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Apply the de nitions of the z parameters, given by (1.10) through (1.13), to the CE h-parameter circuit of Fig. 6-1(a) to nd values for the z parameters in the equivalent circuit of Fig. 6-29 in terms of the CE h parameters. Ans: z11 hie hfe hre =hoe ; z12 hre =hoe ; z21 hfe =hoe ; z22 1=hoe Use the z-parameter model of Fig. 6-29 to calculate (a) the current-gain ratio Ai iL =ii and voltage-gain ratio Av vL =vi for the ampli er of Fig. 3-10(a). Ans: Ai RC RB z21 = RC RL RB z11 z22 RC kRL z12 z21 ; Av z21 RB RL kRC =f z22 RC kRL RB Ri z11 RB Ri g (b) the

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

For the CE ampli er of Fig. 3-17 with values as given in Problem 6.18, nd (a) the input resistance Ri and (b) the output resistance Ro . Ans: a 24:26 ; b 2:154 k A CE transistor ampli er is operating in the active region, with VCC 12 V and Rdc 2 k . If the collector characteristics are given by Fig. 3-9(b) and the quiescent base current is 30 A, determine (a) hfe and (b) hoe . Ans: a 190; b 83:33 S In the circuit of Fig. 6-30, hre 10 4 ; hie 200 ; hfe 100, and hoe 100 S. (a) Find the power gain as Ap jAi Av j, the product of the current and voltage gains. (b) Determine the numerical value of RL that maximizes the power gain. Ans: a h2 =j hoe RL 1 hoe hie hre hfe hie R 1 j; b 14:14 k L fe

On the other hand, some devices don t fit in this classification, such as a monitor with memory mapped video..

public class TestPolynomial { public static void main(String[] args) { Polynomial p = new Polynomial(3, -8, 0, 0, 2, 1); Polynomial q = new Polynomial(0, 5, 6, 9); System.out.println("p: " + p); System.out.println("p.degree(): " + p.degree()); System.out.println("q: " + q); System.out.println("q.degree(): " + q.degree()); System.out.println("p.plus(q): " + p.plus(q)); System.out.println("q.plus(p): " + q.plus(p)); System.out.println("p.plus(q).degree(): " + p.plus(q).degree()); Polynomial z = new Polynomial(0); System.out.println("z: " + z); System.out.println("z.degree(): " + z.degree()); System.out.println("p.plus(z): " + p.plus(z)); System.out.println("z.plus(p): " + z.plus(p)); System.out.println("p: " + p); Polynomial t = new Polynomial(8.88, 44); System.out.println("t: " + t); System.out.println("t.degree(): " + t.degree()); } }

Fig. 6-30

The output is:

The EF ampli er of Fig. 3-26(a) utilizes a Si transistor with negligible leakage current and 59. Also, VCC 15 V; VL 3 V (VL is the dc component of vL ), and RE 1:5 k . Calculate (a) RB ; b the output impedance Zo , and (c) the input impedance Zin . Ans: a 339 k ; b 1:185 k ; c 50:98 k

Programmed I/O Simple character-at-a-time (byte-at-a-time) I/O is sometimes called programmed I/O to distinguish it from interrupt-driven I/O or direct memory access (DMA), which we will discuss shortly With programmed I/O, the driver checks to see if the device is available (not busy) If so, the driver will write a character to the buffer on the controller, and command the transfer to begin with a set control; clear flag (different computers may name these status bits differently) Set control tells the device to start, and clear flag makes the device appear busy The driver then loops, testing for the flag to be set When the byte is transferred, the controller will clear the control bit and set the flag bit The driver will find the flag bit is set, and that will indicate that the transfer is complete and the device is again available to accept another character.

The ampli er of Fig. 6-31 has an adjustable emitter resistor RE , as indicated, with 0 1. Assume that hre hoe % 0 and Cc ! 1, and nd expressions for (a) the current-gain ratio Ai iL =is , (b) the voltagegain ratio Av vL =vs , and (c) the input impedance Zin . Ans: hfe RB RL ; RC RL RB hie hfe 1 RE hfe RC RB RL ; b Av RC RL fRS RB RS RB hie hfe 1 RE g RB hie hfe 1 RE c Ri RB hie hfe 1 RE a Ai