解决方案搜寻 技术支持

TNY280 Problems with the calculations.

Posted by: Raul Diez on

Hi

I have tried to design a Flyback converter with a TinySwitch-III and the PI Online Suite
The basic parameters are:

Input: 85~265VAC 60Hz
Output: 12VDC, 2A (24W Simple output)

and I had some troubles with my calculations:

1) In the design results the maximun duty cycle (Dmax) is 0.64.
When I tried to calculate this parameter with the formula of application note AN-23 (Page 8, Step 8) the result for the Dmax is about 1.08.
I think this is impossible?

Anyway, I selected a Dmax=0.64, like the design tool results, and I went further with the calculations.

2) When I tried to calculate the primary inductance (Lp) with the formula of the AN-23 (Page 8, Step 17A) the result is
about 741uH, too far from the 1028uH of the design tool.

My question is:
1) What is wrong with my calculations?

I atttached the design (.uds file) and the apllication note scans.
Thanks

PD. Excuse me for my ugly english
Raul Diez.

评论

Submitted by PI-Ayel on 03/09/2021

Hi Raul Diez,

Thank you for reaching out, upon checking your inquiries here are my findings:

1) Yes, you are correct, a Dmax > 1.0 is not possible. The Dmax formula in Page 7 Step 8 of AN23 is used to get the maximum value of DMAX during DCM operation. Getting a value Dmax > 1 suggests that your design should be CCM. Since that's the case, the DMAX formula to be used is the one in Page 8 Step 14.

2) Since it is CCM, solve for the KRP first in Step 15. After that, the primary inductance to be used should be the one in Page 8 Step 17B instead of 17A.

Please try my suggestions and let me know if you were able to get values close to the ones provided in the PI Online Suite.

Submitted by Raul Diez on 03/10/2021

Hi PI-Ayel

Thank you very much for your answer.

I followed your suggestions and tried the calculations again.

The result for the primary inductance Lp now is about 901mH.

Still is far from the design nominal primary inductance(1028mH) but
is very close to the minimun inductance of the design tool (904mH).

My question is:

1) In the AN23, step17 say "Calculate primary inductance Lp" but
this is a formula for the nominal or the minimal inductance?

2) If step17 is a formula for the nominal inductance, what is wrong
with my calculations?

Thanks

I atttached the apllication note scans and a design capture.

PD. Excuse me for my ugly english
Raul Diez.

Submitted by PI-Ayel on 03/10/2021

Hi Raul Diez,

I looked at your calculations and checked the valued. Here are my findings:

1) This PI XLS/PI Expert design tools calculate components using the extreme limits. As such, I suggest you to use IP=0.791 as the peak current instead of 90%ILIMIT = 0.712. This is something that has to be addressed in the app notes in the future.

2) As a result, the KRP calculation will give a different KRP now. KRP must be less than 1. Obtaining a value of KRP > 1 means that there was a mistake in the calculations.

3) You are right, the calculated inductance is the MINIMUM Lp. Disregard the factor of 1/0.9, and again use IP = 0.791 upon calculating the inductance.

See if these steps work well and let me know your feedback. Thanks!

Submitted by Raul Diez on 03/11/2021

Thanks again for your help.

I re-calculate the Krp follow your advice. Maybe my first calculation was wrong...
Anyway, the result now is about 0.44 (Kpr < 1). Follow the step 16, I set the Kpr for 0.6 and
re-calculate the Dmax with the equation on step 15 and the result is 0.68.

Next I re-calculate the Vor back in the step14 but, since is over or 150V, can not go further with the design, and I have to go back to setp 7 and select another tinyswitch?.

But, since in the PI expert the TNY280NP is good for this design and a Dmax of 0.64, somethings going wrong in my calculations... but what?

I atttached the apllication note scan.
Raul Diez.

