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5. 利用相同方法测量稳压二极管,得到相似结果

<img src="../assets/IMG_9033.jpeg" style="zoom:8%;" /><img src="./assets/IMG_9034.jpeg" alt="IMG_9034" style="zoom:8%;" />
<img src="../assets/IMG_9033.jpeg" style="zoom:8%;" /><img src="../assets/IMG_9034.jpeg" alt="IMG_9034" style="zoom:8%;" />



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这份实验报告拿了8分,说*缺少三表法和实验结果记录图片*。缘由不知,后人改进。
<iframe src="../Lab1.pdf" width="100%" height="600px"></iframe>
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# Lab2 功率测量和功率因数提高

3220104119 冯静怡 2024/3/24

## 一、 实验目的

1. 学习日光灯管的工作原理,了解镇流器、启辉器的作用;
2. 了解功率因数的概念,学会对功率因数进行补偿

## 二、 实验要求

1. 保持日光灯两端电压不变,测量I-C,P-C;
2. 作出I^2^、P、cosφ与补偿电容C的关系曲线;
3. 求I^2^-C曲线的有理经验公式;
4. 用P-C曲线求单位电容的等效电导g;
5. 推算电网电压三次谐波分量的大小;
6. 测量灯管、镇流器以及总负载的等效参数;
7. 解释或研究实验过程中遇到的异常现象。

### 三、实验方案与实验原理

​ 连接日光灯管电路为下图所示,利用三表法(功率表,电流表,电压表)测量电路等效阻抗;调节并联电容C的大小,记录U、I、P的数据,计算功率因数,从而得到最佳电容补偿方案。

<img src="../assets/image-20240322224714465.png" alt="image-20240322224714465" style="zoom:40%;" />

​ 由于试验台的瓦特表能够同时测量电流电压及众多参数,包括功率因数,所以本实验电路只需要使用瓦特表一个电表即可。

<img src="../assets/88f523b42ce79a10dcf6b6a57c2e9cd.jpg" alt="88f523b42ce79a10dcf6b6a57c2e9cd" style="zoom:15%;" />

日关灯及镇流器可以等效为一个阻抗为$Z=R+jwL$的元件,计算如下:
$$
\begin{align}
\dot I_总&=\dot I_C+\dot I_L\\
&=-\frac{\dot U}{j\frac1{wC}}+\frac{\dot U}{R+j\omega L}\\
&=Re(\dot I_L)+j[-Im(\dot I_L)+\dot UwC]
\end{align}\\
$$
由上式可知,电容产生的电流可以很好地弥补电感产生的电流,减小总体的输入电流,增大功率因数角;电容的取值有一个特定的范围,若超过该取值,则功率因数过补偿。

若将所有元件表示为电导的形式,则会得到更加简洁的表达式:

<img src="../assets/e65dd9a18bc55a36060f32be862c095.jpg" alt="e65dd9a18bc55a36060f32be862c095" style="zoom:15%;" />

$$
\dot I_{Cx}=I_R+jI_L+gC_x\dot U+jwC_x\dot U\\
\Rightarrow I_{Cx}^2=(I_R+gC_xU)^2+(I_L+wC_xU)^2\\
$$

$I^2$的曲线,可由C_x的相关曲线进行二次拟合,即:
$$
I_{Cx}^2=aC_x^2+bC_x+I_{C0}^2\\
\begin{cases}
a=(g^2+w^2)U^2\\
b=2(gI_R-wI_L)U
\end{cases}
$$
理论上,电路的整体功率不会随着电容的增加而增加,但是实际上,电容越大,功率消耗越大,因此引进参数$g$,相当于电容并联一个与电容大小成正比的电阻,$g$的大小可以用P的一次线性拟合得到:
$$
\because P-P_0=UI_g\\
\therefore P=(gU^2)C_x+P_0
$$

