[Circuit] OP Amp Feedback 推導

電路知識：OP Amp
電子學 交大電子物理系 陳振芳教授

簡介：證明負回授導致虛短路

$$ V_{out} = (V_+ - V_-)\cdot A_o $$

理想運算放大器的特性：

• \(A \to \infty\)
• \(R_{in} \to \infty\)
• \(BW \to \infty\)
• \(R_{out} \to 0\)

Non-inverting 架構：

證明

$$ \begin{align*} V^+ &= V_i \\ V^- &= \frac{V_o\cdot R_1}{R_1+R_2} \\\\ V_o &= (V^+-V^-) \cdot A\\ &=\left(V_i-\frac{V_o\cdot R_1}{R_1+R_2}\right)\cdot A\\ \left(1+\frac{R_1 \cdot A}{R_1+R_2}\right)V_o&=AV_i\\ \frac{V_o}{V_i}&=\frac{A}{1+\frac{R_1 \cdot A}{R_1+R_2}}\\ \lim_{A \to \infty }\frac{V_o}{V_i}&=\frac{R_1+R_2}{R_1}\\\\ \lim_{A \to \infty }V^- &= \frac{V_o\cdot R_1}{R_1+R_2} \\ &= \frac{\frac{R_1+R_2}{R_1}V_i\cdot R_1}{R_1+R_2} \\ &= V_i\\ &= V^+ \end{align*} $$

若 A 有限呢？即可反求 \(V^-\)

$$ \text{Known} \begin{cases} V^+ &= V_i \\ V_o &= (V^+-V^-) \cdot A\\ \end{cases}\\ \begin{align*} V^-&= \frac{V^+ \cdot A-V_o}{A} \\ &=\frac{V_i \cdot A-V_o}{A}\\\\ \frac{V^-}{R_1}&=\frac{V_o}{R_1+R_2}\\ \frac{\frac{V_i \cdot A-V_o}{A}}{R_1}&=\frac{V_o}{R_1+R_2}\\\\ (R_1+R_2)(V_iA-V_o)&=AR_1V_o\\\\ A(R_1+R_2)V_i&=\left [ AR_1+(R_1+R_2) \right ]V_o\\\\ \frac{V_o}{V_i}&=\frac{A(R_1+R_2)}{AR_1+(R_1+R_2)}\\ &=\frac{R_1+R_2}{R_1+\frac{R_1+R_2}{A}}\\ &=\frac{\frac{R_1+R_2}{R_1}}{1+\frac{\frac{R_1+R_2}{R1}}{A}}\\ &=\frac{\frac{R_1+R_2}{R_1}}{1+\frac{1+\frac{R_2}{R1}}{A}}\\\\ \end{align*} $$

