Circuit Theory/Extra Element

The Extra Element technique helps break down one complicated problem into several simpler ones like superposition. However the goal is to find impedance rather than current or voltage.

By removing the element that most complicates the circuit the overall circuit impedance can be obtained.

$Z_{in}=Z_{in}^{\infty }\left({\frac {1+{\frac {Z_{e}^{0}}{Z}}}{1+{\frac {Z_{e}^{\infty }}{Z}}}}\right)$

where

$Z\$ is the impedance chosen as the extra element to be removed

$Z_{in}^{\infty }$ is the input impedance with Z removed (or made infinite)

$Z_{e}^{0}$ is the impedance seen by the extra element Z with the source shorted (or made zero)

$Z_{e}^{\infty }$ is the impedance seen by the extra element Z with the source opened (or made infinite)

Computing these three terms may seem like extra effort, but they are often easier to compute than the overall input impedance.

Example

4 resistors, 1 capacitor example

set up for node analsysis

node analysis in mupad ..code .. need to solve for vs/current to compare answers

Consider the problem of finding $Z_{in}$ for the circuit in Figure 1 using the EET (note all component values are unity for simplicity). If the capacitor (gray shading) is denoted the extra element then

$Z={\frac {1}{s}}$

Removing this capacitor from the circuit we find

• 
  capacitor removed, looking at the input (source input)
• 
  capacitor removed, reorganizing circuit so can see input impedance $Z_{in}^{\infty }$

$Z_{in}^{\infty }=2\|1+1={\frac {5}{3}}$

Calculating the impedance seen by the capacitor with the input shorted we find

• 
  shorted source, finding $Z_{e}^{0}$
• 
  shorted source redrawn

$Z_{e}^{0}=1\|(1+1\|1)={\frac {3}{5}}$

Calculating the impedance seen by the capacitor with the input open we find

• 
  open source, finding $Z_{e}^{\infty }$
• 
  open source redrawn

$Z_{e}^{\infty }=2\|1+1={\frac {5}{3}}$

Therefore using the EET, we find

$Z_{in}={\frac {5}{3}}\left({\frac {1+{\frac {3}{5}}s}{1+{\frac {5}{3}}s}}\right)$

Note that this problem was solved by calculating three simple driving point impedances by inspection.