Operational Amplifier

Circuit Schematic

Although the schematic of an Op-amp looks complicated, its operation can be simply modeled by a voltage amplifier with three parameters including input resistance $r_{in}$ (usually large), output resistance $r_{out}$ (usually small), and the open-circuit gain $A$ (usually very large), as show in the figure:

The internal voltage souce depends on both the inverting input $v^-$ and the non-inverting input $v^+$:

$$v_{internal}=A_d (v^+ - v^-)+A_c \frac{1}{2}(v^+ + v^-)$$

where $A_d$ is the differential-mode gain and $A_c$ is the common-mode gain. In general it is desired that $A_d\rightarrow \infty$ and $A_c\rightarrow 0$, i.e., the output is only proportional to the difference of the inputs. The common-mode rejection ratio (CMRR) is defined as the ratio between differential-mode gain and common-mode gain:

$$CMRR=20\;\log_{10} (\frac{A_d}{A_c})$$

We consider the following two typical Op-Amp circuits to show how to carry out circuit analysis.

• Voltage follower: The input is connected to the positive input while the output is directly connected to the negative input (100% negative feedback). The parameters of this circuit can be found by the model shown in the figure.

  

  • Open-circuit voltage gain: Assume an ideal source voltage $v_s$ ($R_s=0$) is applied to the input of the circuit and the output port is open circuit $R_L=\infty$. Then applying KVL to the loop, we get
    
    $$v_s=(r_{in}+r_{out})i_{in}+Ar_{in}i_{in}=[(A+1)r_{in}+r_{out}]i_{in}$$
    
    
    Note that the internal voltage source is $A(v^+-v^-)=r_{in}i_{in}$. The output voltage is:
    
    $$v_{out}=Ar_{in}i_{in}+r_{out}i_{in}=(Ar_{in}+r_{out})i_{in}$$
    
    
    and the open-circuit voltage gain is:
    
    $$A_{oc}=\frac{v_{out}}{v_s}=\frac{Ar_{in}+r_{out}}{(A+1)r_{in}+r_{out}}$$
    
    
    Since $A»1$, $A_{oc}\approx 1$ is approximately unity.
  • Input resistance: Now a load $R_L$ is connected to the output port. Applying KVL to the two loops we get:
    
    $$\left\{ \begin{array}{l} v_s=(r_{in}+r_{out})i_{in}-r_{out}i... ...}i_{in}=(r_{out}+R_L)i_{out}-r_{out}i_{in} \right. \end{array}$$
    
    
    Solving these two equations for $i_{in}$ and $i_{out}$ we get
    
    $$i_{in}=v_s/[(A+1)r_{in}+r_{out}\frac{R_L-Ar_{in}}{R_L+r_{out}}]$$
    
    
    and the input resistance is
    

    $$R_{in} = \frac{v_s}{i_{in}} =(A+1)r_{in}+r_{out}\frac{R_L-Ar_{in}}{R_L+r_{out}} =r_{in}\frac{r_{out}+(A+1)R_L}{R_L+r_{out}}+\frac{R_L r_{out}}{R_L+r_{out}}$$

    $$= r_{in}\frac{r_{out}+(A+1)R_L}{R_L+r_{out}}+R_L\vert\vert r_{out}$$

    
    
    Note that $R_{in}$ is affected by the load $R_L$. As usually $R_L » r_{out}$ and $r_{in}»r_{out}$, the above is approximately
    
    $$R_{in}\approx Ar_{in}$$
    
    

  • Output resistance: This is the ratio between the open-circuit voltage and the short-circuit current $R_{out}=v_{oc}/i_{sc}$. First, find open-circuit voltage $v_{oc}$ (with $R_L=\infty$ and $i_{out}=0$):
    
    $$v_s=(R_s+r_{in}+r_{out})i_{in}+Ar_{in}i_{in},\;\;\;\mbox{i.e.,}\;\;\;\; i_{in}=\frac{v_s}{R_s+(A+1)r_{in}+r_{out}}$$
    
    
    The voltage appearing at the output port is
    
    $$v_{oc}=Ar_{in}i_{in}+r_{out}i_{in} =v_s\frac{Ar_{in}+r_{out}}{R_s+(A+1)r_{in}+r_{out}}$$
    
    
    Second, find short-circuit current $i_{sc}=i_{out}$ ($R_L=0$) by applying KVL to the two loops:
    
    $$\left\{ \begin{array}{l} v_s=(R_s+r_{out}+(A+1)r_{in})i_{in}... ..._{in}i_{in}=r_{out} i_{out}-r_{out} i_{in} \right. \end{array}$$
    
