Operational Amplifier Circuits

An operational amplifier can be model as shown in the figure:

For simplicity of analysis, we make the following assumptions:

• $r_-, r_+, r_i \rightarrow \infty$, therefore input currents $i^+=i^-\approx 0$.
• $r_o \rightarrow 0$,
• $A \rightarrow \infty$,
• $v_i=v^-v^+ \rightarrow 0$, i.e., $v^- \approx v^+$,
• $v_o=-A v_i$ is finite.

Various linear, first order, second order and higher order systems can be constructed with operational amplifiers as the building blocks.

• Inverter

  

  As the input resistances are huge, the input current is negligible $i=i_1+i_2 \approx 0$ and $v^- \approx v^+=0$, we have
  

  $$i_1+i_2= \frac{v_1}{R_1}+\frac{v_o}{R_f}=0$$
  
  
  i.e.,
  
  $$\frac{v_o}{v_1}=-\frac{R_f}{R_1}\stackrel{\triangle}{=}-k$$
  
  

• Summer-inverter

  

  As the input current is negligible and $v^- \approx v^+=0$, we have
  

  $$\sum_{j=1}^n \frac{v_j}{R_j}+\frac{v_o}{R_f} \approx 0, \;\;\... ... v_o=-\sum_{j=1}^n \frac{R_f}{R_j}v_j = - \sum_{j=1}^n k_j v_j$$
  
  
  where $k_j \stackrel{\triangle}{=}\frac{R_f}{R_j}$ ($j=1,\cdots,n$) are the $n$ coefficients.

• Summer with different signs

  

  Define $v\stackrel{\triangle}{=}v^+ \approx v^-$ and note $i_1 \approx 0$ and $i_2 \approx 0$, we have these simultaneous equations
  

  $$\left\{ \begin{array}{ll} (v_2-v)/R_2+(v_{out}-v)/R_f=0 & (a) \\ (v_1-v)/R_1+(v_0-v)/R_0=0 & (b) \end{array} \right.$$
  
  
  We solve (b) for $v$ to get
  
  $$v=\frac{R_1}{R_1+R_0} v_0 + \frac{R_0}{R_1+R_0}v_1$$
  
  
  and substitute it into (a) to get
  
  $$v_{out}=(\frac{R_f}{R_2}+1)[\frac{R_1}{R_1+R_0} v_0+\frac{R_0}{R_1+R_0} v_1]-\frac{R_f}{R_2}v_2 =k_0v_0+k_1v_1-k_2v_2$$
  
  
  where
  
  $$k_0=(\frac{R_f}{R_2}+1)\frac{R_1}{R_1+R_0},\;\;\;\; k_1=(\frac{R_f}{R_2}+1)\frac{R_0}{R_1+R_0},\;\;\;\; k_2=\frac{R_f}{R_2}$$
  
  

• Integrator

  

  The resistors in previous discussion can be generalized to impedances. Similar to the inverter,
  

  $$H(j\omega)=\frac{V_o(j\omega)}{V_i(j\omega)}=-\frac{Z_C(j\omega)}{Z_R(j\omega)} =\frac{-1}{j\omega CR}$$
  
  
  If we define
  
  $$\xi \stackrel{\triangle}{=}\frac{1}{RC}$$
  
  
  we have an integrator with frequency response function
  
  $$H(j\omega)=-\frac{\xi}{j\omega}$$
  
  
  If all appearances of $j\omega$ are replaced by $s$, then the frequency response function $H(j\omega)$ becomes $H(s)$, the transfer function, which is the description of the system in Laplace domain:
  
  $$H(s)=-\frac{\xi}{s}$$
  
  

• First-order system

  

  From the diagram, we get
  

  $$y_0=x+ky,\;\;\;\;\;y=-\frac{\xi}{j\omega}y_0, \;\;\;\mbox{i.e.}\;\;\; y_0=-\frac{j\omega}{\xi} y$$
  
  
  Equating the two expressions of $y_0$, we get
  
  $$x+ky=-\frac{j\omega}{\xi} y$$
  
  
  or
  
  $$H(j\omega)=\frac{y}{x}=\frac{1}{-k-j\omega/\xi}=-\frac{\xi}{\xi k+j\omega}$$
  
  
  In Laplace domain, the transfer function is
  
  $$H(s)=-\frac{\xi}{\xi k+s}$$
  
  

• Second order system

  

  From the diagram, we get
  

  $$\left\{ \begin{array}{ll} y_2=-\xi_2y_1/s,\;\;\;i.e.,\;\;\;... .../\xi_1\xi_2 \\ y_0=k_0 x+k_1y_1+k_2y_2 \end{array} \right.$$
  
  
  Substituting the first two equations into the last one, we get
  
  $$\frac{s^2}{\xi_1\xi_2} y_2=k_0x+k_1(-\frac{s}{\xi_2})y_2+k_2y_2$$
  
  
  from which we obtain the transfer function as
  
  $$H(s)=\frac{y_2}{x}=\frac{k_o\xi_1\xi_2}{s^2+k_1\xi_1s-\xi_1\xi_2k_2}$$
  
  
  This is a second order system. In particular, if we let $\xi_1=\xi_2=\xi$, we get
  
  $$H(s)=k_0\frac{\xi^2}{s^2+k_1 \xi s-k_2\xi^2}$$
  
  
  Comparing this with the canonical 2nd order system transfer function
  
  $$H(s)=\frac{\omega_n^2}{s^2+2\zeta \omega_n s+\omega_n^2}$$
  
  
  we see that we could let $\xi=\omega_n$ and $k_1=2\zeta$. Moreover, $k_2<0$, i.e., the feedback from the output should be negative. $k_0$ is a constant scalar which can take any value.

• Higher order systems

  Similarly, higher order systems can be built with more integrators, as shown here for a third order system: