7/14/2021 9:31 AM  
Joined: 10/5/2020 Last visit: 9/11/2021 Posts: 3 Rating: (0) 
Hello, I need help with understanding of datasheets for 3phase SITOP power supplies, for example: 6EP14373BA00 (24 V / 40 A) Input current / at rated value of input voltage 400 V: 2.2 A supplied active power typical 960 W Efficiency at Vout rated, Iout rated, approx. 90 % So it gives us a total of 1067 W of energy at input. No matter how I try, calculated power never equals this from datasheet, unless assumption of power factor ~= 0,75, but it's a bit too low I believe. Someone knows the answer :) ? 
7/17/2021 7:36 AM  
Joined: 5/28/2008 Last visit: 1/19/2022 Posts: 4360 Rating: (814) 
Hello, Output power = 960 W = 40 A * 24 V And as per your explanation and available data in Siemens datasheets, you were expecting that input rated current to be less than 2.2A. Please note that there are some other factors that should affect the input current like harmonics. Please note that we can not consider the threephase power supply as a pure sinewave load. And to exactly know the value of difference, you should get the internal circuit of the power supply which Siemens or any other manufacture will never provide. Hope this helps. 
Ayman Elotaify 

7/19/2021 12:05 PM  
Joined: 4/28/2015 Last visit: 1/24/2022 Posts: 13 Rating: (1) 
Hello, you have primary side > Szu = U * I * root (3) = 400V * 1.73 * 2.2A = 1524VA if you know the cos phi you can also calculate > Pzu = U * I * root (3) * cos phi. But you can calculate the cos phi: Secondary side actual DC voltage > Sab = Pab = (cos phi = 1) So > Pab = 24V * 40A = 960W eta = efficiency = 0,9 (look data sheet) eta = Pab / Pzu Pzu = 960W / 0.9 = 1066W Cos phi = Pzu / Szu = 1066W / 1524VA = 0.699 Now all the data are known and are also in accordance with the data sheet. And the 6EP14373BA00 consumes an apparent current of 2.2A per phase. 1524VA are taken from the power grid. There are 960W DC available at the output. Greetings Georg 
7/19/2021 1:42 PM  
Joined: 10/5/2020 Last visit: 9/11/2021 Posts: 3 Rating: (0) 
Thank you for your answer. I know, no manufacturer will provide exact schemas of their product, and I don't expect perfect match of my calculations with data provided in datasheet, but I was just wondering what's behind these differences.
Thank's for the answer too. I agree, this way calculations match provided data, but isn't the cos = 0,7 way too low? I think that explanation with harmonics makes more sense for me. Thank's for help! 
7/19/2021 2:29 PM  
Joined: 4/28/2015 Last visit: 1/24/2022 Posts: 13 Rating: (1) 
Hello, the cos phi of 0.7 is a realistic value. This value arises from the fact that the sinus voltage is always present on the module. But the current only flows impulsively. Current only flows if the instantaneous value of the sinusoidal voltage, before the rectifier, is higher than the intermediate circuit DC voltage, after the rectifier. this creates a pulsating current. This model of the power supply does not yet have an active cos phi correction circuit. Compare the 6EP33367SB003AX0. The PSU6200 has a cos phi of >0.9. Greeting Georg 
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