At time t = 0, the V driver is turned on and the VD is subjected to back pressure and turned off. The power supply E supplies power to the load, the load voltage U₀ = E, and the load current i₀ rises in an exponential manner (Figure 1).

At t = t₁, V is turned off and the diode VD is freewheeling. The load voltage U₀ is approximately zero, and the load current decreases exponentially (Figure 2, Figure 3). The large inductor L is usually connected in series so that the load current is continuous and the ripple is small [2] .

When the current is continuous, the average load voltage is:

U 0 = t o n t o n + t o f f E = t o n T E = α E (2-1)

α―Duty Ratio.

The maximum value of U₀ is E, and if α is reduced, U₀ will decrease with it reducing the chopper circuit, also known as the Buck converter.

The average load current is:

I 0 = U 0 − E m R (2-2)

When the current is interrupted, the average value of U₀ will be raised and it is generally undesirable [2] .

Assume that the L value is large and the C value is also large (Figure 8). The

inductor current i₁ and the capacitor voltage, i.e., the load voltage U₀, are basically constant. There are two modes [3] :

1) When S is on, the power supply U supplies power to the L via S, and the current is i₁. At the same time, C maintains the output voltage constant and supplies power to the load R.

2) When S is off, the energy of L is released to the load, and the current is i₂. The polarity of the load voltage is positive and negative, and the polarity of the power supply voltage is reversed. This circuit is also referred to as a reverse polarity chopper circuit.

In steady state, the integral of the voltage U_L in a period T is zero:

∫ 0 T u L d t = 0 (4-1)

When V is in the on state, U_L = E; and when S is in the off state, U_L = −U₀. then:

E ⋅ t o n = U 0 ⋅ t o f f (4-2)

Output voltage:

U 0 = t o n t o f f E = t o n T − t o n E = α 1 − α E (4-3)

By changing α, the output voltage can be either higher than the supply voltage or lower than the supply voltage. When 0 < α < 1/2, the output voltage is lower than the supply voltage, and when 1/2 < α < 1, the output voltage is higher than the supply voltage. Therefore, it is called a buck-boost chopper circuit or a buck-boost converter [2] .

In the steady state, the integral of the current of the capacitor C in one cycle T is zero, that is:

∫ 0 T i c d t = 0 (5-1)

In the equivalent circuit, the time for the S point to connect to the B point is t_on, So the product of capacitance current and time is I₂ * t_on. The time for S to connect A point is t_off, so the product of capacitance current and time is I₁ * t_off. As a result, it can be obtained:

I 2 t o n = I 1 t o f f (5-2)

When the capacitance C is very strong, the capacitance voltage U_C pulsation is sufficiently small, and the relationship between the output voltage U₀ and the input voltage E can be calculated as follows [4] :

When the S point is connected to B point, B point voltage U_B = 0, A point voltage U_A = −U_C;

When the S point is connected to A point, U_B = U_C, U_A = 0.

Therefore, the average value of U_B is [3] :

U B = t o f f T U C (5-3)

E = U B = t o f f T U C (5-4)

In addition, the average voltage of the voltage of the A point is:

U A = − t o n T U C (5-5)

U 0 = t o n T U C (5-6)

Then we can get the relationship between the output voltage U₀ and the power supply voltage E:

U 0 = t o n t o f f E = t o n T − t o n E = α 1 − α E (5-7)

Parameter settings:

Cycle: 0.0005 s; Duty cycle: 20%, 50%; Duration: 0.005 s; Supply voltage VDc: 12 V; Inductance L: 0.00045H; Load resistance R: 1 Ω; Capacitance C: 0.0026F; Algorithm: ode45.

The simulation diagram of the buck chopper circuit is shown in Figures 11-13 respectively represent the simulation waveforms with duty cycles of 20% and 50%. The waveforms in Figure 12 and Figure 13 are represented from top to bottom as follows: pulse waveform, load voltage U_R, inductor voltage U_L, inductor current I_L, thyristor current, and supply current. By analyzing the waveform diagram, it can be seen that the simulation waveform is consistent with the theoretical analysis [5] .

Parameter settings:

Cycle: 0.001 s; Duty cycle: 20%, 50%; Duration: 0.01 s; Supply voltage VDc: 12 V; Inductance L: 0.001H; Load resistance R: 1 Ω; Capacitance C: 0.0026F; Algorithm: ode 23.

The simulation diagram of the boost chopper circuit is shown in Figure 11-13 respectively represent the simulation waveforms with duty cycles of 20% and 50%. The waveforms in Figures 14-16 are represented from top to bottom as follows: pulse waveform, load voltage U_R, inductor voltage U_L, inductor current

I_L, thyristor current, and supply current. By analyzing the waveform diagram, it can be seen that the simulation waveform is consistent with the theoretical analysis.

Parameter settings:

Cycle: 0.001 s; Duty cycle: 20%, 80%; Duration: 0.02 s; Supply voltage VDc: 12V; Inductance L: 5e−3; Load resistance R: 50 Ω; Capacitance C: 5e−5; Algorithm: ode45.

The simulation diagram of the boost and buck chopper circuit is shown in Figures 17-19 respectively represent the simulation waveforms with duty cycles of 20% and 80%. The waveforms in Figure 18 and Figure 19 are represented from top to bottom as follows: pulse waveform, input current, output current, output voltage. By analyzing the waveform diagram, it can be seen that the simulation waveform is consistent with the theoretical analysis.

Parameter settings:

Cycle: 0.001 s; Duty cycle: 20%, 80%; Duration: 0.02 s; Supply voltage VDc: 12V; Inductance L: 0.08H; Load resistance R: 50 Ω; Capacitance C: 1e−5; Algorithm: ode45.

The simulation diagram of the Cuk chopper circuit is shown in Figures 20-22 respectively represent the simulation waveforms with duty cycles of 20% and 80%. The waveforms in Figures 20-22 are represented from top to bottom as follows: pulse waveform, input current, output current, output voltage. By analyzing the waveform diagram, it can be seen that the simulation waveform is consistent with the theoretical analysis.