19.8 Computing Two-Port Parameters Using PSpice

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19.8 Computing Two-Port Parameters Using PSpice

Hand calculation of the two-port parameters may become difficult when the two-port is complicated. We resort to PSpice in such situations. If the circuit is purely resisti ve, PSpice dc analysis may be used; otherwise, PSpice ac analysis is required at a specific frequency. The key to using PSpice in computing a particular two-port parameter is to remember how that parameter is defined and to constrain the appropriate port variable with a 1-A or 1-V source while using an open or short circuit to impose the other necessary constraints. The following two examples illustrate the idea.

Find the h parameters of the network in Fig. 19.49. Example 19.15

Solution:

From Eq. (19.16),

$$\mathbf{h}_{11} = \frac{\mathbf{V}_1}{\mathbf{I}_1} \Big|_{\mathbf{V}_2 = 0}, \qquad \mathbf{h}_{21} = \frac{\mathbf{I}_2}{\mathbf{I}_1} \Big|_{\mathbf{V}_2 = 0}$$

showing that h11 and h21 can be found by setting V2 = 0. Also by setting I1 = 1 A, h11 becomes V1∕1 while h21 becomes I2∕1. With this in mind, we draw the schematic in Fig. 19.50(a). We insert a 1-A dc current

6 Ω

4i x

• ‒

5 Ω

Figure 19.50

For Example 19.15: (a) computing h11 and h21, (b) computing h12 and h22.

source IDC to take care of I1 = 1 A, the pseudocomponent VIEWPOINT to display V1 and pseudocomponent IPROBE to display I2. After saving the schematic, we run PSpice by selecting Analysis/Simulate and note the values displayed on the pseudocomponents. We obtain

$$\mathbf{h}_{11} = \frac{\mathbf{V}_1}{1} = 10 \ \Omega, \qquad \mathbf{h}_{21} = \frac{\mathbf{I}_2}{1} = -0.5$$

Similarly, from Eq. (19.16),

Practice Problem 19.15 Obtain the h parameters for the network in Fig. 19.51 using PSpice.

$$\mathbf{h}_{12} = \frac{\mathbf{V}_1}{\mathbf{V}_2} \Big|_{\mathbf{I}_1 = 0}, \qquad \mathbf{h}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} \Big|_{\mathbf{I}_1 = 0}$$

indicating that we obtain h12 and h22 by open-circuiting the input port (I1 = 0). By making V2 = 1 V, h12 becomes V1∕1 while h22 becomes I2∕1. Thus, we use the schematic in Fig. 19.50(b) with a 1-V dc voltage source VDC inserted at the output terminal to take care of V2 = 1 V. The pseudocomponents VIEWPOINT and IPR OBE are inserted to display the values of V1 and I2, respectively. (Notice that in Fig. 19.50(b), the 5- Ω resistor is ignored because the input port is open-circuited and PSpice will not allow such. We may include the 5- Ω resistor if we replace the open circuit with a very large resistor, say, 10 MΩ.) After simulating the schematic, we obtain the values displayed on the pseudocomponents as shown in Fig. 19.50(b). Thus,

$$\mathbf{h}_{12} = \frac{\mathbf{V}_1}{1} = 0.8333, \qquad \mathbf{h}_{22} = \frac{\mathbf{I}_2}{1} = 0.1833 \text{ S}$$

Answer: h11 = 4.238 Ω, h21 = −0.6190, h12 = −0.7143, h22 = −0.1429 S.

Figure 19.52 For Example 19.16.

Example 19.16 Find the z parameters for the circuit in Fig. 19.52 at ω = 106 rad/s.

Solution:

Notice that we used dc analysis in Example 19.15 because the circuit in Fig. 19.49 is purely resistive. Here, we use ac analysis at f = ω∕2π = 0.15915 MHz, because L and C are frequency dependent.

