[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：

• $A\to \mathrm{\infty }$
• $R_{in}\to \mathrm{\infty }$
• $BW\to \mathrm{\infty }$
• $R_{out}\to 0$

Non-inverting 架構：

證明

$$\begin{array}{rl}{V}^{+}& =V_i\\ V^{-}& =\frac{V_o\cdot R_1}{R_1+R_2}\\ \\ V_o& =(V^{+}-V^{-})\cdot A\\ & =(V_i-\frac{V_o\cdot R_1}{R_1+R_2})\cdot A\\ (1+\frac{R_1\cdot A}{R_1+R_2})V_o& =A{V}_i\\ \frac{V_o}{V_i}& =\frac{A}{1+\frac{R_1\cdot A}{R_1+R_2}}\\ \underset{A\to \mathrm{\infty }}{lim}\frac{V_o}{V_i}& =\frac{R_1+R_2}{R_1}\\ \\ \underset{A\to \mathrm{\infty }}{lim}{V}^{-}& =\frac{V_o\cdot R_1}{R_1+R_2}\\ & =\frac{\frac{R_1+R_2}{R_1}{V}_i\cdot R_1}{R_1+R_2}\\ & =V_i\\ & =V^{+}\end{array}$$

若 A 有限呢？即可反求 $V^{-}$

$$\begin{gathered} \text{Known}\{\begin{array}{ll}{V}^{+}& =V_i\\ V_o& =(V^{+}-V^{-})\cdot A\end{array} \\ \begin{array}{rl}{V}^{-}& =\frac{V^{+}\cdot A-V_o}{A}\\ & =\frac{V_i\cdot A-V_o}{A}\\ \\ \frac{V^{-}}{R_1}& =\frac{V_o}{R_1+R_2}\\ \frac{\frac{V_i\cdot A-V_o}{A}}{R_1}& =\frac{V_o}{R_1+R_2}\\ \\ (R_1+R_2)(V_{i}A-V_o)& =A{R}_1{V}_o\\ \\ A(R_1+R_2)V_i& =[A{R}_1+(R_1+R_2)]V_o\\ \\ \frac{V_o}{V_i}& =\frac{A(R_1+R_2)}{A{R}_1+(R_1+R_2)}\\ & =\frac{R_1+R_2}{R_1+\frac{R_1+R_2}{A}}\\ & =\frac{\frac{R_1+R_2}{R_1}}{1+\frac{\frac{R_1+R_2}{R1}}{A}}\\ & =\frac{\frac{R_1+R_2}{R_1}}{1+\frac{1+\frac{R_2}{R1}}{A}}\\ \end{array} \end{gathered}$$

