What is the solution to the chopper circuit? See completely understand

1. What is a chopper circuit?

The chopper circuit originally refers to the "smashing" of a part of the sine wave for some purpose in the power application. (For example, when the voltage is 50V, the electronic component is used to cut off the rear 50~0V part, the output voltage 0.) Later borrowed from the DC-DC switching power supply, mainly in the process of voltage regulation of the switching power supply, the original straight line of power, the line "æ–©" into a piece of pulse.

2, chopper circuit classification a

--- Buck circuit: Step-down chopper, whose output average voltage Uo is smaller than the input voltage Ui, and the output voltage is the same as the input voltage.

b

--- Boost circuit: The boost chopper has an output average voltage Uo greater than the input voltage Ui, and the output voltage is the same as the input voltage.

d

--- Buck-Boost circuit: buck or boost chopper, whose output average voltage Uo is greater than or less than the input voltage Ui, the output voltage is opposite to the input voltage polarity, and the inductor is transmitted.

e

--- Cuk circuit: buck or boost chopper, whose output average voltage Uo is greater or less than the input voltage Ui, the output voltage is opposite to the input voltage polarity, and the capacitance is transmitted.

This article mainly explains the principle of the boosted Boost circuit.

3, boost chopping (Boost) circuit

The boost circuit is shown in the figure below. Assume that the inductor L value and the capacitor C value are both large. The working principle is analyzed below.

a. When V is connected, E is charged to L, the charging current is always Ii, and C supplies power to the load. Because the C value is large, the output voltage is always Uo, and the time of V-pass is Ton, which is accumulated in this stage. The energy level is EIiTon.

b. When V is off, E and L jointly charge C and supply power to the load R. When the time Toff of V is set, the energy released by the inductor L at this stage is (Uo - E)IiToff.

c. If the steady state is reached, in one cycle T, the energy accumulated in L and the energy released should be equal, then

EIiTon = (Uo - E)IiToff

Uo - E = ETon / Toff

Uo = E(1 + Ton / Toff)

Uo = E(Ton + Toff) / Toff = E(T / Toff)

d. Since (T / Toff) is greater than or equal to 1, the output voltage of this circuit is higher than the power supply voltage, so the circuit is called a boosted Boost circuit.

e, (T / Toff) is called the boost ratio, and adjusting its size can change the Uo size.