Dear Sreekanth,

The duty cycle is the ratio of a  pulse width to the total cycle time Tau/ T. Every MOS transistor needs an on time and an off time where the device is transferred from the off state to the on state and vice verse. In the on state there is the on power losses which is Ion Von and the OFF power loss which is Ioff Voff. The on and the off power , the switching power is Ion Voff/2.The off power is normally neglected.

Then the total power energy loss in one cycle

EPC= Von Ion Ton + Ion Voff Tsw/2,

where Tsw is switching on time assumed equal switching off time.

For efficient operation the first term must be greater than the second term

That is  Von Ion Ton > Ion Voff Tsw/2, The power consumption during the on period which is the useful period must be greater that losses at the switching transitions which is not useful in the sense of operation of a switch.

It follows that 

Von Ton > Voff Tsw/2,

or 2 Von/ Voff > Tsw/ Ton,  equation 1, the duty cycle is (Ton+ 2Tsw)/ T,

equation 2.

Best wishes.

Sir,

Thank you for your well explained answer. In the equation duty cycle=(Ton+ 2Tsw)/ T,

we can get Tsw from Mosfet datasheet(typical switch off time), T is the switching period operation in a given converter. But Ton varies with our requirement in the converter. Then what value of Ton should be taken?

Dear Sreekanth,

To answer your new question, The value of Ton is determined by required Duty ratio,

For the DC to DC converter to operate correctly the duty ratio D= The input voltage Vi / The output voltage. So, it determines the voltage conversion ratio of the converter

Since D is known, Tsw and T are known one can determine Ton from the definition equation of the duty ratio.

After that one has to verify the validity of equation 1. That is one has to verify the inequality : 2 Von/ Voff > Tsw/ Ton, sice Von, Voff= VDD , Tsw and Ton are all known.

As a numerical example assume that the required duty ratio is .5, The cycle time= 100 mictosecond for f=10 kHz, Tsw= 1 microsecond, Von= 1 volt, VDD= 100 volt,

Then The left hand side of the inequality 2 Von/ Voff = .02,

And the right hand side= Tsw/ Ton= 1/ ( 50-2)= 1/48, 

The inequality is not valid, so one has to readjust the performance parameters to force the inequality. This may be accomplished by decreasing Tsw by decreasing the time constant of the charging circuit constant of the MOS transistor. Even one can use a more fast switching transistor. On can use a transistor with smaller Von,

One can reduce the the frequency of operation of the circuit. For the same D one increases the absolute value of Ton.

Wish you success.

Sir,

Thank you for your answer. In  first reply, In equation for EPC, why only one switching loss is considered(IonVoffTsw/2), instead of both switching on and off loss (IonVoffTsw)? considering your numerical example, for Von =1 volt and Voff =100V, Tsw=1microsecond and for f=10kHz, it is not advisable to operate the switch  below 0.5 duty ratio. But for the circuit with 0.7 duty cycle operation, equation (1) will satisfied, no problem will  arise. Here output voltage of converter will change. Am I correct sir?

Dear reekanth,

I am very happy with your fruitful discussion on this important matter. Because i want to shorten my comment as possible, i summed up the two sitching tomes the on and the off switching time. It is so that the voltage decreases from Voff to and the current rices from zero to Ion, during time Tsw. Therefore the energy consumption will be (Voff/2) ((Ion/2) Tsw, And same value for the other switching event. Therefore the overall switching energy loss will be Voff Ion/2.

Concerning the second part of your question, you are right. You are exact.

Best wishes 

Dear Zekry sir,

I am extremely happy about the discussion with you. The way you are explaining simple things is very good. I am learning a lot. Sir, I find  that total switching losses  = (VoffIon/2)(tswon)+(VoffIon/2)(tswoff) = VoffIontsw in "Fundamentals power electronics" textbook by R. Ericson. I am attaching the images of pages of the book. This seems to be different from your answer. Could you please check the attachments.

Deer Sreekanth,

I assumed resistive load while in the displayed book, an inductive load is assumed.

In case of inductive loads the switching losses is doubled.

wish you success.

Thank you very much Zekry sir. I got clear idea. 

minimum 46-50 % duty cycle. and maximum 90-97%.