This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 CHAPTER 13 Exercises E13.1 The emitter current is given by the Shockley equation: = 1 exp T BE ES E V v I i For operation with 1 exp have we , >> >> , and we can write exp Solving for , we have mV 4 . 718 10 10 ln 26 ln 14 2 = = V 2816 . 4 5 7184 . = = = CE BC 9804 . 51 50 1 = = + = mA 804 . 9 = = C A 1 . 196 = = B E13.2 = 1 0.9 9 0.99 99 0.999 999 E13.3 mA 5 . = = 95 . / = = 19 / = = E13.4 The base current is given by Equation 13.8: = = 1 026 . exp 10 961 . 1 1 exp ) 1 ( 16 which can be plotted to obtain the input characteristic shown in Figure 13.6a. For the output characteristic, we have = provided that 2 V. 0.2 ely approximat CE v For V, 0.2 C i falls rapidly to zero at . = The output characteristics are shown in Figure 13.6b. E13.5 The load lines for V 0.8and V 8 . in = are shown: As shown on the output load line, we find V. . 1 and V, 5 V, 9 min max CEQ V 3 E13.6 The load lines for the new values are shown: As shown on the output load line, we have V. . 3 and V, 7 V, 8 . 9 min max CE CEQ V 4 E13.7 Refer to the characteristics shown in Figure 13.7 in the book. Select a point in the active region of the output characteristics. For example, we could choose the point defined by mA 5 . 2 and V 6 = = C CE i v at which we find A. 50 = B Then we have . 50 / = = (For many transistors the value found for depends slightly on the point selected.) E13.8 (a) Writing a KVL equation around the input loop we have the equation for the input load lines: 8000 ) ( 8 . in = + BE t The load lines are shown: Then we write a KCL equation for the output circuit: = + 3000 9 The resulting load line is: From these load lines we find 5 , A 5 A, 24 A, 48 min max B BQ I V 3 . 8 , V 3 . 5 V, 8 . 1 min max CE CEQ V (b) Inspecting the load lines, we see that the maximum of v in corresponds to min which in turn corresponds to min . Because the maximum of in corresponds to minimum , the amplifier is inverting. This may be a little confusing because CE takes on negative values, so the minimum value has the largest magnitude....
View
Full
Document
 Spring '07
 Haris

Click to edit the document details