若 A \(\gg 1+\frac{R_2}{R1}\) 成立 \(\frac{R_1+R_2}{R_1} ~ \Rightarrow\) 得證虛短路

為何保證一定收斂呢？若從 \(V_o \Rightarrow V^-\) 一步一步求

$$ \text{Known} \begin{cases} V^+ &= V_i \\ V_o &= (V^+-V^-) \cdot A\\ \end{cases}\\\\ \begin{align*} V^- &= \frac{V^+ \cdot A-V_o}{A} \\ &=\frac{V_i \cdot A-V_o}{A}\\ &=V_i-\frac{V_o}{A}\\\\ \end{align*} $$ $$ \begin{align*} t=0 \Rightarrow V_{o,t=0} &=V_{o,t=0}\\ V^-_{t=0} &= V_i -\frac{V_{o,t=0}}{A}\\ t=1 \Rightarrow V_{o,t=1} &= V^-_{t=0} + \frac{V^-_{t=0}}{R_1} \cdot R_2\\ &=V_i -\frac{V_{o,t=0}}{A} + \frac{V_i -\frac{V_{o,t=0}}{A}}{R_1} \cdot R_2\\ &=V_i -\frac{V_{o,t=0}}{A} + \frac{V_i -\frac{V_{o,t=0}}{A}}{R_1} \cdot R_2\\ &=V_i\left (1 + \frac{R_2}{R_1} \right )-V_{o,t=0}\left ( \frac{1}{A}+ \frac{\frac{1}{A}}{R_1} \cdot R_2\right )\\ &=V_i\left (1 + \frac{R_2}{R_1} \right )-V_{o,t=0}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A}\\ V^-_{t=1} &= V_i -\frac{V_{o,t=1}}{A}\\ &= V_i -\frac{V_i\left (1 + \frac{R_2}{R_1} \right )-V_{o,t=0}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A}}{A}\\ t=2 \Rightarrow V_{o,t=2} &= V^-_{t=1} + \frac{V^-_{t=1}}{R_1} \cdot R_2\\ &=V_i\left (1 + \frac{R_2}{R_1} \right )-V_{o,t=1}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A}\\ &=V_i\left (1 + \frac{R_2}{R_1} \right )-\frac{\left[ V_i\left (1 + \frac{R_2}{R_1} \right )-V_{o,t=0}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A} \right]\left ( 1+ \frac{R_2}{R_1} \right )}{A}\\ V^-_{t=2} &= V_i -\frac{V_{o,t=2}}{A}\\ &= V_i -\frac{V_i\left (1 + \frac{R_2}{R_1} \right )-\frac{\left[ V_i\left (1 + \frac{R_2}{R_1} \right )-V_{o,t=0}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A} \right]\left ( 1+ \frac{R_2}{R_1} \right )}{A}}{A}\\ \end{align*} $$ $$ \vdots\\ $$

若 A \(\gg 1+\frac{R_2}{R1} ~ \Rightarrow\) 得證虛短路

但這個推導是有問題的，因為不符合實際情況
感覺上就像是有了 \(V_o\) 然後反向操作 OP，才有了 \(V^-\)

那若從 \(V^- \Rightarrow V_o\) 推導呢？

$$ \text{Known} \begin{cases} V^+ &= V_i \\ V_o &= (V^+-V^-) \cdot A\\ \end{cases}\\\\ \begin{align*} V_o &= (V^+-V^-) \cdot A\\ &=(V_i-V^-) \cdot A\\\\ \end{align*} $$ $$ \begin{align*} t=0 \Rightarrow V^-_{t=0} &= V^-_{t=0}\\ V_{o,t=0}&=(V_i-V^-_{t=0}) \cdot A \\ t=1 \Rightarrow V^-_{t=1} &= V_{o,t=0}\frac{R_1}{R_1+R_2}\\ V_{o,t=1}&=(V_i-V^-_{t=1}) \cdot A \\ &=\left ( V_i-V_{o,t=0}\frac{R_1}{R_1+R_2} \right )\cdot A \\ t=2 \Rightarrow V^-_{t=2} &= V_{o,t=1}\frac{R_1}{R_1+R_2}\\ V_{o,t=2}&=(V_i-V^-_{t=2}) \cdot A \\ &=(V_i-V_{o,t=1}\frac{R_1}{R_1+R_2}) \cdot A \\ &=\left [V_i-\left ( V_i-V_{o,t=0}\frac{R_1}{R_1+R_2} \right )\cdot A \cdot \frac{R_1}{R_1+R_2}\right ] \cdot A \\ &=V_i \cdot A\left ( 1- A \frac{R_1}{R_1+R_2} \right )+V_{o,t=0}\cdot A^2 \left (\frac{R_1}{R_1+R_2} \right )^2 \end{align*} $$ 若是照以上推導，會出現來來回回振盪，永不收斂。
為何？
因電阻的分壓反應時間遠小於 OP 的反應輸出
當 \(V_o\) 有一些些變化時，\(V^-\) 會立即隨之變化

為何從 \(V^- \Rightarrow V_o\) 推導看起來會正確的原因？
因為等同於求當下的真實狀態，什麼樣的 \(V^-\) 會造成目前的 \(V_o\)
然後這個 \(V^-\) 又會立即反應在 \(V_o\) 上

