Chapter 10 - CHAPTER 10 Exercises E10.1 Solving Equation 10.1 for the saturation current and substituting values we have Is = iD exp(vD nVT 1 10 4

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1 CHAPTER 10 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 i I vn V = = Then for 0.650 v = V, we have 15 exp( / ) 1 9.502 10 exp(0.650/0.026) 1 0.6841 mA DD iI n =− = × ×   = Similarly for 0.700 = V, 4.681 = mA. E10.2 The approximate form of the Shockley Equation is exp( / ) = . Taking the ratio of currents for two different voltages, we have 11 12 22 exp( / ) exp ( )/ exp( / ) iv vv == Solving for the difference in the voltages, we have: ln( / ) Vii ∆= Thus to double the diode current we must increase the voltage by 0.026ln(2) 18.02 mV = and to increase the current by an order of magnitude we need 0.026ln(10) 59.87 mV = E10.3 The load line equation is . SS Ri =+ 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 (b) 1.2 V 13.8 mA (c) 0.91 V 4.5 mA

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2 E10.4 Following the methods of Example 10.4 in the book, we determine that: (a) For 1200 , 600 , and 12 V. TT L RR V =Ω = = (b) For 400 , 300 , and 6 V. = The corresponding load lines are:
3 At the intersections of the load lines with the diode characteristic we find (a) 9.4 V LD vv =− ; (b) 6.0 V . E10.5 Writing a KVL equation for the loop consisting of the source, the resistor, and the load, we obtain: 15 100( ) D ii 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 = (b) 10 V; = (c) 5 V. = Notice that the regulator is effective only for values of load current up to 50 mA. E10.6 Assuming that 1 and 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 equation around the left-hand loop we obtain 1 10 = V, which is not consistent with the assumption that 1 is off.

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4 E10.7 Assuming that D 1 and 2 are both on results in this equivalent circuit: Writing a KVL equation around the outside loop, we find that the voltage across the 4-k resistor is 7 V and then we use Ohm’s law to find that i 1 equals 1.75 mA. The voltage across the 6-k resistance is 3 V so x is 0.5 mA. Then we have 21 1.25 DD ii =− = mA, which is not consistent with the assumption that 2 is on. E10.8 (a) If we assume that 1 is off, no current flows, the voltage across the resistor is zero, and the voltage across the diode is 2 V, which is not consistent with the assumption. If we assume that the diode is on, 2 V appears across the resistor, and a current of 0.5 mA circulates clockwise which is consistent with the assumption that the diode is on. Thus the diode is on. (b) If we assume that 2 is on, a current of 1.5 mA circulates counterclockwise in the circuit, which is not consistent with the assumption. On the other hand, if we assume that 2 is off we find that 2 3 v where as usual we have referenced 2 positive at the anode.
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This note was uploaded on 11/02/2010 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Institute of Technology.

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Chapter 10 - CHAPTER 10 Exercises E10.1 Solving Equation 10.1 for the saturation current and substituting values we have Is = iD exp(vD nVT 1 10 4

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