13.8 PSpice Analysis of Magnetically Coupled Circuits

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13.8 PSpice Analysis of Magnetically Coupled Circuits

PSpice analyzes magnetically coupled circuits just lik e inductor cir cuits except that the dot convention must be followed. In PSpice Schematic, the dot (not sho wn) is al ways next to pin 1, which is the lefthand terminal of the inductor when the inductor with part name L is placed (horizontally) without rotation on a schematic. Thus, the dot or pin 1 will be at the bottom after one 90 ° counterclockwise rotation, since rotation is al ways about pin 1. Once the magnetically coupled inductors are arranged with the dot convention in mind and their value attributes are set in henries, we use the coupling symbol K_LINEAR to define the coupling. For each pair of coupled inductors, take the following steps:

• 1. Select Draw/Get New Part and type K_LINEAR.
• 2. Hit or click OK and place the K_LINEAR symbol on the schematic, as shown in Fig. 13.51. (Notice that K_LINEAR is not a component and therefore has no pins.)
• 3. DCLICKL on COUPLING and set the v alue of the coupling coefficient k.
• 4. DCLICKL on the boxe d K (the coupling symbol) and enter the reference designator names for the coupled inductors as v alues of Li, i = 1, 2, …, 6. For example, if inductors L20 and L23 are coupled, we set L1 = L20 and L2 = L23. L1 and at least one other Li must be assigned values; other Li’s may be left blank.

In step 4, up to six coupled inductors with equal coupling can be specified.

For the air -core transformer, the partname is XFRM_LINEAR. It can be inserted in a circuit by selecting Draw/Get P art Name and then typing in the part name or by selecting the part name from the analog.slb library. As shown typically in Fig. 13.52(a), the main attributes of the linear transformer are the coupling coef ficient k and the inductance v alues L1 and L2 in henries. If the mutual induc tance M is specified, its value must be used along with L1 and L2 to calculate k. Keep in mind that the v alue of k should lie between 0 and 1.

For the ideal transformer , the part name is XFRM_NONLINEAR and is located in the breako ut.slb library. Select it by clicking Draw/ Get Part Name and then typing in the part name. Its attrib utes are the coupling coefficient and the numbers of turns associated with L1 and L2, as illustrated typically in Fig. 13.52(b). The value of the coef ficient of mutual coupling k = 1.

PSpice has some additional transformer configurations that we will not discuss here.

Figure 13.51

K_Linear for defining coupling.

COUPLING = 0.5 L1_VALUE = 1 mH L2_VALUE = 25 mH (a)

Figure 13.52 (a) Linear transformer XFRM_LINEAR, (b) ideal transformer XFRM_NONLINEAR.

Use PSpice to find i1, i2, and i3 in the circuit displayed in Fig. 13.53.

For Example 13.13.

Solution:

The coupling coefficients of the three coupled inductors are determined as follows:

$$k_{12} = \frac{M_{12}}{\sqrt{L_1 L_2}} = \frac{1}{\sqrt{3 \times 3}} = 0.3333$$ $$k_{13} = \frac{M_{13}}{\sqrt{L_1 L_3}} = \frac{1.5}{\sqrt{3 \times 4}} = 0.433$$ $$k_{23} = \frac{M_{23}}{\sqrt{L_2 L_3}} = \frac{2}{\sqrt{3 \times 4}} = 0.5774$$

The operating frequency f is obtained from Fig. 13.53 as ω= 12 π=  2 πf → f = 6 Hz.

The schematic of the circuit is portrayed in Fig. 13.54. Notice ho w the dot convention is adhered to. For L2, the dot (not shown) is on pin 1 (the left-hand terminal) and is therefore placed without rotation. For L1, in order for the dot to be on the right-hand side of the inductor, the inductor must be rotated through 180 °. For L3, the inductor must be rotated through 90° so that the dot will be at the bottom. Note that the 2-H in ductor (L4) is not coupled. To handle the three coupled inductors, we use three K_LINEAR parts provided in the analog library and set the following attributes (by double-clicking on the symbol K in the box):

Figure 13.54 Schematic of the circuit of Fig. 13.53.

designators of the inductors on the schematic.

