Op-Amp Circuits

To simplify the analysis of the Op-Amp circuits, we further make the following assumptions:

• The huge input resistance $r_{in}$ can be treated as infinity $r_{in}\rightarrow \infty$.
• The input current drawn by an op-amp is samll ( $10^{-9}\sim10^{-12}\;A$), and could be approximated to be zero $i^+=i^-=0$.
• The small output impedance $r_{out}$ can be treated as 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 $r_{out}$).
• Based on the fact that $v^-\approx v^+$, we could assume $v^-=v^+$, i.e., the virtual ground assumption.
• The bandwidth is large ($1 \sim 20MHz$).

Based on these approximations, an Op-Amp can be further simplified as the modeled shown on the right of the figure below, based on which the analysis of op-amp circuits can be much simplified, as shown in the following examples.

• Voltage follower (buffer)

  

  
  

  $$V_{out}=v^-\approx v^+=V_{in}$$
  
  
  As the output $V_{out}=V_{in}$ is the same as the input, why can't we replace this op-amp circuit by a piece of wire?

• Inverter

  

  Current into the op-amp is negligible, and $V^-\approx V^+=0$. Applying KCL to the node of $v^-$, we have
  

  $$\frac{V_{in}}{R_1}+\frac{V_{out}}{R_f}=0,\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;\; V_{out}=-\frac{R_f}{R_1}V_{in}$$
  
  

  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:
  

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

  

• Non-Inverting Amplifier

  

  
  

  $$V_{in}=V^+\approx V^-=V_{out}\frac{R_1}{R_1+R_f},\;\;\;\;\mbo... ...frac{R_1+R_f}{R_1}V_{in}=\left(1+\frac{R_f}{R_1}\right) V_{in}$$
  
  

• Summer-inverter

  

  Apply KCL to $V^-$:
  

  $$\sum_{k=1}^n \frac{V_k}{R_k}+\frac{V_{out}}{R_f}=0,\;\;\;\;\;... ...{k=1}^n \frac{V_k}{R_k} =- \sum_{k=1}^n \frac{R_f}{R_k} \;V_k$$
  
  

• Differential amplifier

  

  Define $V=V^-\approx V^+$, we get:
  

  $$\frac{V_1-V}{R_1}+\frac{V_{out}-V}{R_2}=0,\;\;\;\;\mbox{i.e.}... ... V_{out}=-\frac{R_2}{R_1}V_1+\left(1+\frac{R_2}{R_1}\right) V$$
  
  
  But as
  
  $$V\approx V^+=\frac{R_4}{R_3+R_4}V_2$$
  
  
  therefore
  
  $$V_{out}=-\frac{R_2}{R_1}V_1+\left(1+\frac{R_2}{R_1}\right)\frac{R_4}{R_3+R_4}V_2$$
  
  

  Consider some special cases:

  • If $R_1=R_3$ and $R_2=R_4$, we get
    
    $$V_{out}=\frac{R_2}{R_1}\;(V_2-V_1)$$
    
    

  • If $R_4=\infty$ (open circuit, and $R_3$ can be any value), then $V=V_2$ and we get
    
    $$V_{out}=-\frac{R_2}{R_1}V_1+\left(1+\frac{R_2}{R_1}\right)V_2$$
    
    
    This is a combination of inverter and a non-inverter amplifiers.
  • If $R_1=R_4=\infty$, then $V_{out}=V_2$, this is the follower.
  • If $R_4=0$ ($R_3=\infty$), then we get the inverter
    
    $$V_{out}=-\frac{R_2}{R_1}$$
    
    

  • If $R_3=0$ ($R_4=\infty$) and $V_1=0$, we get the non-inverter:
    
    $$V_{out}=\left(1+\frac{R_2}{R_1}\right)V_2$$
    
    