## 四、实验过程

1. 将调节旋钮逆时针旋转到底,归零。
2. 在断电的情况下按照上图连接电路,先不将电容接进电路中。
3. 对电路通电,调节自耦变压器,点亮灯管。
4. 当灯管亮起,并在180V左右一闪一闪,说明灯管本身无损坏,将电压调节至220V,准备开始进行实验。
5. 先记录无电容的情况下,即电容C=0,情况下的电表读数;后调节电容的大小,记下各个点的读数,实验过程中,电压的大小在缓慢改变,变化幅度在$\pm0.3V$​左右。

### 数据记录

| C | U | I | W | Var | VA | PF | phi |
| ------- | --------- | --------- | -------- | ------- | -------- | ------- | -------- |
| 0 | 220.3 | 0.289 | 26.4 | 58 | 63.6 | 0.417 | 65.3 |
| 1 | 220 | 0.229 | 26.4 | 42.7 | 50.3 | 0.526 | 58.2 |
| 2 | 219.9 | 0.174 | 26.4 | 26.9 | 38.2 | 0.69 | 46.2 |
| 3 | 220.2 | 0.14 | 26.4 | 11.7 | 30.6 | 0.863 | 30.3 |
| 3.2 | 220.4 | 0.137 | 26.6 | 8.5 | 30.1 | 0.884 | 27.7 |
| 3.4 | 220.2 | 0.135 | 26.6 | 5.2 | 29.7 | 0.898 | 26 |
| **3.7** | **220.3** | **0.134** | **26.6** | **0.4** | **29.5** | **0.9** | **25.3** |
| 3.8 | 220.3 | 0.135 | 26.6 | 1 | 29.7 | 0.896 | 333.6 |
| 3.9 | 220.3 | 0.136 | 26.5 | 2.6 | 29.8 | 0.891 | 333.2 |
| 4 | 220.6 | 0.142 | 26.8 | 7.3 | 31.2 | 0.854 | 328.5 |
| 5 | 220 | 0.177 | 26.5 | 22.2 | 38.9 | 0.684 | 313 |
| 6 | 220.2 | 0.232 | 26.6 | 38.1 | 51.3 | 0.518 | 301.3 |
| 7 | 220.1 | 0.292 | 26.6 | 53 | 63.8 | 0.417 | 294.5 |
| 8 | 220.1 | 0.352 | 26.6 | 67.3 | 77.3 | 0.345 | 290.1 |

### 绘制图线

在Matlab中绘制$I^2-C,P-C,\cos\varphi -C$,如图所示:

![](./assets/untitled.bmp)

将三条图线使用三个笛卡尔坐标系重合的方式,绘制在同一张图中:

* 可以看到$I^2$和功率因数,变化趋势一致,在某一电容值位置分别取到最小值和最大值。
* 功率变化非常奇怪,似乎没有规律,主要原因是,功率变化并不明显,图表的y坐标变化范围为[26.3,26.8]使得较小的变化被放大。
* 经实验可知,当$C=3.7\mu F$,能够取到最大功率因数补偿(在C的精确度为0.1uF的情况下)

### 二次曲线拟合

对$I^2-C$曲线由理论分析部分可知,可进行二次拟合,拟合如下:

<img src="../assets/3.bmp" style="zoom:55%;" />

二次曲线拟合结果: $y = 0.005297x^{2} - 0.03704 x + 0.08334$​

通过Matlab计算最小值得到:

```matlab
% 定义幂函数
fun = @(x)0.005297*x^2 - 0.03704*x + 0.08334;
% 使用 fminsearch 求解局部最小值
x0 = 0; % 初始猜测值
x_min = fminsearch(fun, x0);
y_min = fun(x_min);
disp(['极小值点 x = ', num2str(x_min)]);
disp(['极小值 y = ', num2str(y_min)]);
```

```shell
>> findmin
极小值点 x = 3.4963
极小值 y = 0.018588
```

* 即,当电容取$3.4963\mu F$时,能够取到电流最小值0.136A,则计算得到功率因数为$\cos\varphi\approx\frac{26.5}{220\times 0.136}=0.88$​,此时功率因数补偿最有效。

