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Question 17.11: The two-port network Na in Figure 17–16 is an amplifier with......

The two-port network N_{a} in Figure 17–16 is an amplifier with h-parameters h_{11a} = 500  Ω, h_{12a} = 0, h_{21a} = –500, and h_{22a} = 0.1 S. The two-port N_{b} is a passive resistance circuit that provides feedback around the amplifier. The purpose of this example is to find the voltage gain of the overall network first without feedback (R = 0) and then with feedback (R = 1 kΩ).

17.16
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To get a two-port description of this connection we use z-parameters since [z] =[\pmb{z}_{a}]+[\pmb{z}_{b}] in a series connection. We are given the h-parameters of N_{a}, namely h_{11a} = 500  Ω, h_{12a} = 0, h_{21a} = -500, h_{22b} = 0.1 S, and \Delta_{ha} = h_{11a}h_{22a}  –  h_{12a}h_{21a} = 50. To get the impedance matrix [\pmb{z}_{a}] we use the [h] to [z] conversion in Table 17–2.

[\pmb{z}_{a}]=\begin{bmatrix}\frac{Δ_{ha}}{h_{22a}}&\frac{h_{12a}}{h_{22a}}\\\frac{-h_{12a}}{h_{22a}}&\frac{1}{h_{22a}} \end{bmatrix} =\begin{bmatrix} 500& 0 \\5000& 10 \end{bmatrix} Ω

Given the matrix [\pmb{z}_{a}], the z-parameters of N_{a} are seen to be z_{11a} = 500  Ω, z_{12a} = 0, z_{21a} = 5000 Ω, and z_{22a} = 10 Ω. Network N_{b} is a simple shunt resistor R  so it is easy to see that its z-parameters are. z_{11b} = z_{12b} = z_{21b} = z_{22b} = R. Table  17–3 expresses the voltage gain T_{V} in terms of z-parameters as T_{V} = z_{21}/z_{11}.  Again, since [z] = [\pmb{z}_{a}]+[\pmb{z}_{b}] in a series connection, the overall voltage gain is

T_{V} =\frac{z_{21}}{z_{11}} =\frac{z_{21a}  +  z_{21b}}{z_{11a}  +  z_{11b}} =\frac{5000  +  R}{500  +  R}

Without feedback (R = 0) the voltage gain is T_{V} = 5000/500 = 10. With  feedback (R = 1000 Ω) the gain is T_{V} = 6000/1500 = 4.

T A B L E    17–2     TWO-PORT PARAMETER CONVERSION TABLE

DESIRED PARAMETERS GIVEN  PARAMETERS
[Z] [Y] [H] [T]
[z] \begin{bmatrix} z_{11}& z_{12}\\z_{21}& z_{22} \end{bmatrix} \begin{bmatrix} \frac{y_{22}}{Δ_{y}}&\frac{-y_{12}}{Δ_{y}}\\\frac{-y_{21}}{Δ_{y}}&\frac{y_{11}}{Δ_{y}} \end{bmatrix} \begin{bmatrix} \frac{Δ_{h}}{h_{22}}&\frac{h_{12}}{h_{22}}\\\frac{-h_{21}}{h_{22}}&\frac{1}{h_{22}}\end{bmatrix} \begin{bmatrix} \frac{A}{C}&\frac{Δ_{t}}{C}\\\frac{1}{C}&\frac{D}{C}\end{bmatrix}
[y] \begin{bmatrix} \frac{z_{22}}{Δ_{z}}&\frac{-z_{12}}{Δ_{z}}\\\frac{-z_{21}}{Δ_{z}}&\frac{z_{11}}{Δ_{z}}\end{bmatrix} \begin{bmatrix} y_{11}&y_{12}\\y_{21}& y_{22} \end{bmatrix} \begin{bmatrix} \frac{1}{h_{11}}&\frac{-h_{12}}{h_{11}}\\\frac{h_{21}}{h_{11}}&\frac{Δ_{h}}{h_{11}} \end{bmatrix} \begin{bmatrix} \frac{D}{B}&\frac{-Δ_{t}}{B}\\\frac{-1}{B}&\frac{A}{B} \end{bmatrix}
[h] \begin{bmatrix} \frac{Δ_{z}}{z_{22}}&\frac{z_{12}}{z_{22}}\\\frac{-z_{21}}{z_{22}}&\frac{1}{z_{22}} \end{bmatrix} \begin{bmatrix}\frac{1}{y_{11}}& \frac{-y_{12}}{y_{11}}\\\frac{y_{21}}{y_{11}}&\frac{Δ_{y}}{y_{11}} \end{bmatrix} \begin{bmatrix} h_{11}& h_{12}\\h_{21}& h_{22} \end{bmatrix} \begin{bmatrix} \frac{B}{D}&\frac{Δ_{t}}{D}\\\frac{-1}{D}&\frac{C}{D} \end{bmatrix}
[t] \begin{bmatrix}\frac{z_{11}}{z_{21}}&\frac{Δ_{z}}{z_{21}}\\\frac{1}{z_{21}}&\frac{z_{22}}{z_{21}} \end{bmatrix} \begin{bmatrix}\frac{-y_{22}}{y_{21}}&\frac{-1}{y_{21}}\\\frac{-Δ_{y}}{y_{21}}&\frac{-y_{11}}{y_{21}} \end{bmatrix} \begin{bmatrix} \frac{-Δ_{h}}{h_{21}}&\frac{-h_{11}}{h_{21}}\\\frac{-h_{22}}{h_{21}}&\frac{-1}{h_{21}} \end{bmatrix} \begin{bmatrix} A& B\\C& D \end{bmatrix}
Δ_{z} = z_{11}z_{22}  –  z_{12}z_{21}     Δ_{y} = y_{11}y_{22}  –  y_{12}y_{21}      Δ_{h} = h_{11}h_{22}  –  h_{12}h_{21}     Δ_{t} = AD  –  BC

 

T A B L E   17–3   VOLTAGE GAIN AND CURRENT GAIN IN TERMS OF TWO-PORT PARAMETERS

DEFINITION GIVEN  PARAMETERS
[Z] [Y] [H] [T]
T_{V}=\frac{V_{2}}{V_{1}}|_{I_{2}=0} T_{V}=\frac{z_{21}}{z_{11}} T_{V}=\frac{-y_{21}}{y_{22}} T_{V}=\frac{-h_{21}}{Δ_{h}} T_{V}=\frac{1}{A}
T_{I}=\frac{I_{2}}{I_{1}}|_{V_{2}=0} T_{I}=\frac{-z_{21}}{z_{22}} T_{I}=\frac{y_{21}}{y_{11}} T_{I}= h_{21} T_{I}=\frac{-1}{D}

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