  Note 1: It is likely that both inputs are 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
  
  $$V_{out}=\frac{R_2}{R_1}\;(V_2-V_1)=\frac{R_2}{R_1}\;(v_2(t)-v_1(t))$$
  
  
  not affected by the common noise at all, i.e., the differential amplifier can suppress common-mode signal (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 two inputs, e.g., $V_2$ is connected to a constant voltage treated as a reference $V_{ref}$, then the differential amplifier can also be used as a level shifter. As
  

  $$\frac{V_1-V^-}{R_1}=\frac{V^-V_{out}}{R_2}$$
  
  
  we get
  
  $$V_{out}=-\frac{R_2}{R_1}V_1-\left(1+\frac{R_2}{R_1}\right)V^-$$
  
  
  But
  
  $$V^-=V^+=V_{ref}\frac{R_4}{R_3+R_4}$$
  
  
  we have
  
  $$V_{out}=-\frac{R_2}{R_1}V_1-V_{ref}\left(1+\frac{R_2}{R_1}\right)\frac{R_4}{R_3+R_4} =-\frac{R_2}{R_1}V_1-V_{shift}$$
  
  
  where
  
  $$V_{shift}=V_{ref}\frac{R_4(R_1+R_2)}{R_1(R_3+R_4)}$$
  
  
  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.

• Instrumentation Amplifier

  One drawback of the differential amplifier is that its input impedance ($R_3+R_4$) may not be high enough if the output impedance of the previous stage is not low enough. To overcome this problem, two non-inverters with high input resistance can be used each for one of the two inputs to the differential amplifier. The resulting circuit is shown below:

  

  The analysis of this circuit is very simple. As the output impedance of the non-inverter is low, the three op-amp circuit can be considered as three independent circuits. The outputs of the two non-inverters are:
  

  $$V'_1=V_1\left(1+\frac{R_f}{R_1}\right),\;\;\;\;\;\;\;\;\; V'_2=V_2\left(1+\frac{R_f}{R_1}\right)$$
  
  
  The output voltage of the differential amplifier is:
  
  $$V_{out}=\frac{R_4}{R_3}(V'_2-V'_1)=\frac{R_4}{R_3}\left(1+\frac{R_f}{R_1}\right)(V_2-V_1)$$
  
  
  Of course the two resistors $R_1$ can be combined to become $R_0=2R_1$, i.e., $R_1=R_0/2$, then the output can be written as:
  
  $$V_{out}=\frac{R_4}{R_3}\left(1+\frac{2R_f}{R_0}\right)(V_2-V_1)$$
  
  

  Alternatively, we consider the current going from $V'_1$ to $V_'_2$:
  

  $$\frac{V'_1-V_1}{R_f}=\frac{V_1-V_0}{R_1}=\frac{V_0-V_2}{R_1}=\frac{V_2-V'_2}{R_f}$$
  
  
  From the equation of the first two terms we get:
  
  $$V'_1=\left(1+\frac{R_f}{R_1}\right)V_1-\frac{R_f}{R_1}V_0$$
  
  
  From the equation of the second two terms we get:
  
  $$V'_2=\left(1+\frac{R_f}{R_1}\right)V_2-\frac{R_f}{R_1}V_0$$
  
  
  Using the equation of the differential amplifier above, we get the same result as above:
  
  $$V_{out}=\frac{R_4}{R_3}(V'_2-V'_1)=\frac{R_4}{R_3}\left(1+\frac{R_f}{R_1}\right)(V_2-V_1)$$
  
  

• Algebraic summer (inputs of different signs)

  

  Define $V\stackrel{\triangle}{=}V^+ \approx V^-$. Apply KCL to $V^-$ and $V^+$ we get:
  

  $$\frac{V_1-V}{R_1}+\frac{V_2-V}{R_2}+\frac{V_{out}-V}{R_f}=0,\... ...ox{and}\;\;\;\;\;\;\;\; \frac{V_3-V}{R_3}+\frac{V_4-V}{R_4}=0$$
  