* 但同时可以发现,理论上功率因数补偿可以到1;但是根据拟合结果,此时功率因数补偿无法增加到一,可能是由于没有

### P-C曲线拟合求g值

在理论分析中估计P值是在U保持不变的情况下进行,但是实际上U有所变化,因此对数据进行处理,即进行$P/(U/U_{0})^2$处理,使得P在相同的U下进行讨论:

| C | 0 | 1 | 2 | 3 | 3.2 | 3.4 | 3.7 | 3.8 | 3.9 | 4 | 5 | 6 | 7 | 8 |
| ---- | -------- | ---- | -------- | -------- | -------- | ------- | ------- | ------- | -------- | -------- | ---- | ------- | -------- | -------- |
| U | 220 | 220 | 220 | 220 | 220 | 220 | 220 | 220 | 220 | 220 | 220 | 220 | 220 | 220 |
| W | 26.32815 | 26.4 | 26.42402 | 26.35207 | 26.50354 | 26.5517 | 26.5276 | 26.5276 | 26.42787 | 26.65441 | 26.5 | 26.5517 | 26.57583 | 26.57583 |

对P的数值进行描点,并得到一次拟合曲线为:$y=0.02951x+26.43$

<img src="../assets/5.bmp" style="zoom:70%;" />

即:

* $P_0=26.43W$,当没有电容并入时,光日光灯管产生的功率为25.43W
* P中间的一些数值大大远离拟合曲线,原因是这些数据与其余数据不是同一时间测量,是在测完整数值电容之后对图线进行的弥补,所以功率值大大增加。
* 根据理论计算得知:$g=\frac{a}{U^2}=6.09\times 10^{-7} \Omega^{-1}\cdot\mu F^{-1}=0.609\Omega^{-1}\cdot F^{-1}$​

### 讨论三次谐波的影响

由上述结果已经可以推知,交流电源输入成分并不时非常好,含有高次谐波分量,使得电容不能完全补偿功率因数

现观察高次谐波对功率因数的影响:

已知在单50Hz的作用下,电流与电容的结果为
$$
I_{Cx}^2=aC_x^2+bC_x+I_{C0}^2\\
\begin{cases}
a=(g^2+w^2)U^2\\
b=2(gI_R-wI_L)U
\end{cases}
$$
则若电路中有三次谐波,则该式变为
$$
I_{Cx}^2=aC_x^2+bC_x+I_{C0}^2\\
\begin{cases}
a=(g^2+w^2)U_1^2+(g^2+9w^2)U_3^2\\
b=2(gI_R-wI_L)U_1+2(gI_R-3wI_L)U_3\\
\end{cases}
$$
可以计算得三次谐波的大小为:(其中由 $y = 0.005297x^{2} - 0.03704 x + 0.08334$,得到$a=0.005297(A/\mu F^2)$)
$$
U_3^2=\frac{a-(g^2+\omega^2)U^2}{8\omega^2}=658.73V^2\\
U_3=25.67V
$$
使用示波器记录输入电压的CSV数据格式,并使用Matlab对输入的电压数据进行FFT分析:

<img src="../assets/2024-03-24%20(5).png" alt="2024-03-24 (5)" style="zoom:55%;" />



得到如图所示的电压数据,可以看到在除了50Hz的基波之外,3、5、7次谐波均有较大的分量,计算三次谐波分量为:

![](./assets/7.bmp)

根据FFT图像,若一次谐波的有效值大小为220V,则三次谐波的有效值大小应与各自显示的FFT大小成正比,则三次谐波的大小为:$220\times\frac{18830.6}{1538750}=26.9V$,计算结果与上述三次谐波计算结果基本一致,可知该电压有较大的谐波分量。

### 日光灯管参数

日光灯管接入分别如下电路图,使用三表法测量整流器和日关灯管的参数

<img src="../assets/33885734c664fb3157a3eec78b42263.jpg" alt="33885734c664fb3157a3eec78b42263" style="zoom:15%;" /><img src="../assets/ad9e7a6964b4ea0f856752378e754fc.jpg" alt="ad9e7a6964b4ea0f856752378e754fc" style="zoom:15%;" />

实验数据记录如下:

| | U | I | W | Var | VA | PF | phi |
| -------- | ----- | ----- | ---- | ---- | ---- | ----- | ---- |
| 整流器 | 193.6 | 0.288 | 12.4 | 53.6 | 55.3 | 0.225 | 76.9 |
| 日光灯管 | 62.8 | 0.287 | 15 | 2.9 | 18 | 0.832 | 33.7 |

计算整流器和日光灯管在50Hz下的等效阻抗:
$$
Z_整=674.2\angle -76.9^\circ \Omega\\
Z_灯=218.8\angle -33.7^\circ \Omega\\
Z_L=Z_整+Z_灯=847.0\angle-66.7^\circ
$$
则根据理论,C的阻抗大小应该为:
$$
C=\frac1{2\pi f\times\Im(847\angle-66.7^\circ)}=3.45\mu F
$$

## 二、 总结

### 各方法结果统计

| 方法 | 补偿电容大小 | 实验结果说明 |
| ------------------------------ | -------------- | ------------------------------------------------------------ |
| 直接测量法(取电流最小值的点) | $3.7\mu F$左右 | 实验结果精度不高,但是是最为直接的方法,也是最为准确的方法;完全考虑了高次谐波对实验结果的影响 |
| 统计数据,二次曲线拟合法 | $3.4963\mu F$ | 该方法理论与实际数据相结合,通过曲线拟合(统计规律)的方式近似得到电流最小值点。 |
| 测量日光灯管等效参数,理论分析 | $3.45\mu F$ | 该测量方法仅仅考虑了基波的影响,完全没有考虑高次谐波对实验结果造成的影响 |

### 异常现象

1. 电容由大变小时,会发出类似于电火花迸发的声音

原因:电容存储电量为$Q=CU$,当电容变小,两端电压不变,电容快速放电,因而会产生砰砰声,比较危险。

2. 电容无法弥补功率因数至1

原因:存在较大的高次谐波。



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# Lab3 三相电路测量

3220104119 冯静怡

## 实验一:相序测定

<img src="../assets/%E5%BE%AE%E4%BF%A1%E6%88%AA%E5%9B%BE_20240328134031.png" style="zoom:30%;" />

#### A相

1$\mu F$电容

#### B相C相

两个灯泡

<img src="../assets/cf51fc87a23f4075507f8bac112ebe8.jpg" alt="cf51fc87a23f4075507f8bac112ebe8" style="zoom:5%;" />

| | 对地电压 | 相电压 | 判断 |
| ---- | -------- | ------- | ------------------------------- |
| L1 | 118.2 V | 179.3 V | $\dot U_U=U_U\angle 0^\circ$ |
| L2 | 118.8 V | 61.7 V | $\dot U_W=U_U\angle 120^\circ$ |
| L3 | 120.2 V | 144.4 V | $\dot U_V=U_U\angle -120^\circ$ |

## 实验二(1):三相三线制星型对称负载的测量

<img src="../assets/%E5%BE%AE%E4%BF%A1%E6%88%AA%E5%9B%BE_20240328142537.png" style="zoom:30%;" />

<img src="../assets/b45c72d046fca6927c88cca9e2140d4.jpg" alt="b45c72d046fca6927c88cca9e2140d4" style="zoom:15%;" />

#### A相B相C相

2$\mu F$电容,两个灯泡

| | | $U_相$ | $I_相=I_线$ | $U_线$ | $U_线/U_相$ | $P$ |
| ------------------------------- | ---- | ------- | ----------- | ------------ | ----------- | ------ |
| $\dot U_U=U_U\angle 0^\circ$ | L1 A | 116.2 V | 0.181 A | A-C 205.9 V | 1.77 | 18.8 W |
| $\dot U_V=U_U\angle -120^\circ$ | L3 C | 117.7 V | 0.180 A | C-B 207.6 V | 1.76 | 18.9 W |
| $\dot U_V=U_U\angle 120^\circ$ | L2 B | 118.4 V | 0.181 A | B-A 205.4 V | 1.73 | 19.0 W |