    
    Solving these two equation we get:
    
    $$i_{in}=\frac{v_s}{R_s+r_{in}}$$
    
    
    
    
    $$i_{out}=i_{sc}=v_s\frac{Ar_{in}+r_{out}}{r_{out}(R_s+r_{in})}$$
    
    
    Now the output resistance can be found:
    
    $$R_{out}=\frac{v_{oc}}{i_{sc}} =\frac{r_{out}(R_s+r_{in})}{R_s+(A+1)r_{in}+r_{out}}$$
    
    
    Note that $R_{out}$ is affected by the internal resistance $R_s$ of the sourc. Since $(A+1)r_{in}»R_s+r_{out}$ and $r_{in}»R_s$, we have
    
    $$R_{out}\approx \frac{r_{out}}{A}$$
    
    

  To summarize, we see that the voltage follower has unit voltage gain, but much increased input resistance $R_{in}\approx A r_{in}$ and much reduced output resistance $R_{out}\approx r_{out}/A$. Typically, $R_{in}$ is $10^{10} \Omega$ and $R_{out}$ is $10^{-3} \Omega$.

  Example:

  

  The figure on the left shows a circuit represented by an ideal voltate source $V_s$ in series with an internal resistance $R_s$ (Thevenin's theorem), with a load $R_L$. The voltage delivered to the load by this non-ideal source is
  

  $$v_{out}=v_s \frac{R_L}{R_L+R_s}$$
  
  
  The output voltage across the load is only a fraction of the voltage due to the voltage drop across the internal resistance $R_s$. If it is desired for the output voltage to be as close to the source as possible, the internal resistance $R_s$ has to be small while the load resistance $R_L$ has to be large (lighter load). However, given $R_s$ and $R_L$, it is still possible for the output voltage to be very close to the source if a voltage follower is used as a buffer between the source and the load, as shown in the figure on the right. The voltage follower is modeled by its input and output resistances $R_{in}$ and $R_{out}$, as well as its voltage gain $A_{oc}$, and the output voltage can be obtained after two levels of voltage dividers:
  
  $$v_{out}=A_{oc} v_{in} \frac{R_L}{R_{out}+R_L}= A_{oc} v_s \frac{R_{in}}{R_s+R_{in}} \frac{R_L}{R_{out}+R_L}$$
  
  
  As $R_{in}$ is huge, $R_{in}/(R_s+R_{in})\approx 1$, also, as $R_{out}$ is very small, $R_L/(R_{out}+R_L) \approx 1$, and $A_{oc}\approx 1$, therefore $v_{out} \approx v_s$, i.e., one hundred percent of the source voltage is delivered to the load.

• Inverting Amilifier

  

  As the analysis of the circut using full model of the op-amp is very involved, certain approximation is made the simplify the analysis.

  • Open-circuit voltage gain: We assume $r_{in}\rightarrow \infty$. Apply an ideal voltage source $v_s$ ($R_s=0$) to node $v^-=v_s$ ($v^+=0$), and find the input current to be:
    
    $$i_{in}=\frac{v_s-Av_{in}}{R_1+R_f+r_{out}} =\frac{v_s+A(v_s-R_1 i_{in})}{R_1+R_f+r_{out}}$$
    
    
    where $v_{in}=v^+-v^-=-v^-=-(v_s-R_1 i_{in})$. Solving this for $i_{in}$, we get:
    
    $$i_{in}=v_s \frac{A+1}{(A+1)R_1+R_f+r_{out}}$$
    
    
    The output voltage can be found to be:
    
    $$v_{out}=v_s-(R_1+R_f) i_{in} =v_s[1-\frac{(A+1)(R_1+R_f)}{(A... ...+R_f+r_{out}}] =v_s\frac{r_{out}-AR_f}{(A+1)R_1+R_f+r_{out}}$$
    
    
    i.e.,
    
    $$A_{oc}=\frac{v_{out}}{v_s}=\frac{r_{out}-AR_f}{(A+1)R_1+R_f+r_{out}} \approx - \frac{R_f}{R_1}$$
    
    
    Alternatively, if we further approximate $v^-\approx v^+=0$, then we can easily find $v_{out}$ given input $V_s$ by applying KCL at negative input node:
    
    $$\frac{v_s}{R_1}+\frac{v_{out}}{R_f} = 0$$
    
    
    we get the same $A_{oc}=v_{out}/v_s=-R_f/R_1$.

