Simulation about the Fundamentals of the power electronics-2021-08-07

时间：2022-08-20 15:30:01 dts二极管二极管lv

Fundamentals of the power electronics

Case 01: Buck-Boost converter
- **Topology**
- **建模分析**
- - **描述**
  - **建模分析**
  - - **问题a**
    - **问题b**
    - **问题c**
  - **Parameters**
  - **Simulation**
  - **仿真波形**
  - 注：python提取psim数据的原始代码

利用psim和python模拟书中的例题，严重计算是否合理。
Psim主要用于仿真，python用于分析仿真后的波形。
PSIM的版本为9.0.3.400

Case 01: Buck-Boost converter

Analysis and design of a buck–boost converter: a practical implementation using a transistor and diode is shown in Fig. 2.
a) Find the dependence of the equilibrium output voltage V and inductor current I on the duty ratio D, input voltage Vg, and load resistance R. You may assume that the inductor current ripple and capacitor voltage ripple are small.
b) Plot your results of part (a) over the range 0 ≤ D ≤ 1.
c) Dc design: for the specifications
Vg = 30V V = ?20V
R = 4Ω fs = 40 kHz
(i) Find D and I
(ii) Calculate the value of L that will make the peak inductor current ripple Δi equal to ten percent of the average inductor current I.
(iii) Choose C such that the peak output voltage ripple Δv is 0.1 V.
d) Sketch the transistor drain current waveform iT (t) for your design of part c). Include the effects of inductor current ripple. What is the peak value of iT ? Also sketch iT (t) for the case when L is decreased such that Δi is 50% of I. What happens to the peak value of iT in this case?
e) Sketch the diode current waveform iD(t) for the two cases of part (d).

建模分析

问题a

Fig.2 Buck-Boost converter

当开关管Q₁二极管关闭时，等效电路如图3所示。

Fig.3 Q~1~开通，D~1~等效电路关闭时如下电路表达式可根据小纹波假设获得。

$v_L=L\frac{di_L(t)}{dt}=v_g$
$i_c=-\frac{v}{R}$
$i_T=i_L$
$i_D=0$
当开关管Q₁处于关断状态时，此时二极管开通，等效电路如图4所示。

Fig.4 Q~1~关断，D~1~开通时等效电路根据小纹波假设，可以得到如下电路表达式。

$v_L=L\frac{di_L(t)}{dt}=v$
$i_c=-i_D-\frac{v}{R}$
$i_T=0$
$i_D=i_L$
定义开关管Q₁导通的时间为D*T_S，则可以得到电感电压波形如图5所示。

Fig.5 电感电压波形根据电感伏秒平衡原理可以得到如下表达式。

$\frac{v_g}{L}DT_s + \frac{v}{L}D'T_s = 0$
由此可以得到输入输出满足
$\frac{v}{v_g} = -\frac{D}{D'}$
同理可以得到电容电流波形，如图6所示。

Fig.6 电容电流波形根据电容充电平衡表达式(A-S平衡)可得：

$-\frac{v}{RC}DT_s + \frac{(-i_D-\frac{v}{R})}{C}D'T_s = 0$
由此可得
$i_D=-\frac{v}{R}\frac{D}{D'}$
至此，问题a已经分析完毕

问题b

Fig.7 电容电流波形

问题c

D = 0.4
I = 8.33A

Fig.8 电感电流波形电感值主要考虑变换器的纹波，所以可以根据图8进行计算。根据图8可以得到电感电流纹波和电感电流关系满足如下表达式：

$\frac{v_g}{L}DT_s=2\Delta i_L$
所以电感L满足：
$L=\frac{v_gDT_s}{2\Delta i_L}$

Fig.9 电容电压波形电容值主要考虑变换器的电压纹波，所以可以根据图9进行计算。根据图9可以得到电容值与电压纹波的关系满足如下表达式：

$\frac{v}{RC}DT_s=2\Delta v_C$
所以电容C满足：
$C=\frac{vDT_s}{2\Delta v_C}$

基于以上分析，可以求得电感值L = 180uH，电容值C = 250uH。

问题d和e 可以根据仿真结果进行分析，在此不在给出

注：python提取psim数据的原代码

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
# csv_path 表示要读取的csv文件的路径，路径前面需要加r进行转义
# columns 表示对应的数据，和psim中的标记一样，字符串类型
def read_csv(csv_path,columns):
    data = pd.read_csv(csv_path)
    data_columns = data[columns]
    return np.array(data_columns)



if __name__ == '__main__':
    path = r'F:\@inbox\Fundamentals of PE\Case2_1_Buck-Boost.csv'
    column1 = 'Time'
    column2 = 'Vout'
    column3 = 'IT'
    column4 = 'IL'
    col_ID = 'ID'
    col_IC = 'IC'
    Time = read_csv(path,column1)
    Vout = read_csv(path,column2)
    IT = read_csv(path,column3)
    IL = read_csv(path,column4)
    ID = read_csv(path,col_ID)
    IC = read_csv(path,col_IC)
    plt.plot(Time,IC)
    plt.xlim(0,0.02)
    # plt.ylim(-10,10)
    plt.xlabel('Time')
    plt.ylabel('IC')
    plt.show()

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Simulation about the Fundamentals of the power electronics-2021-08-07

Fundamentals of the power electronics

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