In Eq. (19.3), we defined the z parameters as

$$\mathbf{z}_{11} = \frac{\mathbf{V}_1}{\mathbf{I}_1} \bigg|_{\mathbf{I}_2 = 0}, \qquad \mathbf{z}_{21} = \frac{\mathbf{V}_2}{\mathbf{I}_1} \bigg|_{\mathbf{I}_2 = 0}$$

Figure 19.53 For Example 19.16: (a) circuit for determining z11 and z21, (b) circuit for determining z12 and z22.

This suggests that if we let I1 = 1 A and open-circuit the output port so that I2 = 0, then we obtain

$$\mathbf{z}_{11} = \frac{\mathbf{V}_1}{1} \quad \text{and} \quad \mathbf{z}_{21} = \frac{\mathbf{V}_2}{1}$$

We realize this with the schematic in Fig. 19.53(a). We insert a 1-A ac current source IAC at the input terminal of the circuit and two VPRINT1 pseudocomponents to obtain V1 and V2. The attributes of each VPRINT1 are set as AC = yes, MAG = yes, and PHASE = yes to print the magnitude and phase values of the voltages. We select Analysis/Setup/AC Sweep and enter 1 as Total Pts, 0.1519MEG as Start Freq, and 0.1519MEG as Final Freq in the AC Sweep and Noise Analysis dialog box. After saving the schematic, we select Analysis/Simulate to simulate it. We obtain V1 and V2 from the output file. Thus,

$$\mathbf{z}_{11} = \frac{\mathbf{V}_1}{1} = 19.70 \underline{/ 175.7^{\circ}} \,\Omega, \qquad \mathbf{z}_{21} = \frac{\mathbf{V}_2}{1} = 19.79 \underline{/ 170.2^{\circ}} \,\Omega$$

In a similar manner, from Eq. (19.3),

$$\mathbf{z}_{12} = \frac{\mathbf{V}_1}{\mathbf{I}_2} \big|_{\mathbf{I}_1 = 0}, \qquad \mathbf{z}_{22} = \frac{\mathbf{V}_2}{\mathbf{I}_2} \big|_{\mathbf{I}_1 = 0}$$

suggesting that if we let I2 = 1 A and open-circuit the input port,

$$z_{12} = \frac{V_1}{1}$$

and $z_{22} = \frac{V_2}{1}$

This leads to the schematic in Fig. 19.53(b). The only difference between this schematic and the one in Fig. 19.53(a) is that the 1-A ac current source IA C is no w at the output terminal. We run the schematic in Fig. 19.53(b) and obtain V1 and V2 from the output file. Thus,

$$\mathbf{z}_{12} = \frac{\mathbf{V}_1}{1} = 19.70 \underline{\text{/} 175.7^{\circ}} \,\Omega, \qquad \mathbf{z}_{22} = \frac{\mathbf{V}_2}{1} = 19.56 \underline{\text{/} 175.7^{\circ}} \,\Omega$$

Practice Problem 19.16 Obtain the z parameters of the circuit in Fig. 19.54 at f = 60 Hz.

Answer:

$$z_{11} = 3.987 / 175.5^{\circ} \Omega$$

, $z_{21} = 0.0175 / -2.65^{\circ} \Omega$ ,
\n $z_{12} = 0$ , $z_{22} = 0.2651 / 91.9^{\circ} \Omega$ .

19.9 Applications

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 S ection 19.7. In this section, we will consider tw o important application areas of two-port parameters: transistor circuits and synthesis of ladder networks.

19.9.1 Transistor Circuits

The two-port network is often used to isolate a load from the e xcitation of a circuit. For example, the two-port in Fig. 19.55 may represent an amplifier, a filter, or some other netw ork. When the two-port represents an amplifier, expressions for the voltage gain Av, the current gain Ai, the input impedance Zin, and the output impedance Zout can be derived with ease. They are defined as follows:

$$A_v = \frac{V_2(s)}{V_1(s)}$$

(19.62)

$$A_i = \frac{I_2(s)}{I_1(s)}\tag{19.63}$$ $$Z_{\text{in}} = \frac{V_1(s)}{I_1(s)}$$

(19.64)

$$Z_{\text{out}} = \frac{V_2(s)}{I_2(s)} \bigg|_{V_s=0} \tag{19.65}$$

Any of the six sets of tw o-port parameters can be used to deri ve the expressions in Eqs. (19.62) to (19.65). Ho wever, the hybrid (h) parameters are the most useful for transistors; the y are easily measured and are often provided in the manufacturer’s data or spec sheets for transis tors. The h parameters pro vide 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.

