PMP8740 shim inductor calculation

Aneesh, hello and thanks for your continued interest in our reference designs.  Your follow on question has been forwarded to Roberto for response.  Please allow 1-2 business days for a reply. 

Best Regards,

J. Fullilove

Reference Design Operations

Texas Instruments

John,

Thank you for forwarding this request to me.

Aneesh,

The minimum load is actually the output load current at which the converter enters hard switching; this should be kept as small as possible. There is, of course, trade off regarding it:

The higher the value of the shim inductor, the flatter is the efficiency curve at light load, but you loose some % of duty cycle at full load, therefore the efficiency at full load drops a bit.

The lower the shim inductor, the higher is the peak efficiency and full load, but light load will suffer, and also EMI will be worse (consider that when you loose ZVS, you get current spikes during switching).

For this reason, and also in your case, I suggest to take as minimum load for ZVS 15% of nominal, therefore you multiply the denominator by 0.15.

With your values, I calculated the shim inductance = 31.18uH.

By the way, the leakage inductance of the transformer should be subtracted from this value.

Best regards,

Roberto

Dear Robert,

Thank you for the explanation. I understood the concept. I have the following doubts too

1. Iout(min)=0.15 if we want the ZVS action to happen at 15% of output Load right?.

2. The  following values 

Coss=86pF (IPP65R190CFD mosfet), Vin(fb,nom)=400V, IPP= 7.526A, ΔIlout= 12A ,Nps=9.5

I have taken from your 2KW design and i am getting 31.18uH but how you are getting 11.3uH?

Can you just correct me where i did mistake in your calculation?

Regards

Aneesh

Dear Anjana,

I believe there is a mistake in my formula.

The correct formula should have the LOAD(min) multiplied for Ipp inside the brackets and not outside.

Therefore the denominator should be:

[( Ipp * Load(min) - Delta(Ilout) / Nps )]^2.

Please consider also that Delta(Ilout) is the peak output ripple current and not the peak-peak ripple, therefore here is 6A (we considered 20% peak-peak ripple, which is 10% peak ripple, when Iout nominal is 60A).

For this reason, if I consider that Load(min) = 15% (0.15) and we calculate the inductor, it will be too big.

If you reduce the level, where the converter enters ZVS, to 20%, then by considering:

Vin = 400V, Coss = 86 pF, Ipp = 7.526A, Delta(Ilout) = 6A (peak value and not peak-peak), Nps = 9.5, we get:

Ls = 36.06 uH, which is a reasonable value.

As said previously, you have then to subtract the leakage inductance of your transformer.

Sorry for the mistake.

Best regards,

Roberto

Dear Robert,

Thank you for the reply.

According to you the correct formula for the shim inductor is

Ls= 2Coss*((Vin)^2)/[( Ipp * Load(min) - Delta(Ilout) / Nps )]^2 

In the above formula  LOAD(min) multiplied for Ipp inside the brackets. Which results 

Ls= 2*(86E-12)*400*400/[(7.526*0.15)-(6/9.5)]^2

Which is equal to 111.26 uH for 15% load not 11.3 uH 

Kindly clarify 

Regards

Aneesh

Dear Aneesh,

According to my last email, I clarified that by correcting my formula and using 15% minimum load, the inductance value would be too high (in your case you calculated 111uH). If you replace the factor 0.15 with 0.2 (which is the minimum load for ZVS), then you get 36uH.

Please let me know if that answers your question.

Best regards,

Roberto