• Voltage gain: Assume an ideal voltage source $v_s$ is applied to the input port AB. Apply KVL to the loop to get:
  
  $$v_s=i_{in} r_{in} +i_{in} r_{out} +A r_{in} i_{in} =i_{in}[r_{out}+(A+1) r_{in}]$$
  
  
  i.e.,
  
  $$i_{in}=\frac{v_s}{r_{out}+(A+1) r_{in}}$$
  
  
  The voltage across the output port CD is:
  
  $$v_{out}=A r_{in} i_{in}+r_{out}i_{in} =v_s\frac{A r_{in}+r_{out}}{r_{out}+(A+1) r_{in}}$$
  
  
  i.e.,
  
  $$A_{oc}=\frac{A r_{in}+r_{out}}{r_{out}+(A+1) r_{in}} \approx 1$$
  
  

• Input resistance: Assume a load $R_L$ is connnected to the output port CD. Apply KVL to the two loops and get:
  
  $$v_s=(r_{in}+r_{out}+A r_{in})i_{in}-r_{out}i_{out}$$
  
  
  
  
  $$Ar_{in}i_{in}=(r_{out}+R_L)i_{out}-r_{out} i_{in}$$
  
  
  Solving the second equation we get
  
  $$i_{out}=\frac{Ar_{in}+r_{out}}{r_{out}+R_L} i_{in}$$
  
  
  Plugging that into the first equation we get
  
  $$v_s =i_{in}[(r_{in}+r_{out}+A r_{in})-r_{out}\frac{Ar_{in}+r_{out}}{r_{out}+R_L}]$$
  
  
  i.e.,
  
  $$R_{in}=\frac{v_s}{i_{in}}=r_{in}(1+\frac{AR_L}{r_{out}+R_L}) +R_L\vert\vert r_{out} \approx A r_{in}$$
  
  

• Output resistance: Apply a source voltage $v_s$ with an internal resistance $R_s$ to the input port AB, and get
  
  $$v_s=i_{in} [R_s+r_{in}+r_{out} +A r_{in}]$$
  
  
  i.e.,
  
  $$i_{in}=\frac{v_s}{R_s+(A+1)r_{in}+r_{out}}$$
  
  
  The open-circuit output voltage is:
  
  $$v_{oc}=(r_{out}+Ar_{in}) i_{in}=(r_{out}+Ar_{in}) \frac{v_s}{R_s+(A+1)r_{in}+r_{out}}$$
  
  
  Next, we find short-circuit output current. Due to superposition (two voltage sources $v_s$ and $Ar_{in}i_{in}$) we have:
  
  $$i_{sc}=\frac{v_s}{R_s+r_{in}}+Ar_{in}\frac{v_s}{(R_s+r_{in})r_{out}} =v_s\frac{Ar_{in}+r_{out}}{(R_s+r_{in})r_{out}}$$
  
  
  Now we get:
  
  $$R_{out}=\frac{v_{oc}}{i_{sc}}=\frac{(R_s+r_{in})r_{out}}{R_s+(A+1)r_{in}+r_{out}} \approx \frac{r_{out}}{A}$$
  
  
  The approximation is due to the assumption that $A»1$ and both $R_s$ and $r_{out}$ are very small.