Figure 19.55 Two-port network isolating source and load.

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:

$$h_i = h_{11}$$

, $h_r = h_{12}$ , $h_f = h_{21}$ , $h_o = h_{22}$ (19.66)

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 connec tion. Thus, the four h parameters for the common-emitter amplifier are:

$$h_{ie} = \text{Base input impedance}$$

\n

$$h_{re} = \text{Reverse voltage feedback ratio}$$

\n

$$h_{fe} = \text{Base-collector current gain}$$

\n

$$h_{oe} = \text{Output admittance}$$

\n(19.67)

These are calculated or measured in the same w ay as the general h parameters. Typical values are hie = 6 kΩ, hre = 1.5 × 10−4, hfe = 200, hoe = 8 µS. We must keep in mind that these values represent ac characteristics of the transistor, measured under specific circumstances.

Figure 19.56 sho ws the circuit schematic for the common-emitter amplifier and the equivalent hybrid model. From the figure, we see that

$$\mathbf{V}_b = h_{ie}\mathbf{I}_b + h_{re}\mathbf{V}_c$$

\n

$$\mathbf{I}_c = h_{fe}\mathbf{I}_b + h_{oe}\mathbf{V}_c$$

\n(19.68a)\n(19.68b)

Figure 19.56

Common emitter amplifier: (a) circuit schematic, (b) hybrid model.

Consider the transistor amplifier connected to an ac source and a load as in Fig. 19.57. 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. (19.68) in mind. (See Example 19.6.) Recognizing

Transistor amplifier with source and load resistance.

from Fig. 19.57 that Vc = −RLIc and substituting this into Eq. (19.68b) gives

$$\mathbf{I}_c = h_{fe}\mathbf{I}_b - h_{oe}R_L\mathbf{I}_c$$ $$(1 + h_{oe}R_L)\mathbf{I}_c = h_{fe}\mathbf{I}_b \tag{19.69}$$

From this, we obtain the current gain as

$$A_i = \frac{\mathbf{I}_c}{\mathbf{I}_b} = \frac{h_{fe}}{1 + h_{oe}R_L}$$

(19.70)

From Eqs. (19.68b) and (19.70), we can express Ib in terms of Vc:

$$\mathbf{I}_c = \frac{h_{fe}}{1 + h_{oe}R_L}\mathbf{I}_b = h_{fe}\mathbf{I}_b + h_{oe}\mathbf{V}_c$$

or

or

$$\mathbf{I}_{b} = \frac{h_{oe} \mathbf{V}_{c}}{h_{fe}} - h_{fe}$$

(19.71)

Substituting Eq. (19.71) into Eq. (19.68a) and dividing by Vc gives

1. into Eq. (19.68a) and dividing by

$$\mathbf{V}_c$$

gives
\n

$$\frac{\mathbf{V}_b}{\mathbf{V}_c} = \frac{h_{oe}h_{ie}}{\frac{h_{fe}}{1 + h_{oe}R_L} - h_{fe}} + h_{re}$$

\n

$$= \frac{h_{ie} + h_{ie}h_{oe}R_L - h_{re}h_{fe}R_L}{-h_{fe}R_L} \tag{19.72}$$

Thus, the voltage gain is

gain is
\n

$$A_{v} = \frac{\mathbf{V}_{c}}{\mathbf{V}_{b}} = \frac{-h_{fe}R_{L}}{h_{ie} + (h_{ie}h_{oe} - h_{re}h_{fe})R_{L}}$$

\n(19.73)