  
  Solving the 2nd equation for $V$ we get:
  
  $$V=\frac{R_4}{R_3+R_4} V_3 + \frac{R_3}{R_3+R_4} V_4$$
  
  
  and substitute it into the first equation to get
  
  $$V_{out}=-\frac{R_f}{R_1}V_1-\frac{R_f}{R_2}V_2 +\left(\frac{... ...t)\left(\frac{R_4}{R_3+R_4} V_3+\frac{R_3}{R_3+R_4} v_4\right)$$
  
  

  

• A/D converter

  Without feedback, the output of an op-amp is $V_{out}=A(V^-^+)$. As $A$ is large, $V_{out}$ is saturated, equal to either the positive or the negative voltage 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 $V_{in}$ from 0 to 3 volts with resolution 1 V.

  

  Due to the voltage divider, the input voltages to the three op-amps are, respectively, 2.5V, 1.5V and 0.5V. The output of these op-amps are listed below for each of the input voltage levels. A digital logic circuit is then needed to convert the 3-bit output of the op-amps to the two-bit binary representation.

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

• First order system -- integrator and differentiator

  

  Integrator

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

  $$i_R+i_C=\frac{v_i}{R}+C\frac{d v_{out}}{dt}=0, \;\;\;\; \mbox{i.e.,}\;\;\;\;v_{out}=-\frac{1}{\tau} \int v_i dt$$
  
  
  where $\tau \stackrel{\triangle}{=}RC$. 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:
  

  $$i_R+i_C=\frac{v_{out}}{R}+C\frac{d v_i}{dt}=0,\;\;\;\; \mbox{i.e.,}\;\;\;\;v_{out}=-RC \frac{d v_i}{dt}=-\tau \frac{d v_i}{dt}$$
  
  
  In frequency domain, we have:
  
  $$H(j\omega)=-\frac{Z_2(j\omega)}{Z_1(j\omega)}=-\frac{R}{1/j\omega C} =-j\omega \tau$$
  
  

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

  

  Low-pass filter:
  

  $$H(j\omega)=-\frac{Z_2(j\omega)}{Z_1(j\omega)}=-\frac{R_2\;\ve... ...2}{R_1}\frac{1}{1+j\omega R_2C} =-H(0)\frac{1}{1+j\omega \tau}$$
  
  
  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.

  For example, when $\tau=10^{-3}$, $\omega_c=1/\tau=10^3$, the Bode plots are shown below:

  

  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.

  For example, when $\tau=10^{-6}$, $\omega_c=1/\tau=10^6$, the Bode plots are shown below:

  

  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 $\tau_1=R_1C_1$, $\tau_2=R_2C_2$, $\tau_3=R_2C_1$.

  For example, when $\tau_1=10^{-6}$, $\tau_2=10^{-8}$, $\tau_3=10^{-3}$, the Bode plots are shown below:

  

• 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
  

  $$\left\{ \begin{array}{l} Y_3(s)=Y_2(s)/s \Longrightarrow Y_... ... \Longrightarrow Y_0(s)=Y_1(s)s=Y_3(s)s^3 \end{array} \right.$$
  
  
  But we also have
  
  $$Y_0(s)=X(s)-(k_1Y_1(s)+k_2Y_2(s)+k_3Y_3(s))$$
  
  
  i.e.,
  
  $$X(s)=Y_0(s)+k_1Y_1(s)+k_2Y_2(s)+k_3Y_3(s)=(s^3+k_1s^2+k_2s+k_3) Y_3(s)$$
  
  
  we get the transfer function
  
  $$H(s)=\frac{Y_3(s)}{X(s)}=\frac{1}{s^3+k_1s^2+k_2s+k_3}$$
  
  

• Second order system by 2 integrators

  

