In the previous section we saw that impedance parameters may not exist for a two-port network. So there is a need for an alternative means of describing such a network. This need is met by the second set of parameters, which we obtain by expressing the terminal currents in terms of the terminal voltages.
Fig. 1: Determination of the y parameters: (a) finding $y_{11}$ and $y_{21}$, (b) finding $y_{12}$ and $y_{22}$.
In either Fig. 1(a) or (b), the terminal currents can be expressed in terms of the terminal voltages as
$$\begin{array}{l}\mathbf{I}_{1}=\mathbf{y}_{11} \mathbf{V}_{1}+\mathbf{y}_{12} \mathbf{V}_{2} \\\mathbf{I}_{2}=\mathbf{y}_{21} \mathbf{V}_{1}+\mathbf{y}_{22} \mathbf{V}_{2}\end{array}$$
or in matrix form as
$$\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{I}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{y}_{11} & \mathbf{y}_{12} \\\mathbf{y}_{21} & \mathbf{y}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right]=[\mathbf{y}]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right]$$
The $\mathbf{y}$ terms are known as the admittance parameters (or, simply, y parameters) and have units of siemens.
The values of the parameters can be determined by setting $\mathbf{V}_{1}=0$ (input port short-circuited) or $\mathbf{V}_{2}=0$ (output port short-circuited). Thus,
$$\begin{array}{ll} \mathbf{y}_{11}=\left.\frac{\mathbf{I}_1}{\mathbf{V}_1}\right|_{\mathbf{V}_2=0}, & \mathbf{y}_{12}=\left.\frac{\mathbf{I}_1}{\mathbf{V}_2}\right|_{\mathbf{V}_1=0} \\ \mathbf{y}_{21}=\left.\frac{\mathbf{I}_2}{\mathbf{V}_1}\right|_{\mathbf{V}_2=0}, & \mathbf{y}_{22}=\left.\frac{\mathbf{I}_2}{\mathbf{V}_2}\right|_{\mathbf{V}_1=0} \end{array} \tag{1}$$
Since the $y$ parameters are obtained by short-circuiting the input or output port, they are also called the short-circuit admittance parameters. Specifically,
$\mathbf{y}_{11}=$ Short-circuit input admittance
$\mathbf{y}_{12}=$ Short-circuit transfer admittance from port 2 to port 1
$\mathbf{y}_{21}=$ Short-circuit transfer admittance from port 1 to port 2
$\mathbf{y}_{22}=$ Short-circuit output admittance
Following Eq. (1), we obtain $\mathbf{y}_{11}$ and $\mathbf{y}_{21}$ by connecting a current $\mathbf{I}_{1}$ to port 1 and short-circuiting port 2 as in Fig. 1(a), finding $\mathbf{V}_{1}$ and $\mathbf{I}_{2}$, and then calculating
$$\mathbf{y}_{11}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{1}}, \quad \mathbf{y}_{21}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{1}}$$
Similarly, we obtain $\mathbf{y}_{12}$ and $\mathbf{y}_{22}$ by connecting a current source $\mathbf{I}_{2}$ to port 2 and short-circuiting port 1 as in Fig. 1(b), finding $\mathbf{I}_{1}$ and $\mathbf{V}_{2}$, and then getting
$$\mathbf{y}_{12}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{2}}, \quad \mathbf{y}_{22}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}$$
This procedure provides us with a means of calculating or measuring the $y$ parameters. The impedance and admittance parameters are collectively referred to as immittance parameters.
For a two-port network that is linear and has no dependent sources, the transfer admittances are equal $\left(\mathbf{y}_{12}=\mathbf{y}_{21}\right)$. This can be proved in the same way as for the $z$ parameters. A reciprocal network $\left(\mathbf{y}_{12}=\mathbf{y}_{21}\right)$ can be modeled by the $\Pi$-equivalent circuit in Fig. 2(a).
Fig. 2: (a) $\Pi$-equivalent circuit (for reciprocal case only), (b) general equivalent circuit.
If the network is not reciprocal, a more general equivalent network is shown in Fig. $2(\mathrm{~b})$
Example 1: Obtain the $y$ parameters for the $\Pi$ network shown in Fig. $3$.
Fig. 3: For Example 1.
Solution:
METHOD I To find $\mathbf{y}_{11}$ and $\mathbf{y}_{21}$, short-circuit the output port and connect a current source $\mathbf{I}_{1}$ to the input port as in Fig. 4(a). Since the $8\Omega$ resistor is short-circuited, the $2-\Omega$ resistor is in parallel with the $4\Omega$ resistor. Hence,
$$\mathbf{V}_{1}=\mathbf{I}_{1}(4 \| 2)=\frac{4}{3} \mathbf{I}_{1}, \quad \mathbf{y}_{11}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{1}}=\frac{\mathbf{I}_{1}}{\frac{4}{3} \mathbf{I}_{1}}=0.75 \mathrm{~S}$$
By current division,
$$-\mathbf{I}_{2}=\frac{4}{4+2} \mathbf{I}_{1}=\frac{2}{3} \mathbf{I}_{1}, \quad \mathbf{y}_{21}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{1}}=\frac{-\frac{2}{3} \mathbf{I}_{1}}{\frac{4}{3} \mathbf{I}_{1}}=-0.5 \mathrm{~S}$$
To get $\mathbf{y}_{12}$ and $\mathbf{y}_{22}$, short-circuit the input port and connect a current source I $_{2}$ to the output port as in Fig. 4(b).
Fig. 4: For Example 1: (a) finding $y_{11}$ and $y_{21}$, (b) finding $y_{12}$ and $y_{22}$.
