Op-Amp Circuits

The analysis of various op-amp circuits can be further simplified by the following approximation:

• The huge voltage gain $A$ ( $10^5 \sim 10^9$) is considered to be infinity, i.e., for a finite output $v_{out}$, the input is close to zero $v_{in}=v^+ - v^-=v_{out}/A \approx 0$ or $v^+=v^-$;
• The huge input resistance $r_{in}$ ( $10^6\sim 10^{12} \;\Omega$ depending on whether BJT or FET is used) could also be considered to be infinity $r_{in}\rightarrow \infty$;
• The small input current drawn by an op-amp ( $10^{-9}\sim10^{-12}\;A$) is approximately zero $i^+=i^-=0$;
• The small output impedance $r_{out}$ (50 $\Omega$) is approximately zero $r_{out}\approx 0$, i.e., the output $v_{out}$ is not affected by the load $R_L$ (so long as it is much greater than 50 $\Omega$).
• The CMRR is large ($>100\;dB$).
• The bandwidth is large ($1 \sim 20MHz$).

Based on these approximations, the analysis of op-amp circuits can be much simplified than before, as shown in the following examples.

• Inverter

  

  As the input current is negligible, i.e., $i^-=i_1+i_2 \approx 0$, applying KCL to the node of $v^-$, we have
  

  $$i_1+i_2= \frac{v^- -v_i}{R_1}+\frac{v^- -v_o}{R_f}=0$$
  
  
  but also since $v^-\approx v^+=0$, we have
  
  $$\frac{v_o}{v_i}=-\frac{R_f}{R_1}$$
  
  
  In particular, if $R_f=R_1$, we have
  
  $$v_o=-v_i$$
  
  

  In general, $R_1$ and $R_2$ in the inverter can be replaced by two networks (with impedances $Z_1$ and $Z_2$ respectively) containing resistors and capacitors and the analysis of the circuit can be carried out easily in frequency domain:
  

  $$H(j\omega)=\frac{V_o(j\omega)}{V_i(j\omega)}=-\frac{Z_2(j\omega)}{Z_1(j\omega)}$$
  
  
  This is a convenient way to design filters of various frequency characteristics.

  

• Summer-inverter

  

  As the input current $i$ is negligible, we have
  

  $$i=\sum_{j=1}^n \frac{v^-v_j}{R_j}+\frac{v^-v_o}{R_f} \approx 0$$
  
  
  but as $v^-\approx v^+=0$, we also have
  
  $$v_o=-R_f \sum_{j=1}^n \frac{v_j}{R_j} = - \sum_{j=1}^n k_j v_j$$
  
  
  where $k_j \stackrel{\triangle}{=}R_f/R_j$ ( $j=1,2,\cdots,n$) are the $n$ coefficients.

• Summer with different signs

  

  First define
  

  $$v\stackrel{\triangle}{=}v^+ \approx v^-$$
  
  
  and note $i_1 \approx 0$ and $i_2 \approx 0$, we have these simultaneous equations
  
  $$\left\{ \begin{array}{ll} (v-v_1)/R_1+(v-v_o)/R_f=0 & (a) \\ (v-v_2)/R_2+(v-v_3)/R_3=0 & (b) \end{array} \right.$$
  
  
  Solving (b) to get
  
  $$v=\frac{R_2}{R_2+R_3} v_3 + \frac{R_3}{R_2+R_3}v_2$$
  
  
  and substitute it into (a) to get
  
  $$v_o=(\frac{R_f}{R_1}+1)[\frac{R_2}{R_2+R_3} v_3+\frac{R_3}{R_2+R_3} v_2]-\frac{R_f}{R_1}v_1 =-k_1v_1+k_2v_2+k_3v_3$$
  
  
  where
  
  $$k_1=\frac{R_f}{R_1},\;\;\;\;\;\;k_2=(\frac{R_f}{R_1}+1)\frac{... ...2+R_3}, \;\;\;\;\;\;k_3=(\frac{R_f}{R_1}+1)\frac{R_2}{R_2+R_3}$$
  
  

  

• Differential amplifier

  

  As the input currents are zero, the currents flowing through $R_1$ and $R_2$ are equal for both sides, i.e.,
  

  $$\frac{V_1-V^-}{R_1}=\frac{V^-V_o}{R_2},\;\;\;\;\;\; \frac{V_2-V^+}{R_3}=\frac{V^+-0}{R_4}$$
  
