Buck Boost Converteron 24/2/2012 & Updated on 8/8/2018
A typical Buck-Boost converter is shown below. The input voltage source is connected to a solid state device. The second switch used is a diode. The diode is connected, in reverse to the direction of power flow from source, to a capacitor and the load and the two are connected in parallel as shown in the figure above.
The controlled switch is turned on and off by using Pulse Width Modulation(PWM). PWM can be time based or frequency based. Frequency based modulation has disadvantages like a wide range of frequencies to achieve the desired control of the switch which in turn will give the desired output voltage. Time based Modulation is mostly used for DC-DC converters. It is simple to construct and use. The frequency remains constant in this type of PWM modulation. The Buck Boost converter has two modes of operation. The first mode is when the switch is on and conducting.
Mode I : Switch is ON, Diode is OFF The Switch is ON and therefore represents a short circuit ideally offering zero resistance to the flow of current so when the switch is ON all the current will flow through the switch and the inductor and back to the DC input source. The inductor stores charge during the time the switch is ON and when the solid state switch is OFF the polarity of the Inductor reverses so that current flows through the load and through the diode and back to the inductor. So the direction of current through the inductor remains the same.
Let us say the switch is on for a time TON and is off for a time TOFF. We define the time period, T, as and the switching frequency, Let us now define another term, the duty cycle, Let us analyse the Buck Boost converter in steady state operation for this mode using KVL. Since the switch is closed for a time TON = DT we can say that Δt = DT. While performing the analysis of the Buck-Boost converter we have to keep in mind that
- The inductor current is continuous and this is made possible by selecting an appropriate value of L.
- The inductor current in steady state rises from a value with a positive slope to a maximum value during the ON state and then drops back down to the initial value with a negative slope. Therefore the net change of the inductor current over any one complete cycle is zero.
Mode II : Switch is OFF, Diode is ON In this mode the polarity of the inductor is reversed and the energy stored in the inductor is released and is ultimately dissipated in the load resistance and this helps to maintain the flow of current in the same direction through the load and also step-up the output voltage as the inductor is now also acting as a source in conjunction with the input source. But for analysis we keep the original conventions to analyse the circuit using KVL.
Let us now analyse the Buck Boost converter in steady state operation for Mode II using KVL. Since the switch is open for a time we can say that . It is already established that the net change of the inductor current over any one complete cycle is zero. We know that D varies between 0 and 1. If D > 0.5, the output voltage is larger than the input; and if D < 0.5, the output is smaller than the input. But if D = 0.5 the output voltage is equal to the input voltage.
A circuit of a Buck-Boost converter and its waveforms is shown below. The inductance, L, is 50mH and the C is 100µF and the resistive load is 50Ω. The switching frequency is 1 kHz. The input voltage is 100 V DC and the duty cycle is 0.5. The voltage waveforms are as shown above and the current waveforms are as shown in the figure below.
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