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The ArrayStack class implemented in Example 5.3 can be tested the same way the Stack class is tested in Example 5.1 on page 103. The push(), peek(), and pop() calls should work the same way. The println() calls cannot be executed for the ArrayStack class because it has no toString() method. But that s proper, because a stack really should not allow access to any of its elements other than the one on top. A LINKED IMPLEMENTATION The main alternative to an indexed implementation is a linked implementation, using a linked list. (See 3.) EXAMPLE 5.4 A LinkedStack Implementation

3-17, VCC 15 V; RC 2 k ; RF 150 k ; ICEO % 0, and the transistor is a Si device (a) Find an expression for the sensitivity factor S (b) Use S to predict the change in quiescent collector current due to a change in from 50 to 100 Ans: a S RF RC VCC VBEQ = RF 1 RC 2 ; b S 3:432 10 5 ; ICQ 1:71 mA (compare with Example 54) In the CB ampli er of Fig 3-23, VCC 15 V; VEE 5 V; RE 3 k ; RC 7 k ; and 50 (a) Find an expression for the sensitivity factor S (b) Evaluate S assuming the transistor is a Si device Ans: a S VEE VBEQ = 1 2 RE ; b S 5:51 10 7 (very low sensitivity, but see Problem 58) The circuit of Fig.

switch( yearInSchool ) { case 1: System.out.println( "Oberlies Hall" ); break; case 2: System.out.println( "Kiriazides Hall" ); break; case 3: System.out.println( "Glisson Dorm" ); break; case 4: System.out.println( "Valk Hall" ); break; default: System.out.println( "illegal year" ); } If a student is a freshman, the student will be assigned to Oberlies Hall; if the student is a sophomore, the student will be assigned to Kiriazides Hall; etc. Notice the break statements. The break statement says to exit the switch statement. If the break statement is missing, the execution of the switch statement will fall through to the next case. Sometimes you will want that to happen, but certainly not always. Forgetting to insert the break statement appropriately is a common programming error. Here s an example using the switch statement to allow falling through from one condition to the next in a helpful way: switch( case case case case case case case case case case case case dayOfChristmas ) { 12: System.out.println( 11: System.out.println( 10: System.out.println( 9: System.out.println( 8: System.out.println( 7: System.out.println( 6: System.out.println( 5: System.out.println( 4: System.out.println( 3: System.out.println( 2: System.out.println( 1: System.out.println( break; default: System.out.println(

1 2 3 4

5-1 has the values given in Problem 511; assume that the initial values of ICBO and VBEQ are for 258C (a) Find an expression for the value of ICQ at any temperature T2 ! 258C if the transistor is a Si device (b) Evaluate the expression for ICQ at T2 1258C Ans: a ICQ b ICQ VBB 0:7 0:002 T2 25 0:5 10 6 RB RE 2 T2 25 =10 ; RB = RE 18:97 mA.

The constant-base-current-biased ampli er of Fig. 5-7 contains a Si transistor. Let VCC 15 V; RC 2:5 k ; RE 500 ; Rb 500 k ; and 100. At 258C, ICBO 0:5 A and VBEQ 0:7 V. (a) Find the exact change in ICQ if the temperature changes to 1008C. (b) Use the stability factors developed in Problem 5.13 to predict ICQ for a temperature increase to 1008C. Ans: a ICQ ICQ2 ICQ1 10:864 2:645 8:219 mA; b ICQ 8:22 mA In the constant-base-current-biased ampli er of Fig. 5-7, the Si transistor is characterized by ICBO 0:5 A and VBEQ 0:7 V at 258C. (a) Find an expression for ICQ at any temperature T2 ! 258C. (b) Evaluate ICQ at 1008C if VCC 15 V; RC 2:5 k ; RE 500 ; Rb 500 k ; and 100. Ans: a ICQ b ICQ VCC 0:7 0:002 T2 25 0:5 10 6 Rb RE 2 T2 25 =10 ; Rb = RE 10:864 mA

public class LinkedStack<E> implements Stack<E> { private Node<E> head = new Node<E>(); // dummy node private int size;