Three IPRINT pseudocomponents are inserted in the appropriate branches to obtain the required currents i1, i2, and i3. As an AC singlefrequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts =  1, Start Freq =  6, and Final Freq =  6. After saving the schematic, we select Analysis/Simulate to simulate it. The output file includes:

FREQ	IM(V_PRINT2)	IP(V_PRINT2)
6.000E+00	2.114E-01	-7.575E+01
FREQ	IM(V_PRINT1)	IP(V_PRINT1)
6.000E+00	4.654E-01	-7.025E+01
FREQ	IM(V_PRINT3)	IP(V_PRINT3)
6.000E+00	1.095E-01	1.715E+01

From this we obtain

$$\mathbf{I}_1 = 0.4654 \underline{\smash{\big)}\xspace - 70.25^{\circ}}$$

\n

$$\mathbf{I}_2 = 0.2114 \underline{\smash{\big)}\xspace - 75.75^{\circ}}, \qquad \mathbf{I}_3 = 0.1095 \underline{\smash{\big)}\xspace 17.15^{\circ}}$$

Thus,

$$i_1 = 0.4654 \cos (12 \pi t - 70.25^\circ) \text{ A}$$ $$i_2 = 0.2114 \cos(12 \pi t - 75.75^\circ) \text{ A}$$ $$i_3 = 0.1095 \cos(12 \pi t + 17.15^\circ) \text{ A}$$

Find io in the circuit of Fig. 13.55, using PSpice.

For Practice Prob. 13.13.

Answer: 2.012 cos(4t + 68.52°) A.

Find V1 and V2 in the ideal transformer circuit of Fig. 13.56, using PSpice.

Practice Problem 13.13

Example 13.14

Solution:

• 1. Define. The problem is clearly defined and we can proceed to the next step.
• 2. Present. We have an ideal transformer and we are to find the input and the output voltages for that transformer. In addition, we are to use PSpice to solve for the voltages.
• 3. Alternative. We are required to use PSpice. We can use mesh analysis to perform a check.
• 4. Attempt. As usual, we assume ω = 1 and find the corresponding values of capacitance and inductance of the elements:

$$j10 = j\omega L \qquad \Rightarrow \qquad L = 10 \text{ H}$$ $$-j40 = \frac{1}{j\omega C} \qquad \Rightarrow \qquad C = 25 \text{ mF}$$

Figure 13.57 shows the schematic. For the ideal transformer, we set the coupling factor to 0.99999 and the numbers of turns to 400,000 and 100,000. The two VPRINT2 pseudocomponents are connected across the transformer terminals to obtain V1 and V2. As a single-frequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq =  0.1592, and Final Freq =  0.1592. After saving the schematic, we select Analysis/Simulate to simulate it. The output file includes:

FREQ		VM($N_0003,$N_0006) VP($N_0003,$N_0006)
	1.592E-01 9.112E+01	3.792E+01
FREQ		VM($N_0006,$N_0005) VP($N_0006,$N_0005)
	1.592E-01 2.278E+01	-1.421E+02

This can be written as

$$V_1 = 91.12 \times 37.92^{\circ}$$

V and $V_2 = 22.78 \times 142.1^{\circ}$ V

5. Evaluate. We can check the answer by using mesh analysis as follows:

Loop 1

$$-120\sqrt{30^{\circ}} + (80 - j40)I_1 + V_1 + 20(I_1 - I_2) = 0$$

Loop 2

$$20(-I_1 + I_2) - V_2 + (6 + j10)I_2 = 0$$

Figure 13.57 The schematic for the circuit in Fig. 13.56.

Reminder: For an ideal transformer, the inductances of both the primary and secondary windings are infinitely large.

But

$$V_2 = -V_1/4$$

and $I_2 = -4I_1$ . This leads to
\n $-120/30^\circ + (80 - j40)I_1 + V_1 + 20(I_1 + 4I_1) = 0$
\n $(180 - j40)I_1 + V_1 = 120/30^\circ$
\n $20(-I_1 - 4I_1) + V_1/4 + (6 + j10)(-4I_1) = 0$
\n $(-124 - j40)I_1 + 0.25V_1 = 0$ or $I_1 = V_1/(496 + j160)$

Substituting this into the first equation yields

$$(180 - j40)V_1/(496 + j160) + V_1 = 120/30^{\circ}$$ $$(184.39/-12.53^{\circ}/521.2/17.88^{\circ})V_1 + V_1$$ $$= (0.3538/-30.41^{\circ} + 1)V_1 = (0.3051 + 1 - j0.17909)V_1 = 120/30^{\circ}$$ $$V_1 = 120/30^{\circ}/1.3173/-7.81^{\circ} = 91.1/37.81^{\circ} V \qquad \text{and}$$ $$V_2 = 22.78/-142.19^{\circ} V$$

Both answers check.

6. Satisfactory? We have satisfactorily answered the problem and checked the solution. We can now present the entire solution to the problem.

Obtain V1 and V2 in the circuit of Fig. 13.58 using PSpice.

For Practice Prob. 13.14.

Answer: V1 = 153⧸ 2.18° V, V2 = 230.2⧸ 2.09° V.