  • Input resistance: Again apply an ideal voltage source $v_s$ ($R_s=0$) to the circuit and the input resistance $R_{in}$ is found as the ratio of $v_s$ to the input current $i_{in}$. Apply KCL to the node of $v^-$ to get
    
    $$\frac{v^-v_s}{R_1}+\frac{v^-}{r_{in}}+\frac{v^-+Av^-}{R_f+r_{out}}=0$$
    
    
    Solving for $v^-$ to get
    
    $$v^-=v_s\frac{r_{in}(R_f+r_{out})}{(R_1+r_{in})(R_f+r_{out})+(A+1)R_1r_{in}}$$
    
    
    The input current is
    

    $$i_{in} = \frac{v_s-v^-}{R_1}=\frac{v_s}{R_1}[1-\frac{r_{in}(R_f+r_{out})}{(R_1+r_{in})(R_f+r_{out})+(A+1)R_1r_{in}}]$$

    $$= v_s\frac{R_f+r_{out}+(A+1)r_{in}}{(R_1+r_{in})(R_f+r_{out})+(A+1)R_1r_{in}}$$

    
    
    Therefore
    
    $$R_{in}=\frac{v_s}{i_{in}}=\frac{[R_f+r_{out}+(A+1)R_1]r_{in}+R_1(R_f+r_{out})}{R_f+r_{out}+(A+1)r_{in}}$$
    
    
    Usually, $r_{out}\approx 0$, we get
    
    $$R_{in}\approx \frac{v_s}{i_{in}}=\frac{[R_f+(A+1)R_1]r_{in}+R_1R_f}{R_f+(A+1)r_{in}}$$
    
    
    Moreover, as $A»1$, we have
    
    $$R_{in}\approx R_1$$
    
    

  • Output resistance: Here we assume $r_{in}\rightarrow \infty$ to simplify the analysis. First, to find short-circuit output current, we apply KCL to the output node to get:
    
    $$i_{sc}=\frac{-A v^-}{r_{out}}+\frac{v^-}{R_f} =v^-\frac{r_{out}-AR_f}{r_{out}R_f}$$
    
    
    but as $v^-=v_s R_f/(R_s+R_1+R_f)$, we have
    
    $$i_{sc}=v^-\frac{r_{out}-AR_f}{r_{out}R_f} =v_s\frac{R_f}{R_... ...f}{r_{out}R_f} =v_s\frac{r_{out}-AR_f}{(R_s+R_1+R_f)r_{out}}$$
    
    
    Next we find the open-circuit output voltage:
    
    $$v_{oc}=\frac{v^-(-Av^-)}{R_f+r_{out}} r_{out}-Av^- =v^- \frac{r_{out}-AR_f}{R_f+r_{out}}$$
    
    
    But $v^-$ is related to $v_s$ by:
    
    $$\frac{v_s-v^-}{R_s+R_1}=\frac{v^-(-Av^-)}{R_f+r_{out}}$$
    
    
    which can be solved for $v^-$
    
    $$v^-=v_s \frac{R_f+r_{out}}{(A+1)(R_s+R_1)+R_f+r_{out}}$$
    
    
    i.e.,
    
    $$v_{oc}=v^- \frac{r_{out}-AR_f}{R_f+r_{out}} =v_s \frac{r_{out}-AR_f}{(A+1)(R_s+R_1)+R_f+r_{out}}$$
    
    
    Now we get
    
    $$R_{out}=\frac{v_{oc}}{i_{sc}} =\frac{(R_s+R_1+R_f)r_{out}}{(... ...+R_f+r_{out}} \approx \frac{(R_s+R_1+R_f)r_{out}}{A(R_s+R_1)}$$
    
    
    The approximation is due to the assumption that $A»1$. In particular, when $R_s=0$, we have
    
    $$R_{out}\approx \frac{R_1+R_f}{R_1} \frac{r_{out}}{A}$$
    
    

• Non-Inverting Amilifier (Homework)

  

  Find the three parameters of this non-inverting amplifier:

  • input resistance $R_{in}$,
  • output resistance $R_{out}$,
  • open-circuit voltage gain $A_{oc}$.
  Assume $R_L=\infty$ in all three cases. Moreover, assume $r_{in}=\infty$ for $R_{out}$ and $A_{oc}$, and $r_{out}=0$ for $R_{in}$. Note that when $R_f=0$, this non-inverting amplifier will become a voltage follower, in terms of all three parameters. Verify your results by checking if this is the case.

  Answer