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

$$\begin{gathered} \text{Known}\{\begin{array}{ll}{V}^{+}& =V_i\\ V_o& =(V^{+}-V^{-})\cdot A\end{array} \\ \begin{array}{rl}{V}^{-}& =\frac{V^{+}\cdot A-V_o}{A}\\ & =\frac{V_i\cdot A-V_o}{A}\\ & =V_i-\frac{V_o}{A}\\ \end{array} \end{gathered}$$ $$\begin{array}{rl}t=0\Rightarrow V_{o,t=0}& =V_{o,t=0}\\ V_{t=0}^{-}& =V_i-\frac{V_{o,t=0}}{A}\\ t=1\Rightarrow V_{o,t=1}& =V_{t=0}^{-}+\frac{V_{t=0}^{-}}{R_1}\cdot R_2\\ & =V_i-\frac{V_{o,t=0}}{A}+\frac{V_i-\frac{V_{o,t=0}}{A}}{R_1}\cdot R_2\\ & =V_i-\frac{V_{o,t=0}}{A}+\frac{V_i-\frac{V_{o,t=0}}{A}}{R_1}\cdot R_2\\ & =V_i(1+\frac{R_2}{R_1})-V_{o,t=0}(\frac{1}{A}+\frac{\frac{1}{A}}{R_1}\cdot R_2)\\ & =V_i(1+\frac{R_2}{R_1})-V_{o,t=0}\frac{(1+\frac{R_2}{R_1})}{A}\\ V_{t=1}^{-}& =V_i-\frac{V_{o,t=1}}{A}\\ & =V_i-\frac{V_i(1+\frac{R_2}{R_1})-V_{o,t=0}\frac{(1+\frac{R_2}{R_1})}{A}}{A}\\ t=2\Rightarrow V_{o,t=2}& =V_{t=1}^{-}+\frac{V_{t=1}^{-}}{R_1}\cdot R_2\\ & =V_i(1+\frac{R_2}{R_1})-V_{o,t=1}\frac{(1+\frac{R_2}{R_1})}{A}\\ & =V_i(1+\frac{R_2}{R_1})-\frac{[V_i(1+\frac{R_2}{R_1})-V_{o,t=0}\frac{(1+\frac{R_2}{R_1})}{A}](1+\frac{R_2}{R_1})}{A}\\ V_{t=2}^{-}& =V_i-\frac{V_{o,t=2}}{A}\\ & =V_i-\frac{V_i(1+\frac{R_2}{R_1})-\frac{[V_i(1+\frac{R_2}{R_1})-V_{o,t=0}\frac{(1+\frac{R_2}{R_1})}{A}](1+\frac{R_2}{R_1})}{A}}{A}\end{array}$$ $$⋮$$

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？

$$\begin{gathered} \text{Known}\{\begin{array}{ll}{V}^{+}& =V_i\\ V_o& =(V^{+}-V^{-})\cdot A\end{array} \\ \begin{array}{rl}{V}_o& =(V^{+}-V^{-})\cdot A\\ & =(V_i-V^{-})\cdot A\\ \end{array} \end{gathered}$$ $$\begin{array}{rl}t=0\Rightarrow V_{t=0}^{-}& =V_{t=0}^{-}\\ V_{o,t=0}& =(V_i-V_{t=0}^{-})\cdot A\\ t=1\Rightarrow V_{t=1}^{-}& =V_{o,t=0}\frac{R_1}{R_1+R_2}\\ V_{o,t=1}& =(V_i-V_{t=1}^{-})\cdot A\\ & =(V_i-V_{o,t=0}\frac{R_1}{R_1+R_2})\cdot A\\ t=2\Rightarrow V_{t=2}^{-}& =V_{o,t=1}\frac{R_1}{R_1+R_2}\\ V_{o,t=2}& =(V_i-V_{t=2}^{-})\cdot A\\ & =(V_i-V_{o,t=1}\frac{R_1}{R_1+R_2})\cdot A\\ & =[V_i-(V_i-V_{o,t=0}\frac{R_1}{R_1+R_2})\cdot A\cdot \frac{R_1}{R_1+R_2}]\cdot A\\ & =V_i\cdot A(1-A\frac{R_1}{R_1+R_2})+V_{o,t=0}\cdot A^2{\left(\frac{R_1}{R_1+R_2}\right)}^2\end{array}$$ 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 $$\begin{array}{rl}t=0\Rightarrow V_{t=0}^{-}& =V_{t=0}^{-}\\ V_{o,t=0}& =V_{o,t=0}\\ \\ t=0^{+}\Rightarrow V_{o,t=0^{+}}& =V_{o,t=0}+△[(V_i-V_{t=0}^{-})\cdot A]\\ V_{t=0^{+}}^{-}& =V_{o,t=0^{+}}\frac{R_1}{R_1+R_2}\\ & =\{V_{o,t=0}+△[(V_i-V_{t=0}^{-})\cdot A]\}\frac{R_1}{R_1+R_2}\\ t=0^{++}\Rightarrow V_{o,t=0^{++}}& =V_{o,t=0^{+}}+△[(V_i-V_{t=0^{+}}^{-})\cdot A]\\ & =\{V_{o,t=0}+△[(V_i-V_{t=0}^{-})\cdot A]\}+△[(V_i-\{V_{o,t=0}+△[(V_i-V_{t=0}^{-})\cdot A]\}\frac{R_1}{R_1+R_2})\cdot A]\\ V_{t=0^{++}}^{-}& =V_{o,t=0^{++}}\frac{R_1}{R_1+R_2}\\ & =\frac{R_1}{R_1+R_2}\{\{V_{o,t=0}+△[(V_i-V_{t=0}^{-})\cdot A]\}+△[(V_i-\{V_{o,t=0}+△[(V_i-V_{t=0}^{-})\cdot A]\}\frac{R_1}{R_1+R_2})\cdot A]\}\end{array}$$ 將以上 $△$ 的部分，全簡化為 $△$ $$\begin{array}{rl}t=0\Rightarrow V_{t=0}^{-}& =V_{t=0}^{-}\\ V_{o,t=0}& =V_{o,t=0}\\ \\ t=0^{+}\Rightarrow V_{o,t=0^{+}}& =V_{o,t=0}+△\\ V_{t=0^{+}}^{-}& =V_{o,t=0^{+}}\frac{R_1}{R_1+R_2}\\ & =\{V_{o,t=0}+△\}\frac{R_1}{R_1+R_2}\\ t=0^{++}\Rightarrow V_{o,t=0^{++}}& =V_{o,t=0^{+}}+△\\ & =\{V_{o,t=0}+△\}+△\\ V_{t=0^{++}}^{-}& =V_{o,t=0^{++}}\frac{R_1}{R_1+R_2}\\ & =\{\{V_{o,t=0}+△\}+△\}\frac{R_1}{R_1+R_2}\end{array}$$ 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