正確推導如下 $$ \begin{align*} t=0 \Rightarrow V^-_{t=0} &= V^-_{t=0}\\ V_{o,t=0}&=V_{o,t=0}\\\\ t=0^+ \Rightarrow V_{o,t=0^+}&=V_{o,t=0}+\bigtriangleup \left[(V_i-V^-_{t=0}) \cdot A\right] \\ V^-_{t=0^+} &= V_{o,t=0^+}\frac{R_1}{R_1+R_2}\\ &=\left\{V_{o,t=0}+\bigtriangleup \left[(V_i-V^-_{t=0}) \cdot A\right]\right\}\frac{R_1}{R_1+R_2}\\ t=0^{++} \Rightarrow V_{o,t=0^{++}}&=V_{o,t=0^+}+\bigtriangleup \left [(V_i-V^-_{t=0^+}) \cdot A\right] \\ &=\left \{ V_{o,t=0}+\bigtriangleup \left[(V_i-V^-_{t=0}) \cdot A\right] \right \}+\bigtriangleup \left[\left(V_i-\left\{V_{o,t=0}+\bigtriangleup \left[(V_i-V^-_{t=0}) \cdot A\right]\right\}\frac{R_1}{R_1+R_2}\right) \cdot A \right ] \\ V^-_{t=0^{++}} &= V_{o,t=0^{++}}\frac{R_1}{R_1+R_2}\\ &=\frac{R_1}{R_1+R_2}\left \{ \left \{ V_{o,t=0}+\bigtriangleup \left[(V_i-V^-_{t=0}) \cdot A\right] \right \}+\bigtriangleup \left[\left(V_i-\left\{V_{o,t=0}+\bigtriangleup \left[(V_i-V^-_{t=0}) \cdot A\right]\right\}\frac{R_1}{R_1+R_2}\right) \cdot A \right ] \right \}\\ \end{align*} $$ 將以上 \(\bigtriangleup\) 的部分，全簡化為 \(\bigtriangleup\) $$ \begin{align*} t=0 \Rightarrow V^-_{t=0} &= V^-_{t=0}\\ V_{o,t=0}&=V_{o,t=0}\\\\ t=0^+ \Rightarrow V_{o,t=0^+}&=V_{o,t=0}+\bigtriangleup \\ V^-_{t=0^+} &= V_{o,t=0^+}\frac{R_1}{R_1+R_2}\\ &=\left\{V_{o,t=0}+\bigtriangleup \right\}\frac{R_1}{R_1+R_2}\\ t=0^{++} \Rightarrow V_{o,t=0^{++}}&=V_{o,t=0^+}+\bigtriangleup \\ &=\left \{ V_{o,t=0}+\bigtriangleup \right \}+\bigtriangleup \\ V^-_{t=0^{++}} &= V_{o,t=0^{++}}\frac{R_1}{R_1+R_2}\\ &=\left \{ \left \{ V_{o,t=0}+\bigtriangleup \right \}+\bigtriangleup \right \}\frac{R_1}{R_1+R_2}\\ \end{align*} $$ 從以上式子，大概可以看出 \(V^-\) 會隨著 \(V_o\) 連動
\(V_o\) 上升 \(V^-\) 也上升； \(V_o\) 下降 \(V^-\) 也下降
直到穩定平衡

簡單模擬 by R

# 電路條件
vi = 5
r1 = 10
r2 = 10
A = 1000

# 初始值
v_ = 0
vo = 0

# 記錄
r_V_ = c()
r_Vo = c()

# 設定 step and 上升多少等份電壓
n = 1000
div = 10000

for(i in 1:n)
{
  vo = vo + A*(vi-v_)/div
  r_V_ = c(r_V_, v_)
  r_Vo = c(r_Vo, vo)
  v_ = vo/(r1+r2) * r1
}

plot(r_Vo, col="red", type="l",ylab="V", ylim = c(0,15))
lines(r_V_, col="green")
legend('topright', c('Vo','V-'), lty=1, col=c('red', 'green'), bty='n', cex=.75)