Attachment 大小
PI-1600x1200@2x_SCALE-2-1SP0340.png 460.53 KB
Submitted by PI-Ayel on 03/11/2021

Hi Raul Diez,

I see. Step 6 is a necessary step when the design is too CCM. The KRP ~ 0.44 is correct, but practically speaking, since it is too CCM this can cause some SOA issues during start-up, and that is why the calculations is "recommending" to use a larger TNY size. However, for the interest of calculation, step 6 can be skipped and you can proceed to Step 7 immediately.

A few more things to note in doing step 7:
- Disregard 1/0.9 in the denominator since we are using IP = ILIMITMIN = 0.791.
- Use fs = (I2f min) / (ILIMITMIN^2). The I2fmin value can be found in the data sheet.
- The resulting value is the LPmin. This may not be exact, but should be close to the value suggested in the design tool.

See if this helps and let me know your findings.

Submitted by Raul Diez on 03/11/2021

PI-Ayel

Thanks again for your help.

- I calculate I2fmin from the datasheet and fs from your formula.
- With this results I re-calculate the Lp min (without the 1/0.9)
and the result is about 928uH. Still is far from the 904 of the
design tool but I think is getting closer.

Something wrong? or maybe can I go further with the calculations
with this result? What do you think about?

I atttached the aplication note scan and the datasheet calculations.
Raul Diez.

Submitted by PI-Ayel on 03/11/2021

Hi Raul Diez,

Yes, your calculation is now correct. I suggest that you use more decimal places in your Dmax=0.626 instead of rounding it, for better precision.

Submitted by Raul Diez on 03/11/2021

PI-Ayel

Thanks again for your help.

I re-calculate with Dmax=0.626 and the minimun inductance now is 906,2uH, pretty close to the design tool (904,7uH).

I go further on the calculations with this result.

Thanks very much for your time.
Raul Diez.

Attachment 大小
PI-1600x1200@2x_SCALE-iDriver-SID11xxKQ.png 439.8 KB
Submitted by Raul Diez on 03/18/2021

Hello again.

I had troubles again with the calculations.

In the gap length calculation I used the Ae from the PC40EE25 datasheets (40 mm^2) and the Al form the PI expert (101 nH/T^2) and I had a imposible result.

What is wrong with my calculations?

I atttached the aplication note scan.
Raul Diez.

Attachment 大小
PI-1600x1200@2x_LinkZero-LP SO-8C.png 442.6 KB
Submitted by Raul Diez on 03/28/2021

Hi.

Something about the calculations?

Raul Diez

Submitted by PI-Ayel on 03/29/2021

Hi Raul,

The ungapped core effective inductance (AL) of EE25 is 2140 nH/turns^2. You may see it in the side panel -> Transformer -> Core Selection. Perhaps what you used in your equation was the gapped core effective inductance (ALG) that is why it gave an impossible result.

Please see if this works.

Submitted by Raul Diez on 03/30/2021

PI-Ayel
Thanks again for your help.
I re-calculate and the gap length is now 0.477, pretty close to the 0.485 design.

I go further and in the primary RMS current Irms the result is 0.58A (0.51A by the design.
Something wrong? or maybe can I go further with the calculations
with this result? What do you think about?

I atttached the datasheet calculations.
Raul Diez.

Attachment 大小
battery.jpg 7.78 KB
Submitted by PI-Ayel on 03/30/2021

Hi Raul,
Can you try increasing the significant figures in your calculated DMAX and KRP?
Also, if it helps, the PI Expert tool is taking into account the VDS of the TinySwitch MOSFET, while the AN-23 is assuming that VDS = 0 that's why we may not get the exact values in our calculations.

Submitted by Raul Diez on 03/30/2021

Hi PI-Ayel
Thanks for your answer.
In the PI expert design the Vds is 4.42A but, how can I insert this parameter in the step 19 formula?.
Raul Diez.

Submitted by PI-Ayel on 03/31/2021

Hi Raul,
The VDS will affect the DMAX calculation.
Dmax = VOR/(VOR+VMIN+VDS). Recalculate this and the other affected variables as well and see if this helps