$$\begin{array}{rl}{V}^{+}& =0\\ V^{-}& =V_i+\frac{(V_o-V_i)}{R_1+R_2}{R}_1\\ \\ V_o& =(V^{+}-V^{-})\cdot A\\ & =\{0-[V_i+\frac{(V_o-V_i)\cdot R_1}{R_1+R_2}]\}\cdot A\\ & =\frac{(V_i-V_o)\cdot R_1\cdot A}{R_1+R_2}-V_i\cdot A\\ & =V_i(\frac{R_1\cdot A}{R_1+R_2}-A)-V_o\frac{R_1\cdot A}{R_1+R_2}\\ \\ (1+\frac{R_1\cdot A}{R_1+R_2})V_o& =(\frac{R_1\cdot A}{R_1+R_2}-A)V_i\\ \frac{V_o}{V_i}& =\frac{\frac{R_1\cdot A}{R_1+R_2}-A}{1+\frac{R_1\cdot A}{R_1+R_2}}\\ \\ \underset{A\to \mathrm{\infty }}{lim}\frac{V_o}{V_i}& =\frac{\frac{R_1}{R_1+R_2}-1}{\frac{R_1}{R_1+R_2}}\\ & =\frac{-R_2}{R_1}\\ \\ \underset{A\to \mathrm{\infty }}{lim}{V}^{-}& =V_i+\frac{(V_o-V_i)}{R_1+R_2}{R}_1\\ & =V_i+\frac{(\frac{-R_2}{R1}{V}_i-V_i)}{R_1+R_2}{R}_1\\ & =V_i+\frac{-R_2{V}_i-R_1{V}_i}{R_1+R_2}\\ & =V_i-V_i\\ & =0\\ & =V^{+}\end{array}$$