  From the diagram, we can get
  

  $$\left\{ \begin{array}{ll} Y_2(s)=-c_2Y_1(s)/s \Longrightarr... ...\\ Y_0(s)=k_0 X(s)+k_1Y_1(s)+k_2Y_2(s) \end{array} \right.$$
  
  
  substituting the first two equations into the last one, we get
  
  $$\frac{s^2}{c_1c_2} Y_2(s)=k_0X(s)+k_1(-\frac{s}{c_2})Y_2(s)+k_2Y_2(s)$$
  
  
  from which we obtain the transfer function as
  
  $$H(s)=\frac{Y_2(s)}{X(s)}=\frac{k_o}{\frac{s^2}{c_1c_2}+\frac{s}{c_2}s-k_2} =\frac{k_oc_1c_2}{s^2+k_1c_1s-c_1c_2k_2}$$
  
  
  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 scalar which can take any value.

• Sallen-Key Topology

  The Sallen–Key topology is an electronic filter topology used to implement second-order active filters that is particularly valued for its simplicity.

  

  We represent the input and output in s-domain as $X(s)$ and $Y(s)$, respectively, and the voltage at node a as $W(s)$, and apply KCL to nodes a and b to get:
  

  $$\frac{W(s)-Y(s)}{Z_3}+\frac{W(s)-Y(s)}{Z_2}+\frac{W(s)-X(s)}{Z_1}=0$$
  
  
  
  
  $$\frac{Y(s)-W(s)}{Z_2}+\frac{Y(s)}{Z_4}=0$$
  
  
  Solving the second equation for $W(s)$ we get
  
  $$W(s)=Y(s)\frac{Z_2+Z_4}{Z_4},\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\; W(s)-Y(s)=Y(s)\frac{Z_2}{Z_4}$$
  
  
  Substituting this into the first equation we get
  
  $$H(s)=\frac{Y(s)}{X(s)}=\frac{Z_3Z_4}{Z_1Z_2+Z_1Z_3+Z_2Z_3+Z_3Z_4}$$
  
  

  Example 1 $Z_1=R_1$, $Z_2=R_2$, $Z_3=1/sC_1$, $Z_4=1/sC_2$, then we get a second order low-pass filter:
  

  $$H(s) = \frac{1/s^2C_1C_2}{R_1R_2+R_1/sC_1+R_2/sC_1+1/s^2C_1C_2} =\frac{1/R_1C_1R_2C_2}{s^2+s(R_1+R_2)/R_1R_2C_1+1/R_1C_1R_2C_2}$$

  $$= \frac{\omega_0^2}{s^2+2\zeta\omega_0\;s+\omega_0^2} =\frac{\omega_0^2}{s^2+\omega_0/Q \;s+\omega_0^2}$$

  
  
  where
  
  $$\omega_0=\frac{1}{\sqrt{R_1R_2C_1C_2}},\;\;\;\;\;\;\; 2\zeta... ..._1C_2}},\;\;\;\;\; Q=\frac{\sqrt{R_1R_2C_1C_2}}{(R_1+R_2)C_2}$$
  
  

  Example 2 $Z_1=1/sC_1$, $Z_2=1/sC_2$, $Z_3=R_1$, $Z_4=R_2$, then we get a second order high-pass filter:
  

  $$H(s) = \frac{R_1R_2}{1/s^2C_1C_2+R_1/sC_1+R_1/sC_2+R_1R_2} =\frac{s^2}{s^2+s\;(C_1+C_2)R_1/R_1R_2C_1C_2+1/R_1R_2C_1C_2}$$

  $$= \frac{s^2}{s^2+2\zeta\omega_0\;s+\omega_0^2} =\frac{s^2}{s^2+\omega_0/Q \;s+\omega_0^2}$$

  
  
  where
  
  $$\omega_0=\frac{1}{\sqrt{R_1R_2C_1C_2}},\;\;\;\;\;\;\; 2\zeta... ...C_2}},\;\;\;\;\;\; Q=\frac{\sqrt{R_1R_2C_1C_2}}{(C_1+C_2)R_1}$$