Lack of details for design data such as component specs and driver current will always lead to unexpected results.
Your load is pretty small, Vout/R= 2.5V/4k7=0.53mA and at 50% Ipk= I=V/Ldt= 1.1V/0.1H*0.5ms = 5.5mA pk
If the inductor is a small size the L/R ratios can be a wide range. This is important. As the duty cycle increases, the current can no longer rise at the rate of dI/dt=V/L if the voltage is reducing internally due to series resistance call DCR.
A 100mH is pretty large and so the DCR also tends to be large ranging from 0.6 to 700 Ohms in small parts. R/L=T needs to be small (<<2% ) compared to the load and also say 100R/100mH = T << 1/f=10ms. So lowering the frequency will increased the output voltage until the IR drop becomes significant or L saturation current is reached and L drops in value.
State your DCR and L Sat current, and datasheet link.
So lower f or change L to a smaller part and put forth some design specs next time.