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🎉auto update by Gmeek action
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Meekdai committed Nov 22, 2023
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Expand Up @@ -93,7 +93,7 @@ <h3>补偿回路</h3>
<p>下图3为温度闭环原理简图</p>
<p><a target="_blank" rel="noopener noreferrer" href="https://github.com/Meekdai/meekdai.github.io/assets/11755104/89fe8c59-92fc-44e4-a136-da81254a0bbd"><img src="https://github.com/Meekdai/meekdai.github.io/assets/11755104/89fe8c59-92fc-44e4-a136-da81254a0bbd" alt="7f3b1e89gy1fzpqd9eljog20fj09ygln" style="max-width: 100%;"></a></p>
<p>图三所示TEC热闭环中为2Hz的交叉补偿的例子。尽量选择高阻值的R3以使积分电容C2尽可能小。然而这种方法也有弊端,它会使得PID的增益变小。因为我们必须在70mHz插入一个零点,我们使用关系:</p>
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="f677ce40468db6667f2ea1d554238c02">$$ 
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="2d0b4583f5eeb65f98e746c85a241272">$$ 
FZ1 = \frac{1}{2π×C2×R3}
$$</math-renderer></p>
<p>前文已经说过,FZ1 = 70mHZ,如果选择R3 =243kΩ,那么C2 就应为9.36μF。我们选择10μF设计。现在我们选择R1 =10kΩ。这使得前端放大器(U2)有足够的增益以减少反射积分器(U1)的错误,同时保持合理的电容器的尺寸。 现在,我们必须插入一个零点以抹去第二个在1Hz的TEC响应曲线极点。因为我们希望得到良好的相位裕度,零插入所需的交叉频率除以至少5或0.4Hz。这给出了一个在交叉频率的一个更好的相位裕度。然后通过放置由R1创建的、至少5倍于交叉频率,或者10Hz的极点终止该零点。这限制了积分器的的增益。 所以,因为: </p>
Expand All @@ -102,7 +102,7 @@ <h3>补偿回路</h3>
$$<br>
 <br>
并且FZ2 = 0.4Hz和R2 =510KΩ,我们可以算出C1 =0.78μF。我们选择C1为1μF。为了计算出R1的大小,我们使用关系:</p>
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="f677ce40468db6667f2ea1d554238c02">$$
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="2d0b4583f5eeb65f98e746c85a241272">$$
F3 = \frac{1}{2π×C1×R1}
$$</math-renderer></p>
<p>因为F3 = 10HZ,C1 =1μF,这样我们可以知道R1 =15.9kΩ。使用10kΩ就可以提供更好的相位裕度。然后,我们必须设置衰减频率在30Hz。因为R3 =243kΩ,FC = 30Hz,并且:<br>
Expand All @@ -117,16 +117,16 @@ <h3>补偿回路</h3>
$$<br>
 <br>
因为C1 = 1μF,FC = 2Hz,那么:</p>
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="f677ce40468db6667f2ea1d554238c02">$$  
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="2d0b4583f5eeb65f98e746c85a241272">$$  
XC = j79.6kΩ 
$$</math-renderer></p>
<p>U1的幅度增益G为: </p>
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="f677ce40468db6667f2ea1d554238c02">$$
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="2d0b4583f5eeb65f98e746c85a241272">$$
G = | R3 / XC | 
$$</math-renderer></p>
<p>由于R3 =243kΩ,XC =j79.6kΩ,G = 3.05或9.7分贝。要想对此增益进行全面的分析,就不应忽视R1、R2、C2和C3,此时G = 3.11或9.8分贝,从而验证了我们的假设。现在,我们必须在前端再提供20.3分贝的增益以完成2Hz处的增益交叉。 <br>
前端增益部分有两个功能:它减少积分器U1产生的误差,和采集从热敏电阻传来的温度信息。 由R4、R5和U2(图3)所设定的直流增益需要足够高以防止该误差信号被PID补偿电路消除。由于R4 =10kΩ,R5 =100kΩ,那么前端增益是11或20.8分贝,足以压倒9.7分贝的PID补偿部分。从热敏电阻传来的温度读数以误差信号的形式被传递给PID控制部分。该误差信号表示的是实际温度和所需温度(设定值)之间的差异。U2输出的误差信号可以按照下式计算:</p>
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="f677ce40468db6667f2ea1d554238c02">$$
<p><math-renderer class="js-display-math" style="display: block" data-static-url="https://github.githubassets.com/static" data-run-id="2d0b4583f5eeb65f98e746c85a241272">$$
ERROR=1.5V×( \frac{R_T×(R4+R5)}{R4×(10KΩ+R_T)})×V_{SET}×( \frac{R5}{R4})
$$</math-renderer></p>
<p>这里的1.5V是通过10kΩ的电阻器连接到热敏电阻的参考值,$R_T$是电阻 热敏电阻,$V_{SET}$是电压设定点。</p>
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