For a step-down DC chopper with a resistive load, when the duty cycle is increased the average value of the output voltage-

Concept:

The average load voltage, \({V_0} = D\left( {{V_{in}} - {V_d}} \right)\)

\(V_o \propto D\)

So, when D increases, \(V_o \) also increases.

\(V_o\) is the average load voltage.

D is the duty cycle.

\(V_{in}\) is the input voltage.

\(V_d\) is the voltage drop across the switch in the ON condition.

Additional Information 

The efficiency of a chopper is given by,

\(\eta = \frac{{{P_o}}}{{{P_{in}}}} \times 100\)

Where Po is the output power delivered to load and it is given by,

\({P_o} = \frac{{V_{or}^2}}{R}\)

Vor is the RMS value of the output voltage

R is the load resistance

Pin is the input power and it is given by,

Pin = Vin I0

Vin is the input voltage

I0 is the average load current

The average load current, \({I_o} = \frac{{{V_0}}}{R}\)

The RMS value of load voltage, \({V_{rms}} = \sqrt D \left( {{V_{in}} - {V_d}} \right)\)

D is the duty cycle.

Vd is the voltage drop across the switch in ON condition.