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Chapter 10

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

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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

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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
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

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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.
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