Substituting Vc = −RLIc into Eq. (19.68a) gives

$$\mathbf{V}_b = h_{ie}\mathbf{I}_b - h_{re}R_L\mathbf{I}_c$$

or

$$\frac{\mathbf{V}_b}{\mathbf{I}_b} = h_{ie} - h_{re} R_L \frac{\mathbf{I}_c}{\mathbf{I}_b}$$

(19.74)

Replacing Ic∕Ib by the current gain in Eq. (19.70) yields the input impedance as

$$Z_{\rm in} = \frac{V_b}{I_b} = h_{ie} - \frac{h_{re}h_{fe}R_L}{1 + h_{oe}R_L}$$

(19.75)

The output impedance Zout is the same as the Thevenin equivalent at the output terminals. As usual, by removing the voltage source and placing a

Figure 19.58 Finding the output impedance of the amplifier circuit in Fig. 19.57.

1-V source at the output terminals, we obtain the circuit in Fig. 19.58, from which Zout is determined as 1∕Ic. Because Vc = 1 V, the input loop gives

$$h_{re}(1) = -\mathbf{I}_b(R_s + h_{ie}) \qquad \Rightarrow \qquad \mathbf{I}_b = -\frac{h_{re}}{R_s + h_{ie}} \qquad (19.76)$$

For the output loop,

$$\mathbf{I}_c = \mathbf{h}_{oe}(1) + h_{fe}\mathbf{I}_b \tag{19.77}$$

Substituting Eq. (19.76) into Eq. (19.77) gives

) into Eq. (19.77) gives
\n

$$\mathbf{I}_c = \frac{(R_s + h_{ie})h_{oe} - h_{re}h_{fe}}{R_s + h_{ie}}$$

\n(19.78)

From this, we obtain the output impedance Zout as 1∕Ic; that is,

the output impedance

$$Z_{\text{out}}$$

as $1/I_c$ ; that is,

$$Z_{\text{out}} = \frac{R_s + h_{ie}}{(R_s + h_{ie})h_{oe} - h_{re}h_{fe}}$$

(19.79)

Consider the common-emitter amplifier circuit of Fig. 19.59. Determine Example 19.17 the voltage gain, current g ain, input impedance, and output impedance using these h parameters:

hie = 1 kΩ, hre = 2.5 × 10−4, hfe = 50, hoe = 20 µS

Find the output voltage Vo.

For Example 19.17.

Solution:

1. Define. In an initial look at this problem, it appears to be clearly stated. However, when we are asked to determine the input impedance and the voltage gain, do they refer to the transistor or the circuit? As far as the current gain and the output impedance are concerned, the y are the same for both cases.

We ask for clarification and are told that we should calculate the input impedance, the output impedance, and the voltage gain

for the circuit and not the transistor. It is interesting to note that the problem can be restated so that it becomes a simple design problem: Given the h parameters, design a simple amplifier that has a gain of −60.

• 2. Present. Given a simple transistor circuit, an input voltage of 3.2 mV, and the h parameters of the transistor, calculate the output voltage.
• 3. Alternative. There are a couple of ways we can approach the prob lem, the most straightforw ard being to use the equi valent circuit shown in Fig. 19.57. Once you ha ve the equivalent circuit you can use circuit analysis to determine the answer. Once you have a solu tion, you can check it by plugging in the answer into the circuit equations to see if they are correct. Another approach is to simplify the right-hand side of the equi valent circuit and w ork backward to see if you obtain approximately the same answer . We will use that approach here.
• 4. Attempt. We note that R ^s = 0.8 k Ω and R ^L = 1.2 k Ω. We treat the transistor of Fig. 19.59 as a two-port network and apply Eqs. (19.70) to (19.79).

$$h_{ie}h_{oe} - h_{re}h_{fe} = 10^3 \times 20 \times 10^{-6} - 2.5 \times 10^{-4} \times 50$$ $$= 7.5 \times 10^{-3}$$ $$A_v = \frac{-h_{fe}R_L}{h_{ie} + (h_{ie}h_{oe} - h_{re}h_{fe})R_L} = \frac{-50 \times 1200}{1000 + 7.5 \times 10^{-3} \times 1200}$$ $$= -59.46$$

