As shown in Figures 1a and 1b, the synchronous buck converter IC can be used in an inverted buck-boost configuration by simply modifying the buck converter schematic. The inverting buck-boost converter generates the negative output voltage as follows: VOUT= -D/(1-D) x VIN
Figure 1. Using a Buck Regulator IC as an Inverting Buck-Boost Converter
The operation of the inverting buck-boost converter is shown in Figures 2a and 2b. During TON (Q1: turn-on; Q2: turn-off), the inductor stores energy; while during TOFF (Q1: off; Q2: turn-on), the inductor charges the output capacitor.
Figure 2. Reversed buck-boost operation
Inverted buck-boost configuration of the falling voltage IC's maximum VIN and IOUT
When using a buck regulator IC in an inverting configuration, you need to reduce the converter's maximum input voltage and maximum output current range. We illustrate these concepts using the LM5017-based inverting buck-boost circuit in Figure 3. In this configuration, the IC's bias ground (RTN pin) is connected to the negative output voltage (-10V). The input (VIN) and return (RTN) voltages of the IC are:
VIN, RTN = VIN + |VOUT|
Therefore, the maximum input voltage is: VIN(MAX) = VIN,RTN (MAX) - |VOUT|
Since the inductor current only supplies the output capacitor during TOFF (Figure 2b), the output current is: IOUT = IL1 (1-D)
Where D is the duty cycle. The maximum output current is related to the switch current limit, as shown in the following equation:
iL1(peak) = iSW(peak) = IL1+ΔIL1/2 = IOUT/(1-D) + ΔIL1/2
Where ΔIL1 is the peak-to-peak inductor current ripple that peaks at maximum VIN(MAX):
Figure 3 is a circuit diagram of the complete RP buck converter based on the LM5017. Due to its wide VIN (100V) rating, the LM5017 can operate over a wide input voltage rail in reverse phase buck-boost applications.
Figure 3. 10V~60V Input to -10V Output, 300mA Inverting Buck-Boost Application Circuit
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