Buck converter simulation in orcad
Even if it's probably to late, there are some things to think about in your circuit. 

In the beginning, when the voltage of the emitter of Q1 is zero volts and you enable the 5V clock signal, you have 5V over the base-emitter junction. The base-emitter junction behaves like a diode and at 5V the current that is flowing into the base will be extremely big (in the real world, your transistor will be destroyed).

But let's ignore that for now. 

Now a current is flowing and charging the capacitor C1. 
Let's say the capacitor is charged to 3V: Now the base-emitter junction voltage is not 5V like before, but only 2V, which in turn leads to a reduced current flowing through the transistor. 
With less current, the charging process will be come slower. If the output voltage becomes even higher, the current becomes even less, and at a certain point, the system will reach equilibrium and stop doing anything. 

In short: The transistor you want to use as a switch, does not work as a switch. 

You should always check if your transistor is used in the correct operation region. For a bipolar transitor as a switch, you want the saturation region for the "on" state, and cut-off region for the "off" state.