What is buck-boost (inverter) converter?
A buck-boost converter is an energy-efficient DC-DC (direct current) converter that steps down and inverts the voltage from positive to negative voltages. The name is “buck” because the output is less than the input voltage (e.g., -10V output is less than +3.3V input). It has a “boost” because the output’s magnitude can be higher than the input voltage. Overall, the output voltage is negative. Of course, an inverting voltage can also be obtained with it. The control and regulation of output voltage are obtained using pulse-width modulation (PWM). It uses lossless components like inductors, capacitors, and switches to achieve zero energy loss.
Buck boost converter circuit diagram
The fundamental configuration of a buck-boost converter comprises the following elements:
- Inductor (L): The inductor stores and releases energy throughout switching cycles. Its primary function is storing energy when the switch is turned on. When the switch is turned off, its energy will be transferred into the load and capacitor by turning on the diode.
- Switch (Q1): The switch controls the current flow into the inductor. The time duration for which the switch is turned on decides how much of the inductor’s energy is built.
- Diode (D): It acts as an automatic switch. It automatically turns off and isolates the input and output nodes when the inductor charges. It turns on automatically when the switch is turned off, and the inductor releases energy.
- Output capacitor (C): The output capacitor is used to smooth out the ripple at the load.
Operating principle of buck-boost converter
The basic operation of a boost converter involves two main states: the switch is ON, and the switch is OFF.
Switch in ON
During the ON state, a logic high is applied to the gate of the switch (assuming the switch is an N-channel MOSFET). The input voltage is applied to the inductor, causing current to flow through it and storing energy in its magnetic field. The diode’s cathode is shorted to the input voltage, and the cathode is at the output voltage. Since the output voltage is less than the input voltage, the diode is turned off. This means there is no energy supply to the load. The output capacitor solely provides the load current.
During this time interval, Ton, the inductor current rises linearly because a constant voltage (Vin) is applied. The change in inductor current can be written as:
$$\Delta{}I_{L}(on)=\cfrac{V_{in}}{L}T_{on}$$
Switch is OFF
When the switch is opened (OFF state), the inductor’s current still flows in the same direction. It turns on the diode by decreasing the voltage at the cathode of the diode. Negative charges (electrons) will accumulate at the cathode if the diode does not turn ON. The diode turns ON, allowing the current to flow through the load and the capacitor. This also charges the capacitor so that it has enough energy for the next turn-off cycle.
Since the voltage across the inductor has flipped direction, the current through it will linearly reduce for the time duration Toff. The change in current can be represented as :
$$\Delta{}I_L(off)=\cfrac{V_{out}-V_D}{L}T_{off}$$
Where VD is the diode drop. For simplicity, the diode drop is assumed to be zero. So,
$$\Delta{}I_L(off)=\cfrac{V_{out}}{L}T_{off}$$
Buck boost converter equation and duty cycle
To maintain the same average current to a fixed load, the change in current through the inductor during the turn-on period has to be the same as the change in the turn-off period. It means ΔIL(on)=ΔIL(off):
$$|\Delta{}I_L(on)|=|\Delta{}I_L(off)|$$
$$\because{} V_{out}<0$$
$$\therefore{} \cfrac{V_{in}}{L}T_{on}=-\cfrac{V_{out}}{L}T_{off}$$
$$\cfrac{V_{out}}{V_{in}}=-\cfrac{T_{on}}{T_{off}}$$
Let’s denote the ratio of Ton and Ton+Toff = Tperiod as duty cycle (D) :
$$D=\cfrac{T_{on}}{T_{on}+T_{off}}$$
Therefore, the buck-boost converter’s equation in terms of duty cycle (D):
$$\cfrac{V_{out}}{V_{in}}=-\cfrac{D}{1-D}$$
Example of buck-boost converter using duty cycle relation
What is the expected output voltage of a buck-boost converter if the duty cycle is 40% (D=0.4) and the input voltage is 10V?
$$V_{out}=-\cfrac{0.4}{1-0.4}\cdot{}10\,\text{V}=-6.66\,\text{V}$$
Buck boost (inverting) topology waveform