若 A 有限呢？即可反求 $V^{-}$

$$\begin{gathered} \text{Known}\{\begin{array}{ll}{V}^{+}& =0\\ V_o& =(V^{+}-V^{-})\cdot A\end{array} \\ \begin{array}{rl}{V}^{-}& =\frac{V^{+}\cdot A-V_o}{A}\\ & =\frac{-V_o}{A}\\ \\ \frac{V_i-V^{-}}{R_1}& =\frac{V^{-}-V_o}{R_2}\\ \frac{V_i-\frac{-V_o}{A}}{R_1}& =\frac{\frac{-V_o}{A}-V_o}{R_2}\\ V_i{R}_2+\frac{V_o}{A}{R}_2& =\frac{-V_o}{A}{R}_1-V_o{R}_1\\ R_2{V}_i& =-(\frac{R_1+R_2}{A}+R_1)V_o\\ \frac{V_o}{V_i}& =\frac{R_2}{-(\frac{R_1+R_2}{A}+R_1)}\\ & =\frac{-\frac{R_2}{R_1}}{(\frac{\frac{R_1+R_2}{R_1}}{A}+1)}\\ & =\frac{-\frac{R_2}{R_1}}{(\frac{1+\frac{R_2}{R_1}}{A}+1)}\\ \end{array} \end{gathered}$$

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

$$\begin{gathered} \text{Known}\{\begin{array}{ll}{V}^{+}& =0\\ V_o& =(V^{+}-V^{-})\cdot A\end{array} \\ \begin{array}{rl}{V}^{-}& =\frac{V^{+}\cdot A-V_o}{A}\\ & =-\frac{V_o}{A}\\ \end{array} \end{gathered}$$ $$\begin{array}{rl}t=0\Rightarrow V_{o,t=0}& =V_{o,t=0}\\ V_{t=0}^{-}& =-\frac{V_{o,t=0}}{A}\\ t=1\Rightarrow V_{o,t=1}& =V_{t=0}^{-}+\frac{V_{t=0}^{-}-V_i}{R_1}\cdot R_2\\ & =-\frac{V_{o,t=0}}{A}+\frac{-\frac{V_{o,t=0}}{A}-V_i}{R_1}\cdot R_2\\ & =-V_i\frac{R2}{R1}-V_{o,t=0}(\frac{1}{A}+\frac{\frac{1}{A}}{R_1}\cdot R_2)\\ & =-V_i\frac{R2}{R1}-V_{o,t=0}\frac{(1+\frac{R_2}{R_1})}{A}\\ V_{t=1}^{-}& =-\frac{V_{o,t=1}}{A}\\ & =-\frac{-V_i\frac{R2}{R1}-V_{o,t=0}\frac{(1+\frac{R_2}{R_1})}{A}}{A}\\ t=2\Rightarrow V_{o,t=2}& =V_{t=1}^{-}+\frac{V_{t=1}^{-}}{R_1}\cdot R_2\\ & =-V_i\frac{R2}{R1}-V_{o,t=1}\frac{(1+\frac{R_2}{R_1})}{A}\\ & =-V_i\frac{R2}{R1}-\frac{[-V_i\frac{R2}{R1}-V_{o,t=0}\frac{(1+\frac{R_2}{R_1})}{A}](1+\frac{R_2}{R_1})}{A}\\ V_{t=2}^{-}& =-\frac{V_{o,t=2}}{A}\\ & =-\frac{-V_i\frac{R2}{R1}-\frac{[-V_i\frac{R2}{R1}-V_{o,t=0}\frac{(1+\frac{R_2}{R_1})}{A}](1+\frac{R_2}{R_1})}{A}}{A}\end{array}$$ $$⋮$$