中性点电压差:2.75V

### 实验二(2):三相三线制星型非对称负载的测量

<img src="../assets/53049cf4620358a662ccf406bc090dd.jpg" alt="53049cf4620358a662ccf406bc090dd" style="zoom:15%;" />

#### A相

两个灯泡

#### B相C相

2$\mu$F,两个灯泡

| | | $U_相$ | $I_相=I_线$ | $U_线$ | $U_线/U_相$ | $P$ |
| ------------------------------- | ---- | ------- | ----------- | ------------ | ----------- | ------ |
| $\dot U_U=U_U\angle 0^\circ$ | L1 A | 128.3 V | 0.169 A | A-C 205.7 V | 1.77 | 21.7 W |
| $\dot U_V=U_U\angle -120^\circ$ | L3 C | 94 V | 0.159 A | C-B 206.6 V | 1.76 | 13.6 W |
| $\dot U_V=U_U\angle 120^\circ$ | L2 B | 132.8 V | 0.192 A | B-A 204.9 V | 1.73 | 22.6 W |

中性点电压差:23.98V

## 实验二(3):三相 ** 线制星型 *对称* 负载的测量

<img src="../assets/%E5%BE%AE%E4%BF%A1%E6%88%AA%E5%9B%BE_20240328145556.png" style="zoom:30%;" />

<img src="../assets/c6c1bf48052179627d29ccb6734ce6c.jpg" alt="c6c1bf48052179627d29ccb6734ce6c" style="zoom:15%;" />

| | | U_相 | I_相=I_线 | U_线 | U_线/U_相 | P |
| ------------------------------- | ---- | ------- | --------- | ------------ | --------- | ------ |
| $\dot U_U=U_U\angle 0^\circ$ | L1 A | 116.9 V | 0.182 A | A-C 205.7 V | 1.77 | 18.9 W |
| $\dot U_V=U_U\angle -120^\circ$ | L3 C | 120.0 V | 0.183 A | C-B 206.6 V | 1.76 | 19.5 W |
| $\dot U_V=U_U\angle 120^\circ$ | L2 B | 118.4 V | 0.181 A | B-A 204.9 V | 1.73 | 19.0 W |

| | 电压 | 电流 |
| ---- | ----- | ------ |
| 中线 | 0.03V | 83.6mA |

## 实验三:△对称负载

<img src="../assets/%E5%BE%AE%E4%BF%A1%E6%88%AA%E5%9B%BE_20240328150648.png" style="zoom:40%;" />

| | 输出电压 | 线电流 |
| ---- | -------- | ------- |
| L1 | 115.4 V | 0.453 A |
| L3 | 116.9 V | 0.450 A |
| L2 | 118.3 V | 0.452 A |

| | $U_相=U_线$ | $I_相$ | P | Q |
| ----- | ----------- | ------- | ------ | -------- |
| L1-L3 | 202.3 V | 0.261 A | 42.8 W | 27.4 Var |
| L1-L2 | 202.0 V | 0.253 A | 42.2 W | 27.2 Var |
| L3-L2 | 203.6 V | 0.255 A | 42.7 W | 27.9 Var |
| | | 总功率 | 127.7W | |

### 二瓦表法测量总功率



<img src="../assets/%E5%BE%AE%E4%BF%A1%E6%88%AA%E5%9B%BE_20240328152731.png" style="zoom:25%;" />

| P~1~ | P~2~ | $P_总=P_1+P_2$ | $Q_总=\sqrt3(P_1-P_2)$ |
| ------ | ------ | -------------- | ------------------------ |
| 87.1 W | 40.4 W | 127.5 W | -80.9 Var |

### 一瓦表法测量总功率

<img src="../assets/%E5%BE%AE%E4%BF%A1%E6%88%AA%E5%9B%BE_20240328154035.png" style="zoom:30%;" />

| P | $Q=\sqrt3P$ |
| ------ | ----------- |
| 47.1 W | 81.6 Var |

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