Chapter 10

Chapter 10 - CHAPTER 10 Solutions for Exercises E10.1...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 10 Solutions for Exercises E10.1 Solving Equation 10.1 for the saturation current and substituting values, we have 4 15 exp( / ) 1 10 exp(0.600 / 0.026) 1 9.502 10 A D s T D i I v nV - - = - = - = × Then for 0.650 D v = V, we have 15 exp( / ) 1 9.502 10 exp(0.650 / 0.026) 1 0.6841 mA s T D D i I v nV - = - = × × - = Similarly for 0.700 D v = V, 4.681 D i = mA. E10.2 The approximate form of the Shockley Equation is exp( / ) s T D D i I v nV = . Taking the ratio of currents for two different voltages, we have 1 1 1 2 2 2 exp( / ) exp ( ) / exp( / ) T D D T D D T D D i v nV v v nV i v nV = = - Solving for the difference in the voltages, we have: 1 2 ln( / ) T D D D v nV i i = Thus to double the diode current we must increase the voltage by 0.026ln(2) 18.02 mV D v = = and to increase the current by an order of magnitude we need 0.026ln(10) 59.87 mV D v = = E10.3 The load line equation is . SS D D V Ri v = + The load-line plots are shown on the next page. From the plots we find the following operating points: (a) 1.1 V 9 mA DQ DQ V I = = (b) 1.2 V 13.8 mA DQ DQ V I = = (c) 0.91 V 4.5 mA DQ DQ V I = = 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
E10.4 Following the methods of Example 10.4 in the book, we determine that: (a) For 1200 , 600 , and 12 V. T T L R R V = = = (b) For 400 , 300 , and 6 V. T T L R R V = = = The corresponding load lines are: 2
Background image of page 2
At the intersections of the load lines with the diode characteristic we find (a) 9.4 V L D v v = - 2245 ; (b) 6.0 V L D v v = - 2245 . E10.5 Writing a KVL equation for the loop consisting of the source, the resistor, and the load, we obtain: 15 100( ) L D D i i v = - - The corresponding load lines for the three specified values of i L are shown: At the intersections of the load lines with the diode characteristic, we find (a) 10 V; o D v v = - = (b) 10 V; o D v v = - = (c) 5 V. o D v v = - = Notice that the regulator is effective only for values of load current up to 50 mA. E10.6 Assuming that D 1 and D 2 are both off results in this equivalent circuit: Because the diodes are assumed off, no current flows in any part of the circuit, and the voltages across the resistors are zero. Writing a KVL 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
equation around the left-hand loop we obtain 1 10 D v = V, which is not consistent with the assumption that D 1 is off. E10.7
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 13

Chapter 10 - CHAPTER 10 Solutions for Exercises E10.1...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online