  
  Assumeing $R_1=R_3$ and $R_2=R_4$ and $V^+=V^-$, we subtract the second equation from the first to get:
  
  $$\frac{V_1-V_2}{R_1}=\frac{-V_o}{R_2},\;\;\;\; V_o=-\frac{R_2}{R_1}\;(V_1-V_2)=\frac{R_2}{R_1}\;(V_2-V_1)$$
  
  
  Note 1: It is likely that both inputs are to subjected to some common noise $n(t)$ (such as interference of 60Hz power supply):
  
  $$V_1=v_1(t)+n(t), \;\;\;\;\;V_2=v_2(t)+n(t)$$
  
  
  In this case the output is
  
  $$V_o=\frac{R_2}{R_1}\;(V_2-V_1)=\frac{R_2}{R_1}\;(v_2(t)-v_1(t))$$
  
  
  not affected by the common noise at all, i.e., the differential amplifier can suppress common-mode singal (such as the noise signal $n(t)$) while amplify the differential-mode signal (such as $v_1(t)$ and $v_2(t)$).

  Note 2: If one of the input, e.g., $V_2$ is connected to a constant voltage treated as a reference $V_{ref}$, then the differential amplifier can also be used aa a level shifter. As
  

  $$\frac{V_1-V^-}{R_1}=\frac{V^-V_o}{R_2}$$
  
  
  we get
  
  $$V_o=-\frac{R_2}{R_1}V_1-(1+\frac{R_2}{R_1})V^-$$
  
  
  But
  
  $$V^-=V^+=V_{ref}\frac{R_4}{R_3+R_4}$$
  
  
  we have
  
  $$V_o=-\frac{R_2}{R_1}V_1-V_{ref}(1+\frac{R_2}{R_1})\frac{R_4}{R_3+R_4} =-\frac{R_2}{R_1}V_1-V_{shift}$$
  
  
  where
  
  $$V_{shift}=V_{ref}\frac{(R_1+R_2)R_4}{R_1(R_3+R_4)}$$
  
  
  In other words, the output is $-R_2/R_1$ times the input $V_1$, shifted by a constant value $V_{shift}$. This level-shifter circuit can be used to change the DC level of the signal (e.g., removal of DC component) as well as amplifying it.

• Instrument Amplifier (Homework)

  

  Express the output voltage $V_o$ as a function of both inputs $V_1$ and $V_2$. Find the gain $A=V_{out}/(V_1-V_2)$.

  Hint: Analyze the three opamps separately. Assume the voltage at the middle point of $R_1$ is zeor, i.e., the $v^-$ input of each of the two opamps is grounded through $R_2/2$.

  Answer

• A/D converter

  Without feedback, the output of an op-amp is $V_o=A(V^-^+)$. As $A$ is large, $V_o$ is usually saturated, equal to approximately either the positive or the negative power supply, depending on whether or not $V^+$ is greater than $V^-$. These two possible outputs, positive and negative, can be treated as ``1'' and ``0'' of the binary system. The figure shows an A/D converter built by three op-amps to measure voltage $V_i$ from 0 to 3 volts with resolution 1 V.

  

  Due to the voltage dividor, the input voltages to the three opamps are, respectively, 2.5V, 1.5V and 0.5V. The output of these opamps are listed below for each of the voltages:

  Voltage (volts)	0	1	2	3
  Opamps Outputs	000	001	011	111
  Binary Representation	00	01	10	11

  A digital logic circuit is then needed to convert the 3-bit output of the op-amps to the two-bit binary representation.

• First order system -- integrator and differentiator

  

  Integrator

  In time domain, as $v^-=v^+=0$ and $i_R+i_C=0$, we have
  

  $$i_R=-i_C \Longrightarrow \frac{v_i}{R}=-C\frac{d v_C}{dt},\;\... ...\mbox{i.e.,}\;\;\;\;v_o=v_C=-\frac{1}{\tau} \int v_i dt+v_C(0)$$
  
  
  where $v_C(0)$ is the initial voltage across $C$ and $\tau \stackrel{\triangle}{=}RC$. In frequency domain, we have:
  
  $$H(j\omega)=-\frac{Z_2(j\omega)}{Z_1(j\omega)}=-\frac{1/j\omega C}{R} =-\frac{1}{j\omega RC}=-\frac{1}{j\omega \tau}$$
  