Inverting 架構：

證明

$$ \begin{align*} V^+&= 0 \\ V^-&= V_i+\frac{(V_o-V_i)}{R_1+R_2}R_1 \\\\ V_o&= (V^+-V^-) \cdot A\\ &=\left \{ 0-\left[V_i+\frac{(V_o-V_i)\cdot R_1}{R_1+R_2}\right] \right \} \cdot A\\ &=\frac{(V_i-V_o)\cdot R_1 \cdot A}{R_1+R_2}-V_i \cdot A \\ &=V_i \left(\frac{R_1 \cdot A}{R_1+R_2} - A\right )-V_o\frac{R_1 \cdot A}{R_1+R_2}\\\\ \left(1+ \frac{R_1 \cdot A}{R_1+R_2}\right )V_o&=\left(\frac{R_1 \cdot A}{R_1+R_2} - A\right )V_i\\ \frac{V_o}{V_i}&=\frac{\frac{R_1 \cdot A}{R_1+R_2} - A}{1+ \frac{R_1 \cdot A}{R_1+R_2}}\\\\ \lim_{A \to \infty }\frac{V_o}{V_i}&=\frac{\frac{R_1}{R_1+R_2} - 1}{\frac{R_1}{R_1+R_2}}\\ &=\frac{-R_2}{R_1}\\\\ \lim_{A \to \infty }V^-&= V_i+\frac{(V_o-V_i)}{R_1+R_2}R_1 \\ &=V_i+\frac{(\frac{-R_2}{R1}V_i-V_i)}{R_1+R_2}R_1 \\ &=V_i+\frac{-R_2V_i-R_1V_i}{R_1+R_2} \\ &=V_i-V_i\\ &=0\\ &=V^+\\ \end{align*} $$

若 A 有限呢？即可反求 \(V^-\)

$$ \text{Known} \begin{cases} V^+ &= 0 \\ V_o &= (V^+-V^-) \cdot A\\ \end{cases}\\\\ \begin{align*} V^-&= \frac{V^+ \cdot A-V_o}{A} \\ &=\frac{-V_o}{A}\\\\ \frac{V_i-V^-}{R_1}&=\frac{V^--V_o}{R_2}\\ \frac{V_i-\frac{-V_o}{A}}{R_1}&=\frac{\frac{-V_o}{A}-V_o}{R_2}\\ V_iR_2+\frac{V_o}{A}R_2&=\frac{-V_o}{A}R_1-V_oR_1\\ R_2V_i&=-\left ( \frac{R_1+R_2}{A}+R_1 \right )V_o\\ \frac{V_o}{V_i}&=\frac{R_2}{-\left ( \frac{R_1+R_2}{A}+R_1 \right )}\\ &=\frac{-\frac{R_2}{R_1}}{\left ( \frac{\frac{R_1+R_2}{R_1}}{A}+1 \right )}\\ &=\frac{-\frac{R_2}{R_1}}{\left ( \frac{1+\frac{R_2}{R_1}}{A}+1 \right )}\\\\ \end{align*} $$

若 A \(\gg (1+\frac{R_2}{R_1})\) 成立 \(\frac{-R_2}{R_1} ~ \Rightarrow\) 得證虛短路

為何保證一定收斂呢？若從 \(V_o \Rightarrow V^-\)，一步一步求

$$ \text{Known} \begin{cases} V^+ &= 0 \\ V_o &= (V^+-V^-) \cdot A\\ \end{cases}\\\\ \begin{align*} V^- &= \frac{V^+ \cdot A-V_o}{A} \\ &=-\frac{V_o}{A}\\\\ \end{align*} $$ $$ \begin{align*} t=0 \Rightarrow V_{o,t=0} &=V_{o,t=0}\\ V^-_{t=0} &= -\frac{V_{o,t=0}}{A}\\ t=1 \Rightarrow V_{o,t=1} &= V^-_{t=0} + \frac{V^-_{t=0}-V_i}{R_1} \cdot R_2\\ &= -\frac{V_{o,t=0}}{A} + \frac{ -\frac{V_{o,t=0}}{A}-V_i}{R_1} \cdot R_2\\ &=-V_i\frac{R2}{R1}-V_{o,t=0}\left ( \frac{1}{A}+ \frac{\frac{1}{A}}{R_1} \cdot R_2\right )\\ &=-V_i\frac{R2}{R1}-V_{o,t=0}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A}\\ V^-_{t=1} &= -\frac{V_{o,t=1}}{A}\\ &= -\frac{-V_i\frac{R2}{R1}-V_{o,t=0}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A}}{A}\\ t=2 \Rightarrow V_{o,t=2} &= V^-_{t=1} + \frac{V^-_{t=1}}{R_1} \cdot R_2\\ &=-V_i\frac{R2}{R1}-V_{o,t=1}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A}\\ &=-V_i\frac{R2}{R1}-\frac{\left [-V_i\frac{R2}{R1}-V_{o,t=0}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A} \right ]\left ( 1+ \frac{R_2}{R_1} \right )}{A}\\ V^-_{t=2} &= -\frac{V_{o,t=2}}{A}\\ &= -\frac{-V_i\frac{R2}{R1}-\frac{\left [-V_i\frac{R2}{R1}-V_{o,t=0}\frac{\left ( 1+ \frac{R_2}{R_1} \right )}{A} \right ]\left ( 1+ \frac{R_2}{R_1} \right )}{A}}{A}\\ \end{align*} $$ $$ \vdots\\ $$

