Varying V_{REF} from 1.0V up to 2.0V and monitoring change in the output voltage V_{OUT}.

Voltage booster is a circuit that generates higher output voltage from a lower input voltage.
It works as follows. An inductor L1 is placed in series with Vdc and a switching transistor T1.
The bottom end of L1 feeds the output capacitor C1, and load resistor through rectifying diode D1.
The output voltage V, is higher than the DC input Vdc,as can be seen qualitatively thus: When Q1
is on for a time t_{on} D1 is reverse biased and current ramps up linearly in L1 to a peak value

*I _{P} = (V_{DC} * t_{ON}) / L1*

This represents an amount of stored energy:

*E = 0.5 L1 I _{P}^{2}*

where E is in joules, L1, is in henries, and I

Varying V_{REF} voltage on U1 PIN2. Measuring current profile in L1 with voltage drop across resistor R2.
Increasing V_{REF} reduces the time between current spikes in L1.

During the T1 on time, the output current is supplied entirely from C1,
which is chosen large enough to supply the load current for the
time t_{ON}, with the minimum specified droop. When T1 turns off, since the
current in an inductor cannot change instantaneously, the current in L1 reverses in an attempt to maintain
the current constant. Now the no-dot end of L1 is positive with respect
to the dot end, and since the dot end is a t V_{DC}, L1 delivers its stored
energy to C1, and charges it up via D1 to a higher voltage (aboosted-up voltage) than V_{DC}.
The circuit shown does not have a feedback and output voltage is unregulated, but can be set to a specified
value by setting V_{REF} voltage. Note that this design is meant for 'static' output currents,
not for variable current draw designs.
This is not for precision electronics!

If the current through D1 has fallen to zero before the next T1 turnon, all the energy stored in L1
during the last T1 on time has been delivered to the output load and the circuit is said to be operating
in the "discontinuous" mode.
An amount of energy E delivered to a load in a time T represents power.
With E in joules and T in seconds, P is in watts.

P_{L} = 0.5 L I_{P}^{2} / T

P_{DC} = V_{DC} (I_{P} / 2) (T_{OFF} / T)

The total power delivered to the load is then:

P_{t} = P_{L} + P_{DC} = 0.5 L I_{P}^{2} / T + V_{DC} (I_{P} / 2) (T_{OFF} / T)

I_{P} = (V_{DC} T_{ON}) / L1

P_{t} = (V_{DC}^{2} T_{ON}) / (2 T L1) (T_{ON} + T_{OFF}) = (V_{DC}^{2} T_{ON}) / (2 T L1) (kT) = V_{O}^{2} / R

The output voltage is:

V_{O} = V_{DC} √( ( k R T_{ON} ) / (2 L1 ))