Hi,
there are many different SMPS techniques.
But for most of them the folowing is true:
Example:
Let´s say it is a 100kHz DCM flyback SMPS, input voltage 320V DC, inductor primary inductance of 2mH, output: 12V, 1A
12V, 1A means 12W.
100kHz means 10us cycle time
Now calculate the energy that the load draws within one cycle:
12V x 1A x 10us = 120uWs.
(100.000 times per second the inductance has to move 120uWs of energy form primary side to secondary side.)
Now the energy stored in the inductance can be calculated as 0.5 x I x I x L
so we can calculate I: 120uWs = 0.5 x I x I x 2mH --> I = 346mA
(100.000 times per second the inductance needs to be connected to the 320V input voltage ..... exactely as long that the current just reaches 346mA)
But how long does it take for the current to rise to 346mA:
L = I x t / V; 2mH = 0.346A x t / 320V --> t = 2.16us
This means 100.000 times per second the primary switch needs to be ON for exactely 2.16us.
2.16us every 10us means a duty cycle of 21.6%
(This is an ideal calculation. Reality may differ somehow)
****
Klaus