若 A \(\gg 1+\frac{R_2}{R1} ~ \Rightarrow\) 得證虛短路

但這個推導是有問題的，因為不符合實際情況
感覺上就像是有了 \(V_o\) 然後反向操作 OP，才有了 \(V^-\)

那若從 \(V^- \Rightarrow V_o\) 推導呢？

$$ \begin{align*} t=0 \Rightarrow V^-_{t=0} &= V^-_{t=0}\\ V_{o,t=0}&=(0-V^-_{t=0}) \cdot A \\ &=-V^-_{t=0} \cdot A \\ t=1 \Rightarrow V^-_{t=1} &= V_i + (V_{o,t=0}-V_i)\frac{R_1}{R_1+R_2}\\ &=V_i\left ( 1- \frac{R_1}{R_1+R_2}\right )+V_{o,t=0}\frac{R_1}{R_1+R_2}\\ V_{o,t=1}&=-V^-_{t=1} \cdot A \\ &=-\left [ V_i\left ( 1- \frac{R_1}{R_1+R_2}\right )+V_{o,t=0}\frac{R_1}{R_1+R_2} \right ] \cdot A \\ t=2 \Rightarrow V^-_{t=2} &= V_i\left ( 1- \frac{R_1}{R_1+R_2}\right )+V_{o,t=1}\frac{R_1}{R_1+R_2}\\ &=V_i\left ( 1- \frac{R_1}{R_1+R_2}\right )+\left \{ -\left [ V_i\left ( 1- \frac{R_1}{R_1+R_2}\right )+V_{o,t=0}\frac{R_1}{R_1+R_2} \right ] \cdot A \right \}\frac{R_1}{R_1+R_2}\\ V_{o,t=2}&=-V^-_{t=2} \cdot A \\ &=-\left \{ V_i\left ( 1- \frac{R_1}{R_1+R_2}\right )+\left \{ -\left [ V_i\left ( 1- \frac{R_1}{R_1+R_2}\right )+V_{o,t=0}\frac{R_1}{R_1+R_2} \right ] \cdot A \right \}\frac{R_1}{R_1+R_2} \right \} \cdot A \\ \end{align*} $$ 若是照以上推導，會出現來來回回振盪，永不收斂。
為何？
因電阻的分壓反應時間遠小於 OP 的反應輸出
當 \(V_o\) 有一些些變化時，\(V^-\) 會立即隨之變化

為何從 \(V^- \Rightarrow V_o\) 推導看起來會正確的原因？
因為等同於求當下的真實狀態，什麼樣的 \(V^-\) 會造成目前的 \(V_o\)
然後這個 \(V^-\) 又會立即反應在 \(V_o\) 上

