Find the three parameters of this non-inverting amplifier: open-circuit voltage gain $G_{oc}$, input resistance $R_{in}$ and output resistance $R_{out}$.

• Voltage gain:
  • First, we assume $r_{out}=0$, and apply an ideal voltage source $v_s$ ($R_s=0$) to the positive input so that $v^+=v_s$, and denote the voltage across $r_{in}$ by $v_0=v^+-v^-$, i.e., $v^-=v_s-v_0$. Applying KCL to the node of $v^-$, we get

    $$\frac{-v_0}{r_{in}}+\frac{v_s-v_0}{R_1}+\frac{v_s-v_0-Av_0}{R_f}=0$$ (1)

    Solving this for $v_0$, we get:

    $$v_0=v_s \frac{r_{in}(R_f+R_1)}{R_1R_f+r_{in}R_f+r_{in}R_1(A+1)}$$ (2)

    The open-circuit voltage gain is:

    $$G_{oc}=\frac{v_{out}}{v_{in}}\approx \frac{Av_0}{v_s} =\frac{Ar_{in}(R_f+R_1)}{R_1R_f+r_{in}R_f+r_{in}R_1(A+1)}$$ (3)

    As $A\gg 1$ and $r_{in}$ is usually very large, we have

    $$G_{oc}\approx \frac{R_1+R_f}{R_1}$$ (4)

  • Second, assume $r_{in}\rightarrow \infty$ but $r_{out}\ne 0$, we have
    

    $$v_{out} = A(v^+-v^-)\frac{R_f+R_1}{r_{out}+R_f+R_1} =A(v^+-v^-)\frac{R_f+R_1}{r_{out}+R_f+R_1}$$

    $$= A\left(v_s-v_{out}\frac{R_1}{R_1+R_f}\right) \frac{R_f+R_1}{r_{out}+R_f+R_1}$$

    
    Solving for $v_{out}$ we get

    $$v_{out} =v_s \frac{A(R_f+R_1)}{r_{out}+R_f+(A+1)R_1}$$ (5)

    and

    $$G_{oc}=\frac{v_{out}}{v_s}=\frac{A(R_1+R_f)}{r_{out}+R_f+(A+1)R_1} \approx \frac{A(R_1+R_f)}{AR_1}=\frac{R_1+R_f}{R_1}$$ (6)

    The approximation is due to $A\gg 1$ and $r_{out}\ll R_f+R_1$, i.e., $(R_f+R_1)(r_{out}+R_f+R_1)\approx 1$.

  • If we could assume both $r_{in}\rightarrow \infty$ and $r_{out}=0$, we can apply KCL to the $v^-$ node to get

    $$\frac{v^-}{R_1}+\frac{v^--v_{out}}{R_f}=0, \;\;\;\;\; \;\;\;\;\; v_{out}=\frac{R_1+R_f}{R_1} v^-$$ (7)

    But as $v^-\approx v^+=v_s$, we get the same result for $G_{oc}$.

  In particular, when $R_f=0$, $G_{oc}=1$ and the circut becomes the voltage follower.

• Input resistance: We let $v^+=v_s$ (with $R_s=0$) and assume the input current is $i_{in}$, then we have $v^+-v^-=r_{in}i_{in}$, i.e., $v^-=v^+-r_{in}i_{in}=v_s-r_{in}i_{in}$. Applying KCL to the node of $v^-$ we get:

  $$i_{in}-\frac{v_s-r_{in}i_{in}}{R_1}-\frac{v_s-r_{in}i_{in}-A r_{in}i_{in}}{R_f+r_{out}}=0$$ (8)

  Solving this we get

  $$i_{in}=\frac{v_s(R_f+r_{out}+R_1)}{(R_f+r_{out})(R_1+r_{in})+(A+1)r_{in} R_1}$$ (9)

  and
  

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

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

  
  The same result can be obtained if we use loop current method. As $A\gg 1$ and $r_{out} \approx 0$, we get

  $$R_{in} \approx \frac{(A+1)R_1+R_f}{R_1+R_f}r_{in}+R_1\vert\vert R... ...{in}\frac{R_1}{R_1+R_f}+R_1\vert\vert R_f \approx A r_{in}\;\frac{R_1}{R_1+R_f}$$ (10)

  Moreover, when $R_f=0$, $R_{in}\approx A r_{in}$ as in the voltage follower case.

• Output resistance: To simplify the analysis we still assume $r_{in}=\infty$. First, as shown above, the open-circuit output voltage is

  $$v_{oc}=\frac{A(R_f+R_1)}{r_{out}+R_f+(A+1)R_1}v_s$$ (11)

  Second, we find the short-circuit current, i.e., output port is shorted with $v_{out}=0$, we have $v^-=v_{out}R_1/(R_1+R_f)=0$, and $v^+-v^-=v_s$,

  $$i_{sc}=\frac{A(v^+-v^-)}{r_{out}}=\frac{Av_s}{r_{out}}$$ (12)

  Now the output resistance can be obtained as:
  

  $$R_{out} = \frac{v_{oc}}{i_{sc}}=\frac{A(R_f+R_1)v_s}{r_{out}+R_f+(A+1)R_1} \frac{r_{out}}{Av_s}$$

  $$= \frac{(R_1+R_f)r_{out}}{r_{out}+R_f+(A+1)R_1} \approx \frac{r_{out}}{A}\;\frac{R_1+R_f}{R_1}$$

  
  The approximation is due to $A\gg 1$. In particular, when $R_f=0$, $R_{out}=r_{out}/A$, as in the voltage follower case.