Av is the voltage gain of the amplifier = Vo ∕ Vb. To calculate the gain of the circuit we need to find Vo ∕ Vs. We can do this by using the mesh equation for the circuit on the left and Eqs. (19.71) and (19.73).

$$-\mathbf{V}_s + R_s \mathbf{I}_b + \mathbf{V}_b = 0$$

or

$$-V_s + R_s I_b + V_b = 0$$ $$V_s = 800 \frac{20 \times 10^{-6}}{50} - \frac{1}{59.46} V_o$$ $$= -0.03047 V_o.$$

Thus, the circuit gain is equal to −32.82. Now we can calculate the output voltage.

voltage.
\n

$$V_o = \text{gain} \times V_s = -105.09 \underline{/0^{\circ}} \text{ mV}.$$

\n

$$A_i = \frac{h_{fe}}{1 + h_{oe}R_L} = \frac{50}{1 + 20 \times 10^{-6} \times 1200} = 48.83$$

\n

$$Z_{in} = h_{ie} - \frac{h_{re}h_{fe}R_L}{1 + h_{oe}R_L}$$

\n

$$= 1000 - \frac{2.5 \times 10^{-4} \times 50 \times 1200}{1 + 20 \times 10^{-6} \times 1200}$$

\n

$$= 985.4 \Omega$$

You can modify Zin to include the 800-ohm resistor so that

Circuit input impedance = 800 + 985.4 = 1785.4 Ω . (Rs + hie)hoe − hrehfe

= (800 + 1000) × 20 × 10−6 − 2.5 × 10−4 × 50 = 23.5 × 10−3

$$Z_{\text{out}} = \frac{R_s + h_{ie}}{(R_s + h_{ie})h_{oe} - h_{re}h_{fe}} = \frac{800 + 1000}{23.5 \times 10^{-3}} = 76.6 \text{ k}\Omega$$

5. Evaluate. In the equi valent circuit, hoe represents a resistor of 50,000 Ω. This is in parallel with a load resistor equal to 1.2 k Ω. The size of the load resistor is so small relative to the hoe resistor that hoe can be neglected. This then leads to

$$I_c = h_{fe}I_b = 50I_b$$

, $V_c = -1200I_c$ ,

and the following loop equation from the left-hand side of the circuit:

−0.0032 + (800 + 1000)Ib + (0.00025)(−1200)(50)Ib = 0 Ib = 0.0032∕(1785) = 1.7927 µA. Ic = 50 × 1.7927 = 89.64 µA and Vc = −1200 × 89.64 × 10−6

$$= -107.57$$

mV

This is a good approximation to −105.09 mV.

$$Voltage gain = -107.57/3.2 = -33.62$$

Again, this is a good approximation to 32.82.

Circuit input impedance = 0.032∕1.7927 × 10−6 = 1785 Ω

which clearly compares well with the 1785.4 Ω we obtained before.

For these calculations, we assumed that Zout = ∞ Ω. Our calculations produced 72.6 kΩ. We can test our assumption by calculating the equivalent resistance of this and the load resistance.

72,600 × 1200∕(72,600 + 1200) = 1,180.5 = 1.1805 kΩ

Again, we have a good approximation.

6. Satisfactory? We ha ve satisf actorily solv ed the problem and checked the results. We can now present our results as a solution to the problem.

For the transistor amplifier of Fig. 19.60, find the voltage gain, current Practice Problem 19.17 gain, input impedance, and output impedance. Assume that

$$h_{ie} = 6 \text{ k}\Omega$$

, $h_{re} = 1.5 \times 10^{-4}$ , $h_{fe} = 200$ , $h_{oe} = 8 \mu\text{S}$

Answer: −123.61 for the transistor and −4.753 for the circuit, 194.17, 6 kΩ for the transistor and 156 kΩ for the circuit, 128.08 kΩ.