正確推導如下 $$ \begin{align*} t=0 \Rightarrow V^-_{t=0} &= V^-_{t=0}\\ V_{o,t=0}&=V_{o,t=0}\\\\ t=0^+ \Rightarrow V_{o,t=0^+}&=V_{o,t=0}+\bigtriangleup \left[(0-V^-_{t=0}) \cdot A\right] \\ V^-_{t=0^+} &= V_i+\left( V_{o,t=0^+}-V_i \right )\frac{R_1}{R_1+R_2}\\ &= V_i\left ( 1-\frac{R1}{R_1+R_2} \right )+V_{o,t=0^+}\frac{R1}{R_1+R_2}\\ &=V_i\left ( 1-\frac{R1}{R_1+R_2} \right )+\left \{ V_{o,t=0}+\bigtriangleup \left[(0-V^-_{t=0}) \cdot A\right] \right \}\frac{R1}{R_1+R_2}\\\\ t=0^{++} \Rightarrow V_{o,t=0^{++}}&=V_{o,t=0^+}+\bigtriangleup \left [(0-V^-_{t=0^+}) \cdot A\right] \\ &=\left \{ V_{o,t=0}+\bigtriangleup \left[(0-V^-_{t=0}) \cdot A\right] \right \}+\bigtriangleup \left [(0-\left \{ V_i\left ( 1-\frac{R1}{R_1+R_2} \right )+\left \{ V_{o,t=0}+\bigtriangleup \left[(0-V^-_{t=0}) \cdot A\right] \right \}\frac{R1}{R_1+R_2} \right \}) \cdot A\right] \\ V^-_{t=0^{++}} &= V_i\left ( 1-\frac{R1}{R_1+R_2} \right )+V_{o,t=0^{++}}\frac{R1}{R_1+R_2}\\ &=V_i\left ( 1-\frac{R1}{R_1+R_2} \right )+\left \{ \left \{ V_{o,t=0}+\bigtriangleup \left[(0-V^-_{t=0}) \cdot A\right] \right \}+\bigtriangleup \left [(0-\left \{ V_i\left ( 1-\frac{R1}{R_1+R_2} \right )+\left \{ V_{o,t=0}+\bigtriangleup \left[(0-V^-_{t=0}) \cdot A\right] \right \}\frac{R1}{R_1+R_2} \right \}) \cdot A\right] \right \}\frac{R1}{R_1+R_2}\\ \end{align*} $$ 將以上 \(\bigtriangleup\) 的部分，全簡化為 \(\bigtriangleup\) $$ \begin{align*} t=0 \Rightarrow V^-_{t=0} &= V^-_{t=0}\\ V_{o,t=0}&=V_{o,t=0}\\\\ t=0^+ \Rightarrow V_{o,t=0^+}&=V_{o,t=0}+\bigtriangleup \\ V^-_{t=0^+} &=V_i\left ( 1-\frac{R1}{R_1+R_2} \right )+\left \{ V_{o,t=0}+\bigtriangleup \right \}\frac{R1}{R_1+R_2}\\\\ t=0^{++} \Rightarrow V_{o,t=0^{++}}&=\left \{ V_{o,t=0}+\bigtriangleup \right \}+\bigtriangleup \\ V^-_{t=0^{++}} &=V_i\left ( 1-\frac{R1}{R_1+R_2} \right )+\left \{ \left \{ V_{o,t=0}+\bigtriangleup \right \}+\bigtriangleup \right \}\frac{R1}{R_1+R_2}\\ \end{align*} $$ 從以上式子，大概可以看出 \(V^-\) 會隨著 \(V_o\) 連動
\(V_o\) 上升 \(V^-\) 也上升； \(V_o\) 下降 \(V^-\) 也下降
直到穩定平衡

簡單模擬 by R

# 電路條件
vi = 5
r1 = 10
r2 = 10
A = 1000

# 初始值
v_ = 0
vo = 0

# 記錄
r_V_ = c()
r_Vo = c()

# 設定 step and 上升多少等份電壓
n = 1000
div = 10000

for(i in 1:n)
{
  vo = vo + A*(0-v_)/div
  r_V_ = c(r_V_, v_)
  r_Vo = c(r_Vo, vo)
  v_ = vi+(vo-vi)/(r1+r2) * r1
}

plot(r_Vo, col="red", type="l",ylab="V", ylim = c(-10,10))
lines(r_V_, col="green")
legend('topright', c('Vo','V-'), lty=1, col=c('red', 'green'), bty='n', cex=.75)

其餘類似電路，皆可用 Superposition 加上以上兩個基本電路得證虛短路