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？

$$\begin{array}{rl}t=0\Rightarrow V_{t=0}^{-}& =V_{t=0}^{-}\\ V_{o,t=0}& =(0-V_{t=0}^{-})\cdot A\\ & =-V_{t=0}^{-}\cdot A\\ t=1\Rightarrow V_{t=1}^{-}& =V_i+(V_{o,t=0}-V_i)\frac{R_1}{R_1+R_2}\\ & =V_i(1-\frac{R_1}{R_1+R_2})+V_{o,t=0}\frac{R_1}{R_1+R_2}\\ V_{o,t=1}& =-V_{t=1}^{-}\cdot A\\ & =-[V_i(1-\frac{R_1}{R_1+R_2})+V_{o,t=0}\frac{R_1}{R_1+R_2}]\cdot A\\ t=2\Rightarrow V_{t=2}^{-}& =V_i(1-\frac{R_1}{R_1+R_2})+V_{o,t=1}\frac{R_1}{R_1+R_2}\\ & =V_i(1-\frac{R_1}{R_1+R_2})+\{-[V_i(1-\frac{R_1}{R_1+R_2})+V_{o,t=0}\frac{R_1}{R_1+R_2}]\cdot A\}\frac{R_1}{R_1+R_2}\\ V_{o,t=2}& =-V_{t=2}^{-}\cdot A\\ & =-\{V_i(1-\frac{R_1}{R_1+R_2})+\{-[V_i(1-\frac{R_1}{R_1+R_2})+V_{o,t=0}\frac{R_1}{R_1+R_2}]\cdot A\}\frac{R_1}{R_1+R_2}\}\cdot A\end{array}$$ 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 $$\begin{array}{rl}t=0\Rightarrow V_{t=0}^{-}& =V_{t=0}^{-}\\ V_{o,t=0}& =V_{o,t=0}\\ \\ t=0^{+}\Rightarrow V_{o,t=0^{+}}& =V_{o,t=0}+△[(0-V_{t=0}^{-})\cdot A]\\ V_{t=0^{+}}^{-}& =V_i+(V_{o,t=0^{+}}-V_i)\frac{R_1}{R_1+R_2}\\ & =V_i(1-\frac{R1}{R_1+R_2})+V_{o,t=0^{+}}\frac{R1}{R_1+R_2}\\ & =V_i(1-\frac{R1}{R_1+R_2})+\{V_{o,t=0}+△[(0-V_{t=0}^{-})\cdot A]\}\frac{R1}{R_1+R_2}\\ \\ t=0^{++}\Rightarrow V_{o,t=0^{++}}& =V_{o,t=0^{+}}+△[(0-V_{t=0^{+}}^{-})\cdot A]\\ & =\{V_{o,t=0}+△[(0-V_{t=0}^{-})\cdot A]\}+△[(0-\{V_i(1-\frac{R1}{R_1+R_2})+\{V_{o,t=0}+△[(0-V_{t=0}^{-})\cdot A]\}\frac{R1}{R_1+R_2}\})\cdot A]\\ V_{t=0^{++}}^{-}& =V_i(1-\frac{R1}{R_1+R_2})+V_{o,t=0^{++}}\frac{R1}{R_1+R_2}\\ & =V_i(1-\frac{R1}{R_1+R_2})+\{\{V_{o,t=0}+△[(0-V_{t=0}^{-})\cdot A]\}+△[(0-\{V_i(1-\frac{R1}{R_1+R_2})+\{V_{o,t=0}+△[(0-V_{t=0}^{-})\cdot A]\}\frac{R1}{R_1+R_2}\})\cdot A]\}\frac{R1}{R_1+R_2}\end{array}$$ 將以上 $△$ 的部分，全簡化為 $△$ $$\begin{array}{rl}t=0\Rightarrow V_{t=0}^{-}& =V_{t=0}^{-}\\ V_{o,t=0}& =V_{o,t=0}\\ \\ t=0^{+}\Rightarrow V_{o,t=0^{+}}& =V_{o,t=0}+△\\ V_{t=0^{+}}^{-}& =V_i(1-\frac{R1}{R_1+R_2})+\{V_{o,t=0}+△\}\frac{R1}{R_1+R_2}\\ \\ t=0^{++}\Rightarrow V_{o,t=0^{++}}& =\{V_{o,t=0}+△\}+△\\ V_{t=0^{++}}^{-}& =V_i(1-\frac{R1}{R_1+R_2})+\{\{V_{o,t=0}+△\}+△\}\frac{R1}{R_1+R_2}\end{array}$$ 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