Re: Boost Converter Output ?
The duty-cycle formula is more complicated than the simple 1-Vin/Vout used in many calculations.
The key to deducing a more exact formula is understanding that the volts*seconds applied to the inductor during the transistor on time must equal the volts*seconds during the off time.
Assuming continuous current mode operation, we can write:
(Vin-Vind-Vmos)*ton=(Vo+Vind+Vd-Vin)*toff,
where Vin is the input voltage, Vind is the voltage drop across the DC resistance of the inductor, Vo the output voltage, Vmos is the drop across the MOSFET during on time.
Dividing by T, and knowing that ton*T=DC and ton+toff=T, ir follows that:
toff/T=1-DC
Then we can re-write:
DC*(Vo+Vd-Vmos)=(1-DC)*(Vo+Vdrop+Vd-Vin)
And then DC=(Vo+Vd-Vin+Vind)/(Vo+Vd-Vmos)
This reduces to the familiar formula if:
Vd«Vo, Vind«Vo and Vmos«Vo
But the real world is not perfect, so if the duty-cycle is forced to 50% (without feedback to adjust it), taking only the diode drop into consideration, we find that the output voltage would be:
Vo=2*Vin-Vd=14.4V
Taking other factors into consideration helps explain why your friend only obtains 11.5V output (again, this is way lower than I would expect for properly chosen components and correct operation).
The first things to check are the input voltage, gate drive, and inductor DC resistance.