﻿ 一种新的防电击保护自动检测装置的研制

# 一种新的防电击保护自动检测装置的研制A New Kind of Auto Testing Device Developed for Protection against Electric Shock

Abstract: In this paper, a new anti-shock protection automatic detection device is developed. When detecting, the detection device automatically realizes peak voltage off and releases residual energy through the discharge load. The system simultaneously collects discharge voltage and current, calculating residual energy. The detection device adopts the microcontroller STM32F407 as the control core, which is used to realize the initialization of the AD7606 liquid crystal display and the USB interface, the detection control and the acquisition, storage and data operation processing of the relevant measured data. The paper gives the circuit diagram and waveform diagram.

1. 引言

2. 国内外研究情况

3. 工作原理

4. 装置组成

Figure 1. Design diagram of auto testing device

Figure 2. Frequency measurement system

4.1. 隔离与信号调理电路

Figure 3. Voltage channel based on AMC1301

4.2. 电压调理电路

${U}_{peak}=0.25×\frac{{R}_{6}+{R}_{7}+{R}_{8}+{R}_{9}}{{R}_{9}}=381.6\text{\hspace{0.17em}}\text{V}$ (1)

4.3. 电流调理电路

${U}_{peak}=\frac{0.25}{{R}_{1}}=5\text{\hspace{0.17em}}\text{A}$ (2)

Figure 4. Current channel based on AMC1301

Figure 5. Signal conditioning circuit

${k}_{v}=\frac{{R}_{9}}{{R}_{6}+{R}_{7}+{R}_{8}+{R}_{9}}×8.2×\frac{{R}_{12}}{{R}_{10}}=0.02435$ (3)

${k}_{i}={R}_{1}×8.2×\frac{{R}_{12}}{{R}_{10}}=1.8587$ (4)

5. 仿真

5.1. 仿真电路图

5.2. 仿真波形

5.3. 放电过程能量计算

5.3.1. 理论计算值

$\begin{array}{c}{E}_{理论}=0.5{C}_{1}{U}^{2}+0.5{C}_{2}{U}^{2}\\ =0.5×0.47×{10}^{-6}×{310.9998}^{2}\text{\hspace{0.17em}}\text{J}+0.5×4.7×{10}^{-6}×{310.9998}^{2}\text{\hspace{0.17em}}\text{J}\\ =250.0232\text{\hspace{0.17em}}\text{mJ}\end{array}$

5.3.2. 矩形法计算

$\begin{array}{c}{E}_{计算}=\frac{\Delta t}{R}×\left[\underset{i=1}{\overset{n}{\sum }}{U}_{i}^{2}\right]\\ =\frac{6.066×{10}^{-6}}{2000}×\left({310.9998}^{2}+{310.8474}^{2}+{310.7041}^{2}+\cdots \right)\\ =250.6792\text{\hspace{0.17em}}\text{mJ}\end{array}$

5.3.3. 梯形法计算

$\begin{array}{c}{E}_{计算}=\frac{\Delta t}{2R}×\left[{U}_{1}^{2}+2\underset{i=2}{\overset{n}{\sum }}{U}_{i}^{2}\right]\\ =\frac{6.066×{10}^{-6}}{2×2000}×\left[{310.9998}^{2}+2×\left({310.8474}^{2}+{310.7041}^{2}+{310.5560}^{2}+\cdots \right)\right]\\ =250.5254\text{\hspace{0.17em}}\text{mJ}\end{array}$

5.3.4. 辛普森法计算

$\begin{array}{c}{E}_{计算}=\frac{\Delta t}{3R}×\left[{U}_{1}^{2}+4\underset{i=1}{\overset{n}{\sum }}{U}_{2i}^{2}+2\underset{i=1}{\overset{n}{\sum }}{U}_{2i+1}^{2}\right]\\ =\frac{6.066×{10}^{-6}}{3×2000}×\left[{310.9998}^{2}+4×\left({310.8474}^{2}+{310.5560}^{2}+\cdots \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2×\left({310.7041}^{2}+{310.4084}^{2}+\cdots \right)\right]\\ =250.5253\text{\hspace{0.17em}}\text{mJ}\end{array}$

5.3.5. 科特斯法计算

$\begin{array}{c}{E}_{计算}=\frac{2\Delta t}{45R}×\left[7{U}_{1}^{2}+32\underset{i=0}{\overset{n}{\sum }}{U}_{4i+2}^{2}+12\underset{i=0}{\overset{n}{\sum }}{U}_{4i+3}^{2}+32\underset{i=0}{\overset{n}{\sum }}{U}_{4i+4}^{2}+14\underset{i=0}{\overset{n}{\sum }}{U}_{4i+5}^{2}\right]\\ =\frac{2×6.066×{10}^{-6}}{45×2000}×\left[7×{310.9998}^{2}+32×\left({310.8474}^{2}+{310.2607}^{2}+\cdots \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+12×\left({310.7041}^{2}+{310.1131}^{2}+\cdots \right)+32×\left({310.5560}^{2}+{309.9645}^{2}+\cdots \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+14×\left({310.4084}^{2}+{309.8182}^{2}+\cdots \right)\right]\\ =250.5253\text{\hspace{0.17em}}\text{mJ}\end{array}$

5.3.6. 误差计算

6. 结论

Figure 6. Simulation circuit diagram

Figure 7. Simulation waveform diagram

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