Power Tips #5: Buck-boost design uses a buck controller

Figure 1 shows a simplified buck-boost circuit and the switching voltage present on the inductor. The circuit's similarity to a standard buck converter should be immediately apparent. In fact, it is identical to a buck converter, but with the output voltage and ground reversed. This arrangement also works for a synchronous buck converter. However, this is where the similarities to a buck or sync buck converter end, because the circuit operates differently than a buck converter.

The voltages present on the inductor during the FET switching intervals are different than that of a buck. Just as in a buck, it is necessary to balance the volt-microsecond (V-μs) product to prevent the inductor from saturating. While the FET is on, shown as the t_on interval in Figure 1; the full input voltage is impressed across the inductor. This positive voltage on the "dot' side of the inductor causes the current to ramp upward. This causes an on-time V-μs product across the inductor. During the FET off-time (t_off), the inductor's voltage polarity must reverse to maintain current flow, pulling the dot side negative. The inductor current ramps downward and circulates through the load and output capacitor, returning through the diode. The off-time V-μs product across the inductor must be equal to the on-time V-μs product. Since V_in and V_out are fixed, the duty cycle (D) expression: D=V_out/(V_out " V_in) is easily derived. The control circuit determines the correct duty cycle to maintain output voltage regulation. This expression and the waveform in Figure 1 assume continuous conduction mode of operation.

Figure 1: The buck-boost inductor requires balancing its volt-microSecond Product.
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The buck-boost inductor must operate at a current level that is higher than the output load current. This is defined as I_L = I/(1-D) or simply the input plus output current. For a negative output voltage that is equal in magnitude to the input voltage, operating at D = 0.5, the average inductor current is 2X that of the output.