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W08 314 HW03 solutions

# W08 314 HW03 solutions - EECS 314 Winter 2008 HW 03...

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Part 1 First, before using the golden rule, construct a node voltage equation at the V + node. This is done by applying KCL to account for all currents exiting or entering the node, followed by rewriting current values wherever possible using given currents or Ohm’s law: I 500 Ω I 200 Ω i 0 5mA v 0 200 Ω i 0 5mA v 200 Ω i 0 This equation is valid, but is made simpler by applying golden rule one: currents at the input nodes of op-amps are always zero. Substitute zero for i + and rewrite: 5mA v 200 Ω 0 v 1V To f nd the voltage at the non-inverting terminal. Part 2 Construct another node voltage equation as done in Part 1. This time, there are no current sources, and many more explicitly labeled voltage nodes: I 10k Ω I 50k Ω i 0 v 3V 10k Ω v V out 50k Ω i 0 6 v V out i 5 Ω 15V This time, in addition to applying GR 1, apply GR 2 because of the negative feedback: v + = v - . 6 1V V out 0 5 Ω 15V V out 21V EECS 314 Winter 2008 HW 03 Solutions Problem 1

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Part 3 Each current is calculated by: I 10k Ω 1V 3V 10k Ω 0 . 4mA I 50k Ω 1V 21V 50k Ω 0 . 4mA I 91 Ω 21V 0V 91 Ω 230 . 8mA Part 4 Output current is found by: I out I 50k Ω I 91 Ω I out 0 . 4mA 230 . 8mA 231 . 2mA KCL at the ground node should satisfy this equation, but it does not: I 10k Ω I 500 Ω I 200 Ω I 91 Ω 0 0 . 4mA 5mA 5mA 230 . 8mA 231 . 6mA ± 0 EECS 314 Winter 2008 HW 03 Solutions Problem 1
EECS 314 Winter 2008 Homework 3 Solutions Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2008 Alexander Ganago Page 1 of 4 Problem 2 The Big Picture The circuit shown on this diagram is known as an inverting summer, or summing amplifier, a.k.a. adder. This summer is inverting, because the signals are fed into the inverting (-) input terminal of the Op Amp. Its output signal V OUT is a sum of the two input signals with individual gain values: V OUT = K 1 ± V S1 + K 2 V S2 [1] Note that coefficients K 1 and K 2 can be individually – and independently! – changed if the resistances R 1 and R 2 are varied. This is a great advantage of the inverting summer: in its twin brother – the non-inverting summer, where the signals are fed into the non- inverting (+) input terminal of the Op Amp – changing one of the resistors results in a change of both gain values thus independent adjustment of gains is nearly impossible. The inverting summer circuit has many applications; one of the most obvious is the mixer at an audio recording studio, where the signals V S1 and V S2 come from two microphones. In control systems, a similar circuit with 3 inputs can play the role of the adder (such as needed for P + I + D control, which we will discuss later in the course). For conditioning of sensor signals, one input V S1 might be used for the sensor signal itself (to be amplified with the desired gain) while the other input V S2 might be used for compensation of DC offset (see Part 2 of this problem.

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W08 314 HW03 solutions - EECS 314 Winter 2008 HW 03...

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