Design principle, parameter calculation and MATLAB simulation of boost circuit (Boost)

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Boost (Boost) conversion circuit is a single-tube non-isolated DC conversion circuit whose output voltage is greater than or equal to the input voltage. It consists of a DC voltage source, inductor, switch tube, diode, filter capacitor, and load resistor. The boost circuit diagram is shown in Figure 1.

In the last step-down (Buck) conversion circuit, its topology consists of a voltage source, a series switch, and a current source load. The boost conversion circuit is a dual topology of the buck conversion circuit. The boost converter is composed of a current source (a voltage source connected in series with a larger resistor), a parallel switch, and a voltage source load (a parallel capacitor). By controlling the duty cycle of the switch tube, and then controlling the output voltage, the two working conditions of the boost converter circuit are shown in Figure 2 and Figure 3, which represent the on-state of the switch tube and the off-state of the switch tube respectively.

The boost conversion circuit is still divided into three states according to whether the inductor current is continuous: continuous conduction, discontinuous conduction, and critical state. In order to facilitate the analysis of the steady-state characteristics of the boost circuit and simplify the process of deriving the formula, the following assumptions are made:

Switch tubes and diodes are ideal devices, that is, regardless of the tube voltage drop during conduction, they can be turned on or off instantaneously, and no leakage current is generated when they are turned off.
Inductors and capacitors are ideal components. The inductor works in the linear region without saturation, the parasitic resistance is 0, and the equivalent series resistance of the capacitor is also 0.
The ratio of the ripple voltage in the output voltage to the output voltage is very small and can be ignored in myopia.

1. The boost converter is in continuous conduction mode

1) When the switch tube T is turned on, as shown in Figure 2. The diode D is connected to the negative pole of Us, and it is cut off by the reverse voltage. The capacitor C supplies power to the load R, and the polarity is positive and negative. The voltage source is fully loaded on both ends of the inductor L, that is, uL=Us. Under this voltage, the inductor current increases linearly, and the stored magnetic field energy also increases linearly. In one cycle of the switch tube T, the switch tube T is turned on for ton.

After the switch tube T is turned on, the inductor current increases by

Among them, D is the duty ratio, and D=ton/Ts.

2) When the switch tube T is cut off, as shown in Figure 3. The diode is subjected to forward voltage and conducts, and the inductor current flows to the output side through the diode. The magnetic field in the inductor L will change the polarity of the voltage across the inductor to ensure that the inductor current remains unchanged. Therefore, the voltage source Us is connected in series with the inductor voltage uL to supply power to the capacitor and the resistor, and the polarity of both ends of the load R is still positive at the top and negative at the bottom. The inductor voltage uL=Us-Uo<0, the inductor current decreases linearly. In one cycle, the time for the switch tube T to be turned off is Ts-ton.

During the cut-off period of the switch tube T, the decrease of the inductor current is

In the steady state, the increase of the inductor current during the turn-on period of the switch tube is equal to the decrease of the inductor current during the turn-off period, that is

That is Uo=Us/(1-D), because D<0, so 1-D<1, so the output Uo is always greater than the input Us, so the circuit is a boost circuit.

2. The boost converter is in continuous conduction mode (omitted, you can refer to the book knowledge yourself)

3. The boost converter is in critical mode

The boost converter is in the critical state of the inductor current, and the inductor current is equal to twice the power supply current, that is, iL=2Is. The input power and output power of the conversion circuit are respectively

Assuming that the loss is neglected and the input power is equal to the output power, we can get

Simultaneous equations , the critical value of inductance can be obtained

It should be noted that, in practical applications, the actual value of the inductance is generally 1.2-1.3 times the critical value of the inductance.

4. Ripple voltage and capacitance design

In the inductive continuous mode, considering that the diode current will all flow into the capacitor, in each switch tube cycle, the energy Q charged or discharged by the capacitor is

The ripple voltage generated by Q can be expressed as

According to the value of the ripple voltage, the capacitance value C can be calculated as

Next, do two experiments to verify whether the circuit design is correct. The switching tube T is to select 20KHz MOSFET and 10KHz IGBT respectively.

experiment one,

Technical indicators: input voltage 5V, output voltage 15V, ripple voltage requirement 0.2% Uo, load resistance 10 ohms, switching device 20kHz MOSFET.

① Determine the duty cycle