Application of the Two Port Networks to the Transistor Circuits

We have seen how the six sets of network parameters can be used to characterize a wide range of two-port networks. Depending on the way two-ports are interconnected to form a larger network, a particular set of parameters may have advantages over others, as we noticed in Section 26.6 ( Interconnection of Networks ). In this section, we will consider two important application areas of two-port parameters: transistor circuits and synthesis of ladder networks

Transistor Circuits

The two-port network is often used to isolate a load from the excitation of a circuit. For example, the two-port in Fig. $ 1 $ may represent an amplifier, a filter, or some other network.
Fig. 1: Two-port network isolating source and load.
When the two-port represents an amplifier, expressions for the voltage gain $ A_{v} $, the current gain $ A_{i} $, the input impedance $ Z_{\text {in }} $, and the output impedance $ Z_{\text {out }} $ can be derived with ease. They are defined as follows:
$$A_{v} =\frac{V_{2}(s)}{V_{1}(s)} \tag{1}$$ $$A_{i} =\frac{I_{2}(s)}{I_{1}(s)} \tag{2}$$ $$Z_{\text {in }} =\frac{V_{1}(s)}{I_{1}(s)} \tag{3}$$ $$Z_{\text {out }} =\left.\frac{V_{2}(s)}{I_{2}(s)}\right|_{V_{s}=0} \tag{4}$$
Any of the six sets of two-port parameters can be used to derive the expressions in Eqs. (1) to (4). Here, we will specifically use the hybrid parameters to obtain them for transistor amplifiers.
The hybrid $ (h) $ parameters are the most useful for transistors; they are easily measured and are often provided in the manufacturer's data or spec sheets for transistors. The $ h $ parameters provide a quick estimate of the performance of transistor circuits. They are used for finding the exact voltage gain, input impedance, and output impedance of a transistor.
The $ h $ parameters for transistors have specific meanings expressed by their subscripts. They are listed by the first subscript and related to the general $ h $ parameters as follows:
$$h_{i}=h_{11}, \quad h_{r}=h_{12}, \quad h_{f}=h_{21}, \quad h_{o}=h_{22} \tag{5}$$
The subscripts $ i, r, f $, and $ o $ stand for input, reverse, forward, and output. The second subscript specifies the type of connection used: $ e $ for common emitter (CE), $ c $ for common collector (CC), and $ b $ for common base (CB). Here we are mainly concerned with the common-emitter connection. Thus, the four $ h $ parameters for the common-emitter amplifier are:
$$\begin{aligned}h_{i e} &=\text { Base input impedance } \\h_{r e} &=\text { Reverse voltage feedback ratio } \\h_{f e} &=\text { Base-collector current gain } \\h_{o e} &=\text { Output admittance }\end{aligned} \tag{6}$$
These are calculated or measured in the same way as the general $ h $ parameters. Typical values are $ h_{i e}=6 \mathrm{k} \Omega, h_{r e}=1.5 \times 10^{-4}, h_{f e}=200 $, $ h_{o e}=8 \mu \mathrm{S} $. We must keep in mind that these values represent ac characteristics of the transistor, measured under specific circumstances.
circuit schematichybrid model
Fig. 2: Common emitter amplifier: (a) circuit schematic, (b) hybrid model.
Figure $ 2 $ shows the circuit schematic for the common-emitter amplifier and the equivalent hybrid model. From the figure, we see that
$$\begin{array}{ll}\mathbf{V}_{b}=h_{i e} \mathbf{I}_{b}+h_{r e} \mathbf{V}_{c} \quad (7.a)\\\mathbf{I}_{c}=h_{f e} \mathbf{I}_{b}+h_{o e} \mathbf{V}_{c} \quad (7.b)\end{array} $$
Consider the transistor amplifier connected to an ac source and a load as in Fig. 3. This is an example of a two-port network embedded within a larger network. We can analyze the hybrid equivalent circuit as usual with Eq. (7) in mind.
Transistor amplifier with source and load resistance
Fig. 3: Transistor amplifier with source and load resistance.
Recognizing from Fig. 3 that $ \mathbf{V}_{c}=-R_{L} \mathbf{I}_{c} $ and substituting this into Eq. (7.b) gives
$$\mathbf{I}_{c}=h_{f e} \mathbf{I}_{b}-h_{o e} R_{L} \mathbf{I}_{c}$$
