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As a consequence of the above approximations, vL and iL can be appropriately determined by (10.1) and (10.2) when Q is OFF and by (1) and (2) of Example 10.1 when Q is ON (replace VC with V1 ). Diode current iD must be equal to iL when Q is OFF. Since load current I2 is the average value of iD , iC iD I2 . Figure 10-5 displays the resulting waveforms for vQ ; iL ; vL ; iD ; and iC : Based on (10.3) and the vL waveform of Fig. 10-5, V1 DTs V2 V1 1 D Ts

The minimum cardinality for both sides of the relationship will be 1, because each student will certainly take some courses and - each course will be attended by some students On the other hand, if we keep courses in the database that are no longer actively taught for some reason, then the minimum cardinality on the student side of the takes/is-taken-by relationship will be 0 Figure 8-1 shows a data model for the entities and relationships we have been discussing, using one of many standard approaches for graphically representing the entity-relationship diagram Figure 8-1 was created using Microsoft Visio The rectangles represent entities, and the label in the upper portion of an entity rectangle specifies the identifier, or key, for the entity type For dormitories, for example, the dorm name is the identifier; the name of the dorm distinguishes the record of one dorm from that of another.

public class TestContains { static public void main(String[] args) { BinaryTree<String> e = new BinaryTree<String>("E"); BinaryTree<String> g = new BinaryTree<String>("G"); BinaryTree<String> h = new BinaryTree<String>("H"); BinaryTree<String> i = new BinaryTree<String>("I"); BinaryTree<String> d = new BinaryTree<String>("D", null, g); BinaryTree<String> f = new BinaryTree<String>("F", h, i); BinaryTree<String> b = new BinaryTree<String>("B", d, e); BinaryTree<String> c = new BinaryTree<String>("C", f, null); BinaryTree<String> a = new BinaryTree<String>("A", b, c); System.out.printf("a: %s%n", a); System.out.println("a.contains(\"H\") = " + a.contains("H")); System.out.printf("b: %s%n", b); System.out.println("b.contains(\"H\") = " + b.contains("H")); System.out.printf("c: %s%n", c); System.out.println("c.contains(\"H\") = " + c.contains("H")); } }

Rearrangement nds the ideal boost converter voltage gain as GV V2 1 V1 1 D 10:7

Unlike the buck converter, the boost converter gain is not a linear function of D. From (10.7), the ideal gain approaches in nity as D approaches 1. When parasitic resistances of the inductor and capacitor 0 are considered, the actual gain GV departs signi cantly from the ideal gain for values of D > 0:8. (See Problem 10.11.) The common case of continuous current iL exists only if the value of L ! Lc (critical inductance) that results in marginally continuous conduction for iL . For this case, Imin 0 in Fig. 10-5 and iL 0 0. By application of (10.2), iL t Evaluate for t DTs and use (10.7) to nd iL DTs iD DTs Imax For the triangular iD waveform, I2 hiD i 1 1 D Ts V I 2 D 1 D 2 Ts 2 max Ts 2Lc After rearrangement, 10:8

The output is:

The labels in the lower portion of the entity rectangles represent the other attributes of the entity The dorm entity includes information for each dorm about the total number of rooms in the dorm, the number of vacant rooms in the dorm, and the room rental rate for the dorm..

But I2 V2 =RL , which can be equated to the above expression for I2 . Lc 1 D DTs RL 1 D DRL 2 2fs

a: [A, B, D, G, a.contains("H") b: [B, D, G, E] b.contains("H") c: [C, F, H, I] c.contains("H") E, C, F, H, I] = true = false = true

Example 10.3. A boost converter with a 20-kHz switching frequency is operating with a 50 percent duty cycle. The connected load is 7 . Determine the value of critical inductance so that current iL is continuous. By (10.8), Lc 1 D 2 DRL 1 0:5 2 0:5 7 21:9 H 2fs 2 20 103

The subtrees b and c are shown in Figure 11.22. The tree a contains the element H. The subtree b does not contain the element H. The subtree c does contain the element H.

The SMPS circuit of Fig. 10-6 is a buck-boost converter. The value of output voltage for this converter may either be less than or greater than the input voltage, depending on the value of duty cycle D. Unlike the buck and boost converters, the buck-boost converter produces an output voltage with polarity opposite to input voltage V1 . The polarity of v2 and the direction of I2 in Fig. 10-6 are chosen so that V2 hv2 i ! 0 and I2 ! 0.