Turn a buck converter module into a buck-boost converter with only two components
Buck converter theory
The schematic of a buck converter is shown below:
The components in the box, in addition to the feedback loop (not shown), are usually integrated into a single chip.
A regular buck-converter works by switching current into the output capacitor through an inductor. When the transistor in the box is on, current can flow from the supply to the output via the inductor. The voltage across the inductor is \(V_{in}-V_{out}\) and the rate of change of the current is \(\frac{V_{in}-V_{out}}{L}\). When the transistor in the box switches off, the current in the inductor continues to flow, but now through the diode. The voltage across the inductor is \(V_{out}\) and the rate of change of the current is \(\frac{V_{out}}{L}\). Once the output voltage is stable, the circuit is said to have reached a steady state. In the steady state, the decrease in current when the transistor is off must exactly cancel the change in current when the transistor is on. Otherwise, the current in the inductor would on average be continuously increasing or decreasing.
Let \(t\) be the period of the switching frequency and let \(D\) be the duty cycle of the transistor. Thus, the transistor is on for \(Dt\) seconds and off for \(\left(1-D\right)t\) seconds. As stated in the previous paragraph:
$$ Dt\cdot\frac{V_{in}-V_{out}}{L}=\left(1-D\right)t\cdot\frac{V_{out}}{L} $$ $$ DtV_{in}-DtV_{out}=\left(1-D\right)tV_{out} $$ $$ DV_{in}=V_{out} $$ Since the duty cycle \(D\) is constrained to \(0\leq D\leq1\) by definition, the output voltage cannot be higher than the input voltage.Modified circuit
The modified circuit is shown below:
To create the buck-boost circuit a single diode and a single N-channel MOSFET have been added to the circuit. Now, when the transistor in the box is on, the N-channel MOSFET is also on, and current begins to flow through the inductor. However, the voltage across the inductor is \(V_{in}\), as opposed to \(V_{in}-V_{out}\). This allows more energy to be stored in the inductor while the transistors are on. As a result, the output voltage can be higher than the input voltage.
The finished device can be seen below. The MOSFET in the 8-pin package and diode are located above the inductor.
Like with the previous circuit, we can set the increase in current equal to the decrease in current
$$ Dt\cdot\frac{V_{in}}{L} = \left(1-D\right)t\cdot\frac{V_{out}}{L} $$ $$ \frac{D}{1-D}V_{in}=V_{out} $$ For duty cycles greater than 50%, the output voltage should be higher than the input voltage.
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