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In the figure below, i1 = 6.69 A and i2 = 2.21 A. What is the value of the line integral of B→ ? T⋅m This is how to assign a sign to a current used in Ampere's law.

In the figure below, i1 = 6.69 A and i2 = 2.21 A. What is the value of the line integral of B→ ? T⋅m This is how to assign a sign to a current used in Ampere's law.

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In the figure below, i 1 = 6.69 A and i 2 = 2.21 A . What is the value of the line integral of B → ? i T ⋅ m This is how to assign a sign to a current used in Ampere's law.

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Mastering presents homework items assigned by your instructor and works with you to answer them. Homework items typically have an introduction, possibly figures, and one or more parts for you to answer. Part A What is the magic number in (Figure 1)? You could try to guess the magic number but you would probably use up all your tries before getting the correct answer. Notice the View Available Hint(s) link underneath the question statement for this part. Selecting this link will open up a list of hints that will guide you to the correct number. Hint 1 for Part A. How to approach the problem Notice that there are two hints for this question. You are not required to use all of the hints or to use them in order. Each hint has a tagline that describes its contents. Based on the tagline you can decide whether or not a particular hint will be useful to you. Your instructor may choose to give you a bonus for not using hints or to deduct a small penalty for using hints. If you are stuck, using the hints will usually result in a higher score than simply trying to guess because you may lose fewer points for opening a hint than for getting the answer to the main question incorrect. To help you, the magic number is equal to 10⋅x, where x is a number between 1 and 8. Now, open up the second hint to uncover the value of x. Hint 2for Part A. What is x? You are told from the previous hint that the magic number is equal to 10⋅x. For this problem, x is equal to 6. Use this to compute the magic number and enter it below. magic number Part B Multiple-choice questions have a special grading rule determined by your instructor. Assume that your instructor has decided to grade these questions in the following way: If you submit an incorrect answer to a multiple-choice question with n options, you will lose 1/(n-1) of the credit for that question. Just like the similar multiple-choice penalty on most standardized tests, this rule is necessary to prevent random guessing. If a multiple-choice question has five answer choices and you submit one wrong answer before getting the question correct, how much credit will you lose for that part of the question? 100% 50% 33% 25% 20%
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Grading for Numeric Answers Tolerance Some numeric answers to questions need to be exact. For example, the answer to the question "How many days are in a week?" must be exactly 7. In general though, numeric answers are graded as correct if they fall within an acceptable range (or tolerance) of the official correct answer. For example, if the answer to a numeric problem with a tolerance of 2% was 105, then entering any value between approximately 103 and 107 would be graded as correct, as shown in the region shaded green below. The typical grading tolerance in Mastering is 2-3%, although this value may vary (e.g. more lenient) depending on the particular problem. A number line is marked at 100, 105, and 110, where 105 is the properly-rounded answer. A green shaded region extends beyond 105 by 2 percent of this value in each direction. 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Current Mirrors (35 points) This problem introduces current mirrors, which will be covered in more details in future lectures. Consider the circuit shown below (Figure 2). (W/L)1 = 1, (W/L)2 = 5, Rref = 50 kΩ, ignore channel length modulation unless otherwise stated. Assume L1 = L2 = 1 μm. Figure 2 a) (5 points) Assume RL = 10 kΩ. Calculate IL, the current flowing through RL. What is the current mirroring ratio ( IL over Iref , where Iref is the current through Rref )? We want the current mirror to provide a load-independent current as much as possible, so check that M2 is in saturation. b) (5 points) Determine the maximum RL and minimum VOut such that the current mirror still functions with the same current mirroring ratio you calculated in part a). Hint: for the current mirror to function as desired, transistor saturation should be maintained. c) (5 points) Channel length modulation introduces load-dependent errors in the mirroring ratio. Using λ = 0.1 V−1 for L = 1 μm, estimate the percentage error in the mirroring ratio. Here, set RL to its maximum value you got in part b ), and assume that we want the same IL as in parts a) and b). Hint: what should be the new VX and Iref ? d) (4 points) Compute the output resistance, Rout, , looking into the drain of M2. Use λ = 0.1 V−1 and assume the same IL as in parts a) through c). e) (5 points) Now consider redesigning the previous current mirror with source-degenerated transistors as shown in Figure 3. Ignore channel length modulation and body effect ( Assume bulk is tied to source). Transistor sizes remain the same, (WL)1 = 1 and (WL)2 = 5. This circuit has been designed to have the same Iref as in part a). What should the value of Rs2 be if we want the same current mirroring ratio as in a)? f) (5 points) Similarly to part b, determine the maximum RL and minimum VOut such that the current mirror still functions with the same current mirroring ratio as in part e). g) (4 points) Now compute the output resistance, Rout , looking into the drain of M2 (using λ = 0.1 V−1 ). Assume IL as used in parts e) and f ). h) (2 points) Comparing the minimum output voltage and the output resistance of these two current mirrors, comment on the advantages and disadvantages of using source degeneration in a current mirror. Figure 3 Current Mirrors (35 points) This problem introduces current mirrors, which will be covered in more details in future lectures. Consider the circuit shown below (Figure 2). (W/L)1 = 1, (W/L)2 = 5, Rref = 50 kΩ, ignore channel length modulation unless otherwise stated. Assume L1 = L2 = 1 μm. Figure 2 a) (5 points) Assume RL = 10 kΩ. Calculate IL, the current flowing through RL. What is the current mirroring ratio ( IL over Iref , where Iref is the current through Rref )? We want the current mirror to provide a load-independent current as much as possible, so check that M2 is in saturation. b) (5 points) Determine the maximum RL and minimum VOut such that the current mirror still functions with the same current mirroring ratio you calculated in part a). Hint: for the current mirror to function as desired, transistor saturation should be maintained. c) (5 points) Channel length modulation introduces load-dependent errors in the mirroring ratio. Using λ = 0.1 V−1 for L = 1 μm, estimate the percentage error in the mirroring ratio. Here, set RL to its maximum value you got in part b ), and assume that we want the same IL as in parts a) and b). Hint: what should be the new VX and Iref ? d) (4 points) Compute the output resistance, Rout, , looking into the drain of M2. Use λ = 0.1 V−1 and assume the same IL as in parts a) through c). e) (5 points) Now consider redesigning the previous current mirror with source-degenerated transistors as shown in Figure 3. Ignore channel length modulation and body effect ( Assume bulk is tied to source). Transistor sizes remain the same, (WL)1 = 1 and (WL)2 = 5. This circuit has been designed to have the same Iref as in part a). What should the value of Rs2 be if we want the same current mirroring ratio as in a)? f) (5 points) Similarly to part b, determine the maximum RL and minimum VOut such that the current mirror still functions with the same current mirroring ratio as in part e). g) (4 points) Now compute the output resistance, Rout , looking into the drain of M2 (using λ = 0.1 V−1 ). Assume IL as used in parts e) and f ). h) (2 points) Comparing the minimum output voltage and the output resistance of these two current mirrors, comment on the advantages and disadvantages of using source degeneration in a current mirror. Figure 3

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