hmwk 8 sol - EECS 314 Fall 2006 HW 08 Problem 1 Student's...

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EECS 314 Fall 2006 HW 08 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) r The big picture The circuit shown on this diagram is known as an inverting summer, o summing amplifier, a.k.a. adder. Its output signal V OUT is a sum of the two input signals V OUT = K 1 V S1 + K 2 V S2 where the coefficients K 1 and K 2 can be individually changed if the resistances R 1 and R 2 are varied. This 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). Problem (Assume an ideal Op Amp and use the “Golden Rules” in all derivations; show your work on additional pages) Part 1 (10 points) For the circuit shown on the top diagram, write a complete node voltage equation for node A; then simplify the equation by applying the “Golden Rules” and derive the coefficients K 1 and K 2 (since the circuit is based on an inverting amplifier, both are negative). Part 2 (5 points) Use your results from Part 1 and choose resistances R 1 , R 2 and R F to ensure V OUT = (-6) V S1 + (-9) V S2 Use the resistance values, which are multiples of 1 k Ω . Part 3 (10 points) Express the output voltage in the circuit with 3 inputs through the input voltages and resistances; determine the ratio of resistances for using the circuit as an adder V OUT,3 = -(V S1 + K 2 V S2 + K 3 V S3 ) Alexander Ganago Page 1 of 4
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EECS 314 Fall 2006 HW 08 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) Solution (Barry Mullins, adapted from Kevin Klein’s W06 solution) Part 1 We must assume that the currents flowing into the three terminals of the Op-Amp are negligible for this solution. Then we have: 4 1 I I = Rewriting using V/R: F Out A R V V I = 1 And rearranging gives: F A Out R I V V 1 = But, because of the golden rule, V A =0V F Out R I V 1 = Replacing I 1 with I 2 + I 3 gives: 2 2 1 1 S F S F Out V R R V R R V = Where we have used the fact that 2 2 1 1 1 R V R V I S S + = Alexander Ganago Page 2 of 4
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EECS 314 Fall 2006 HW 08 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) Comparing with the formula given in the problem, we see that the coefficients are Part 2 We are given the coefficients as K 1 = -6 and K 2 = -9. Substitute these values in the above equations for the coefficients: 1 1 6 R R K F = = and 2 2 9 R R K F = = A solution to these equations is R F = 54k , R 1 = 9k , R 2 =6k . Part 3 Begin by replacing the op-amp with its equivalent model and redrawing the circuit: First, write a KCL equation at the center node: f I I I I = + + 3 2 1 Now rewrite the currents as V/R: Alexander Ganago Page 3 of 4
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EECS 314 Fall 2006 HW 08 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) f out S S S R V R V R V R V = + + 3 3 2 2 1 1 Now solve for the output voltage: + + = 3 3 2 2 1 1 R R V R R V R R V V f S f S f S out Now compare this result with the equation given in the problem statement.
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hmwk 8 sol - EECS 314 Fall 2006 HW 08 Problem 1 Student's...

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