$$\left(1+h_{o e} R_{L}\right) \mathbf{I}_{c}=h_{f e} \mathbf{I}_{b}$$
From this, we obtain the current gain as
$$A_{i}=\frac{\mathbf{I}_{c}}{\mathbf{I}_{b}}=\frac{h_{f e}}{1+h_{o e} R_{L}} \tag{8}$$
From Eqs. (7.b) and (8), we can express $ \mathbf{I}_{b} $ in terms of $ \mathbf{V}_{c} $ :
$$\mathbf{I}_{c}=\frac{h_{f e}}{1+h_{o e} R_{L}} \mathbf{I}_{b}=h_{f e} \mathbf{I}_{b}+h_{o e} \mathbf{V}_{c}$$
$$\mathbf{I}_{b}=\frac{h_{o e} \mathbf{V}_{c}}{\frac{h_{f e}}{1+h_{o e} R_{L}}-h_{f e}} \tag{9}$$
Substituting Eq. (9) into Eq. (7.a) and dividing by $ \mathbf{V}_{c} $ gives
$$\begin{aligned}\frac{\mathbf{V}_{b}}{\mathbf{V}_{c}} &=\frac{h_{o e} h_{i e}}{\frac{h_{f e}}{1+h_{o e} R_{L}}-h_{f e}}+h_{r e} \\=& \frac{h_{i e}+h_{i e} h_{o e} R_{L}-h_{r e} h_{f e} R_{L}}{-h_{f e} R_{L}}\end{aligned}$$
Thus, the voltage gain is
$$A_{v}=\frac{\mathbf{V}_{c}}{\mathbf{V}_{b}}=\frac{-h_{f e} R_{L}}{h_{i e}+\left(h_{i e} h_{o e}-h_{r e} h_{f e}\right) R_{L}}$$
Substituting $ \mathbf{V}_{c}=-R_{L} \mathbf{I}_{c} $ into Eq. (7.a) gives
$$\mathbf{V}_{b}=h_{i e} \mathbf{I}_{b}-h_{r e} R_{L} \mathbf{I}_{c}$$
$$\frac{\mathbf{V}_{b}}{\mathbf{I}_{b}}=h_{i e}-h_{r e} R_{L} \frac{\mathbf{I}_{c}}{\mathbf{I}_{b}}$$
Replacing $ \mathbf{I}_{c} / \mathbf{I}_{b} $ by the current gain in Eq. (8) yields the input impedance as
$$Z_{\text {in }}=\frac{\mathbf{V}_{b}}{\mathbf{I}_{b}}=h_{i e}-\frac{h_{r e} h_{f e} R_{L}}{1+h_{o e} R_{L}}$$
The output impedance $ Z_{\text {out }} $ is the same as the Thevenin equivalent at the output terminals. As usual, by removing the voltage source and placing a $ 1-\mathrm{V} $ source at the output terminals, we obtain the circuit in Fig. 4, from which $ Z_{\text {out }} $ is determined as $ 1 / \mathbf{I}_{c} $.
Fig. 4: Finding the output impedance of the amplifier circuit in Fig. 3.
Since $ \mathbf{V}_{c}=1 \mathrm{~V} $, the input loop gives
$$h_{r e}(1)=-\mathbf{I}_{b}\left(R_{s}+h_{i e}\right) \quad \Longrightarrow \quad \mathbf{I}_{b}=-\frac{h_{r e}}{R_{s}+h_{i e}} \tag{10}$$
For the output loop,
$$\mathbf{I}_{c}=h_{o e}(1)+h_{f e} \mathbf{I}_{b} \tag{11}$$
Substituting Eq. (10) into Eq. (11) gives
$$\mathbf{I}_{c}=\frac{\left(R_{s}+h_{i e}\right) h_{o e}-h_{r e} h_{f e}}{R_{s}+h_{i e}}$$
From this, we obtain the output impedance $ Z_{\text {out }} $ as $ 1 / \mathbf{I}_{c} $; that is,
$$Z_{\text {out }}=\frac{R_{s}+h_{i e}}{\left(R_{s}+h_{i e}\right) h_{o e}-h_{r e} h_{f e}} \tag{12}$$
Example 1: Consider the common-emitter amplifier circuit of Fig. 5.
(a) Determine the voltage gain, current gain, input impedance, and output impedance using these $ h $ parameters: $$h_{i e}=1 \mathrm{k} \Omega, \quad h_{r e}=2.5 \times 10^{-4}, \quad h_{f e}=50, \quad h_{o e}=20 \mu \mathrm{S}$$ (b) Find the output voltage $ \mathbf{V}_{o} $
Fig. 5: For Example 1.
(a) We note that $ R_{s}=0.8 \mathrm{k} \Omega $ and $ R_{L}=1.2 \mathrm{k} \Omega $. We treat the transistor of Fig. $ 18.59 $ as a two-port network and apply Eqs. (8) to (12).
$$\begin{array}{c}h_{i e} h_{o e}-h_{r e} h_{f e}=10^{3} \times 20 \times 10^{-6}-2.5 \times 10^{-4} \times 50 \\=7.5 \times 10^{-3} \\A_{v}=\frac{-h_{f e} R_{L}}{h_{i e}+\left(h_{i e} h_{o e}-h_{r e} h_{f e}\right) R_{L}}=\frac{-50 \times 1200}{1000+7.5 \times 10^{-3} \times 1200} \\=-59.46 \\A_{i}=\frac{h_{f e}}{1+h_{o e} R_{L}}=\frac{50}{1+20 \times 10^{-6} \times 1200}=48.83 \\Z_{\text {in }}=h_{i e}-h_{r e} A_{i} R_{L}=1000-2.5 \times 10^{-4} \times 48.83 \times 1200=985.4 \Omega \\\left(R_{s}+h_{i e}\right) h_{o e}-h_{r e} h_{f e} \\=(800+1000) \times 20 \times 10^{-6}-2.5 \times 10^{-4} \times 50=23.5 \times 10^{-3} \\Z_{\text {out }}=\frac{R_{s}+h_{i e}}{\left(R_{s}+h_{i e}\right) h_{o e}-h_{r e} h_{f e}}=\frac{800+1000}{23.5 \times 10^{-3}}=76.6 \mathrm{k} \Omega\end{array}$$
(b) The output voltage is
$$\mathbf{V}_{o}=A_{v} \mathbf{V}_{s}=-59.46\left(3.2 \angle 0^{\circ}\right) \mathrm{mV}=0.19 \angle 180^{\circ} \mathrm{V}$$