Step-Up vs. Step-Down Chopper: Key Differences Explained

A chopper acts like a high-speed on/off switch. This page compares step-up choppers (boost converters) and step-down choppers (buck converters), outlining the differences in their operation. Both circuits rapidly connect and disconnect the source from the load, effectively “chopping” the constant DC supply voltage to produce a different output.

Step Up Chopper (Boost Converter)

Figure-1: Circuit of Step Up Chopper

The step-up chopper, or boost converter, increases the input voltage. Its operation is as follows:

• ON State (Chopper Closed): The switch path is closed, and current flows. The inductor, connected in series, stores energy during this period.
• OFF State (Chopper Open): The switch path is open. The inductor current doesn’t stop instantly; it flows through the diode and the load.
• Voltage Boost: This process results in the voltage across the load exceeding the source voltage ($V_s$). This characteristic gives the circuit its name: “step-up” chopper.

The average load voltage ($V_o$) for this boost converter is calculated as:

$V_o = \left( \frac{T}{T_{off}} \right) \cdot V_s = \left( \frac{1}{1-\alpha} \right) \cdot V_s$

Where:

• $\alpha$ is the duty cycle
• $T = T_{on} + T_{off}$

Step Down Chopper (Buck Converter)

Figure-2: Circuit of Step Down Chopper

The step-down chopper, or buck converter, decreases the input voltage. Its operation differs from the boost converter:

• ON State (Chopper Closed): The load voltage equals the source voltage ($V_s$).
• OFF State (Chopper Open): The load current flows through the diode. This effectively short-circuits the load terminals via the diode. Consequently, the load voltage is zero during the OFF period.
• Voltage Reduction: The load current increases during the ON period and decreases during the OFF period, providing a chopped DC voltage at the load. Hence, it’s called a “step-down” chopper.

The average load voltage ($V_o$) for this buck converter is:

$V_o = \left( \frac{T_{on}}{T_{on} + T_{off}} \right) \cdot V_s = \alpha \cdot V_s$

Where:

• $\alpha$ is the duty cycle
• $T = T_{on} + T_{off}$