. It really cleared up a lot of things for me.
Looking forward for your next one. Thanks again.

0 LazyHD over 7 years ago in reply to Bernard Keogh

Expert 1060 points

I have attached the sheet based on my circuit inputs. Any thoughts or feedback are highly appreciated.
In order to keep a safe margin for the output loads, I made the calculations based on 0.8A@12V and 1.5A@5V (17.1W)

For reference, these are the PN I am using:

DA = df08s1
DG = SBRT15U50SP5
DE = BAS20HT1G
CA = CB = ESG226M400AM5AA
LA = SRN1060-471M
CDD = GRM155R61H474KE11D
RCS = RC0402FR-071R3L
COUT = EEV337M035A9PAA
QA = STD11N60DM2
DZ = 1SMB5947BT3G
RS1 = RT1206FRE0766K5L
RS2 = RC1206FR-0716KL
RZ = RC1206FR-074K7L
RNTC = NCU15WF104F60RC
RLC = RC0402FR-07909RL

Thanks in advance

sluc603b.xlsx

0 Bernard Keogh over 7 years ago in reply to LazyHD

TI__Genius 9130 points

The revised spread-sheet is maybe allowing too much design margin for output power. You specified 1.5 A @ 12 V, and the spreadsheet adds 5% margin to the Iocc limit, Pomax = 12 V * 1.575 A = 18.9 W. This could be reduced to get better efficiency. But for now, it should be ok as a starting point.

The spreadsheet recommends max Np/Ns ratio of 5.47 (cell C42). However, the Wurth transformer 750811613 has a turns ratio of 11.11, this is far too high. That will cause the design to hit Dmax and power limit at low line when the bulk cap voltage is < ~160 V for Dmax of 46.5 % (cell C37).

The spreadsheet assumes min bulk cap voltage is 60% of Vinminpk, i.e. 0.6 * 85 * sqrt(2) = 72 V. The transformer turns ratio is chosen to set the reflected voltage at the right level to hit Dmax at this minimum bulk cap level.

I would recommend using the recommended Nps of approx. 5.5. When I sue this value in the spreadsheet, the required mag inductance is approx. 500 uH. So again, this makes the chosen Wurth transformer unsuitable.

I would suggest contacting Wurth to see if they already have a transformer to match these specs, or if they can do a suitable version for your application.

If this answers your questions, please click the "verify answer" button.

Thanks,
Bernard

0 LazyHD over 7 years ago in reply to Bernard Keogh

Expert 1060 points

Will do, very much appreciate it

Power management

Power management forum

UCC28704: Using the calculation sheet with multiple output transformer