  

  Differentiator

  If we swap the resistor and the capacitor, we get in time domain:
  

  $$i_R=-i_C \Longrightarrow \frac{v_o}{R}=-C\frac{d v_i}{dt},\;\... ...{i.e.,}\;\;\;\;v_o=-RC \frac{d v_i}{dt}=-\tau \frac{d v_i}{dt}$$
  
  
  In frequency domain, we have:
  
  $$H(j\omega)=-\frac{Z_2(j\omega)}{Z_1(j\omega)}=-\frac{R}{1/j\omega C} =-j\omega \tau$$
  
  
  where $\tau=RC$.

• First order systems - low-pass and high-pass filters

  

  Low-pass filter:
  

  $$H(j\omega)=-\frac{Z_2(j\omega)}{Z_1(j\omega)}=-\frac{R_2\vert... ...2}{R_1}\frac{1}{1+j\omega R_2C} =-H(0)\frac{1}{1+j\omega \tau}$$
  
  
  where $H(0)=R_2/R_1$, $\tau=R_2C$. Intuitively, when frequency is high, $Z_2(j\omega)$ is small and the effect of negative feedback is strong, therefore the output is low.

  High-pass filter:
  

  $$H(j\omega)=-\frac{Z_2(j\omega)}{Z_1(j\omega)}=-\frac{R_2}{R_1... ..._1C}{1+j\omega R_1C} =-H(0)\frac{j\omega \tau}{1+j\omega \tau}$$
  
  
  where $H(0)=R_2/R_1$, $\tau=R_1C$. Intuitively, when frequency is low $Z_1(j\omega)$ is large and the signal is difficult to pass, therefore the output is low.

  Band-pass filter:

  

  
  

  $$H(j\omega)=-\frac{Z_2(j\omega)}{Z_1(j\omega)} =-\frac{R_2\ver... ... =-\frac{j\omega \tau_3}{(1+j\omega \tau_1)(1+j\omega \tau_2)}$$
  
  
  where $\tau_1=R_1C_1$, $\tau_2=R_2C_2$, $\tau_3=R_2C_1$.

• Higher order systems

  Higher than first order systems can be built with multiple integrators, as shown here for a third order system:

  

  From the diagram, we can get
  

  $$\left\{ \begin{array}{l} y_3=y_2/s \Longrightarrow y_2=y_3s... ...y_1=y_0/s \Longrightarrow y_0=y_1s=y_3s^3 \end{array} \right.$$
  
  
  But we also have
  
  $$y_0=x-(k_1y_1+k_2y_2+k_3y_3)$$
  
  
  i.e.,
  
  $$x=y_0+k_1y_1+k_2y_2+k_3y_3=(s^3+k_1s^2+k_2s+k_3) y_3$$
  
  
  we get the transfer function
  
  $$H(s)=\frac{y_3}{x}=\frac{1}{s^3+k_1s^2+k_2s+k_3}$$
  
  

• Second order system by 2 integrators

  

  From the diagram, we can get
  

  $$\left\{ \begin{array}{ll} y_2=-c_2y_1/s \Longrightarrow y_1... ...2y_2/c_1c_2 \\ y_0=k_0 x+k_1y_1+k_2y_2 \end{array} \right.$$
  
  
  substituting the first two equations into the last one, we get
  
  $$\frac{s^2}{c_1c_2} y_2=k_0x+k_1(-\frac{s}{c_2})y_2+k_2y_2$$
  
  
  from which we obtain the transfer function as
  
  $$H(s)=\frac{y_2}{x}=\frac{k_o}{\frac{s^2}{c_1c_2}+\frac{s}{c_2}s-k_2} =\frac{k_oc_1c_2}{s^2+k_1c_1s-c_1c_2k_2}$$
  
  
  which is a second order system. In particular, if $c_1=c_2=c$, we have
  
  $$H(s)=k_0\frac{c^2}{s^2+c k_1s-k_2c^2}$$
  
  
  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 can let $c=\omega_n$ and $k_1=2\zeta$. Moreover, $k_2<0$, i.e., the feedback from the output should be negative. $k_0$ is a constant scaler which can take any value.