The $4-\Omega$ resistor is short-circuited so that the $2\Omega$ and $8\Omega$ resistors are in parallel.
$$\mathbf{V}_{2}=\mathbf{I}_{2}(8 \| 2)=\frac{8}{5} \mathbf{I}_{2}, \quad \mathbf{y}_{22}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}=\frac{\mathbf{I}_{2}}{\frac{8}{5} \mathbf{I}_{2}}=\frac{5}{8}=0.625 \mathrm{~S}$$
By current division,
$$-\mathbf{I}_{1}=\frac{8}{8+2} \mathbf{I}_{2}=\frac{4}{5} \mathbf{I}_{2}, \quad \mathbf{y}_{12}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{2}}=\frac{-\frac{4}{5} \mathbf{I}_{2}}{\frac{8}{5} \mathbf{I}_{2}}=-0.5 \mathrm{~S}$$
METHOD 2 Alternatively, comparing Fig. $3$ with Fig. 2(a),
$$\begin{array}{c}\mathbf{y}_{12}=-\frac{1}{2} \mathrm{~S}=\mathbf{y}_{21} \\\mathbf{y}_{11}+\mathbf{y}_{12}=\frac{1}{4} \quad \Longrightarrow \quad \mathbf{y}_{11}=\frac{1}{4}-\mathbf{y}_{12}=0.75 \mathrm{~S} \\\mathbf{y}_{22}+\mathbf{y}_{12}=\frac{1}{8} \quad \Longrightarrow \quad \mathbf{y}_{22}=\frac{1}{8}-\mathbf{y}_{12}=0.625 \mathrm{~S}\end{array}$$
as obtained previously.
Example 2: Determine the $y$ parameters for the two-port shown in Fig. $5$.
Fig. 5: For Example 2.
Solution: We follow the same procedure as in the previous example. To get $\mathbf{y}_{11}$ and $\mathbf{y}_{21}$, we use the circuit in Fig. 6(a), in which port 2 is short-circuited and a current source is applied to port 1 .
Fig. 6: Solution of Example 2: (a) finding $y_{11}$ and $y_{21}$, (b) finding $y_{12}$ and $y_{22}$.
At node 1 ,
$$\frac{\mathbf{V}_{1}-\mathbf{V}_{o}}{8}=2 \mathbf{I}_{1}+\frac{\mathbf{V}_{o}}{2}+\frac{\mathbf{V}_{o}-0}{4}$$
But $\mathbf{I}_{1}=\frac{\mathbf{V}_{1}-\mathbf{V}_{o}}{8}$; therefore,
$$\begin{array}{c}0=\frac{\mathbf{V}_{1}-\mathbf{V}_{o}}{8}+\frac{3 \mathbf{V}_{o}}{4} \\0=\mathbf{V}_{1}-\mathbf{V}_{o}+6 \mathbf{V}_{o} \quad \Longrightarrow \quad \mathbf{V}_{1}=-5 \mathbf{V}_{o}\end{array}$$
Hence,
$$\mathbf{I}_{1}=\frac{-5 \mathbf{V}_{o}-\mathbf{V}_{o}}{8}=-0.75 \mathbf{V}_{o}$$
and
$$\mathbf{y}_{11}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{1}}=\frac{-0.75 \mathbf{V}_{o}}{-5 \mathbf{V}_{o}}=0.15 \mathrm{~S}$$
At node 2 ,
$$\frac{\mathbf{V}_{o}-0}{4}+2 \mathbf{I}_{1}+\mathbf{I}_{2}=0$$
or
$$-\mathbf{I}_{2}=0.25 \mathbf{V}_{o}-1.5 \mathbf{V}_{o}=-1.25 \mathbf{V}_{o}$$
Hence,
$$\mathbf{y}_{21}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{1}}=\frac{1.25 \mathbf{V}_{o}}{-5 \mathbf{V}_{o}}=-0.25 \mathrm{~S}$$
Similarly, we get $\mathbf{y}_{12}$ and $\mathbf{y}_{22}$ using Fig. 6(b). At node 1,
$$\frac{0-\mathbf{V}_{o}}{8}=2 \mathbf{I}_{1}+\frac{\mathbf{V}_{o}}{2}+\frac{\mathbf{V}_{o}-\mathbf{V}_{2}}{4}$$
But $\mathbf{I}_{1}=\frac{0-\mathbf{V}_{o}}{8}$; therefore,
$$0=-\frac{\mathbf{V}_{o}}{8}+\frac{\mathbf{V}_{o}}{2}+\frac{\mathbf{V}_{o}-\mathbf{V}_{2}}{4}$$
or
$$0=-\mathbf{V}_{o}+4 \mathbf{V}_{o}+2 \mathbf{V}_{o}-2 \mathbf{V}_{2} \quad \Longrightarrow \quad \mathbf{V}_{2}=2.5 \mathbf{V}_{o}$$
Hence,
$$\mathbf{y}_{12}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{2}}=\frac{-\mathbf{V}_{o} / 8}{2.5 \mathbf{V}_{o}}=-0.05 \mathrm{~S}$$
At node 2 ,
$$\frac{\mathbf{V}_{o}-\mathbf{V}_{2}}{4}+2 \mathbf{I}_{1}+\mathbf{I}_{2}=0$$
or
$$-\mathbf{I}_{2}=0.25 \mathbf{V}_{o}-\frac{1}{4}\left(2.5 \mathbf{V}_{o}\right)-\frac{2 \mathbf{V}_{o}}{8}=-0.625 \mathbf{V}_{o}$$
Thus,
$$\mathbf{y}_{22}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}=\frac{0.625 \mathbf{V}_{o}}{2.5 \mathbf{V}_{o}}=0.25 \mathrm{~S}$$
Notice that $\mathbf{y}_{12} \neq \mathbf{y}_{21}$ in this case, since the network is not reciprocal.