HELP:problem of duty in open loop buck converter

I have a buck converter design that L1=450uH,C1=500uF,R1=10ohm,Vin=20V,Vout=5V,f=10kHz.

I know the duty of the converter can be calculated with the equation as followed:
Duty=Vout/Vin=5/20=0.25
I simulate the circuit with Ansoft/Simplorer.Finally I gain the figure of the voltage of C1 vs time as followed:

From the above figure,we can know the final voltage of C1 and R1 is 4.4V. Why not is 5V?
Thanks!!!!!!!!!

Simple, the diode's voltage drop.

In an actual circuit, you also have to consider the primary switch's losses too.

I guess, the diode has forward voltage?

Thank you! When the pass voltage of diode is configured to be 0V,I found the output voltage really near 5V.
However,the output voltage looks like a sine in sum presented as the followed figure.


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FvM said:

I guess, the diode has forward voltage?

Click to expand...

Thank you!Your suppose is right when I did an experiment.

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schmitt trigger said:

Simple, the diode's voltage drop.

In an actual circuit, you also have to consider the primary switch's losses too.

Click to expand...

Thank for your reply!The switch is ideal,so its loss in my simulation is neglected!

However,the output voltage looks like a sine in sum presented as the followed figure.

Click to expand...

Yes, its the resonance frequency formed by the L and C. Putting in a diode with 0 V forward voltage (ideal switch) reduces the losses and leaves a Q=0.1 resonator.

FvM said:

Yes, its the resonance frequency formed by the L and C. Putting in a diode with 0 V forward voltage (ideal switch) reduces the losses and leaves a Q=0.1 resonator.

Click to expand...

Thank you!So clever you are!