Buck step-down converter

A linear power regulator needs to dissipate a lot of heat to decrease the output voltage with respect to the input. This can be a problem and can be solved by using a switching regulator. The switching regulator functions similar to a PWM device: the output signal is turned off and on very fast, and filtered so the actual output is a stable voltage.

As the output voltage will depend on the load, a feedback circuit is used to regulate the voltage to an exact voltage.

Although it is possible to create such a circuit with discrete components, more practical is the use of a specific IC. Let’s look at the circuit described in the datasheet of such an IC, the LM2596.

We can calculate the output voltage, using the formula from the datasheet:

\[V_{out} = V_{ref} * (1 + \dfrac{R_2}{R_1}) = 1.23V * (1 + \dfrac{4k7Ω}{1kΩ}) = 7.0V\]

As we will keep \(R_1\) fixed at 1kΩ, \(R_2\) will be a variable resistor, so the formula to calculate \(R_2\) is:

\[R_2 = R_1 * (\dfrac{V_{out}}{V_{ref}} - 1) = 1kΩ * (\dfrac{V_{out}}{1.23V} - 1)\]

The circuit uses an LC-filter to filter the output of the circuit to a smooth level. Some ripple remains, as is visible in the graph below.

Power circuits use LC-filters instead of RC-filters. This has two advantages. The first is obviously power: an RC-filter has a power loss over the resistor, an LC-filter doesn’t loose power (or at least: ideal components won’t). The second is beter filtering: an RC-filter is a one-pole filter, an LC-filter is a two-pole filter, so it will do a better job smoothing out the voltage and current ripply.

The schottky diode is necessary to give the inductor a path to ground in the “off” part of the switching circuit. This should be a fast switching diode, that’s the reason for the schottky version of the diode. In the example above an 1N5819 is used (40V, 1A). The datasheet recommends the specific version with regard to the current requirements of the load.