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Error in calculating power losses from direct current of the secondary winding of the LLC converter

Posted by: svarog46 on

Row 126   -Idc=1.5a

Row 133   -Rdc(100)=29.33m-ohm

Row 134   -Ploss both secondary =0.53W

BUT!!!  Ploss both secondary=1.5^2*0.02933*2=0.132W

WHERE IS THE ERROR ?

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Submitted by PI-TOMMY BRADFORD on 04/15/2025

Hello svarog46,

Good morning and thank you for choosing Power Integrations.

Let me check this one with the relevant team internally. I will give an update as soon as I receive feedback from them.

Thank you and have a great day ahead.

Regards,

Tommy

Submitted by svarog46 on 04/17/2025

Thank you. I'm waiting for your reply.

Submitted by PI-TOMMY BRADFORD on 04/21/2025

Hello svarog46,

Good morning.

The formula for the estimated power loss due to DC resistance (DCR_Ploss_Sec1) is as follows:

DCR_Ploss _Sec1 =  (IO1)^2 * DCR_100C_Sec1 * 2

where IO1  (row 14) is the maximum main output current and DCR_100C_Sec1 (row 133) is the estimated resistance per phase at 100 C.

 

DCR_Ploss _Sec1 = (3)^2 * 29.33 mΩ * 2 = 0.52794 W = 0.53 W

 

I hope that this clarifies your inquiry. Let me know if there's anything that I can help you with.

Regards,

Tommy

Submitted by svarog46 on 04/22/2025

Hello Tommy

Thank you for your attention.

A pulsating current flowsthrough eachsecondary half-winding, which contains a constant and an alternating component

If for the alternating component the power is determined by

row 135 is multiplied by row 127 squared and multiplied by two- and the result is clear !

Then why is a different formula givenfordirectcurrent –where is the logic?

Submitted by PI-TOMMY BRADFORD on 04/23/2025

Hi svarog46,

Good day.

Just a quick note: the actual DC current that will flow continuously through each winding at full load is the maximum main output current listed in row 14, which is 3 A. Therefore, this is the parameter used to calculate the estimated power loss due to DC resistance (DCR_Ploss_Sec1).

I hope this clarifies the confusion. Thank you and have a great day ahead.

Regards,

Tommy

Submitted by svarog46 on 04/23/2025

Hello Tommy

I do not understand this calculation, then why write in row 126 that the constant component of the secondary half-winding is 1.5 amperes
if it is not applied, but the load current is applied I looked at several articles there calculating the power as I wrote

Submitted by PI-TOMMY BRADFORD on 04/24/2025

Hello svarog46,

Good morning.

Let me check this one internally and will let you know as soon as there's an update.

Thank you for your understanding.

Regards,

Tommy