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In an ideal op amp the open-loop gain is 103. The op amp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured as v2 = 0V and v3 = 2V where it is assumed that v1 and v2 are input terminals and v3 is the output terminal. The value of the differential (vd) and common-mode (vcm)signal is
  • a)
    Vd = 2 mV and vcm = 1 mv
  • b)
     Vd = 2 mV and vcm = -1 mV
  • c)
    Vd = 2 mV and vcm = 2mV
  • d)
    Vd = 2 mV and vcm = -2mV
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
In an ideal op amp the open-loop gain is 103. The op amp is used in a ...
Vc = 0.5(V1 + V2) and
Vd = V2 – V1.
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Most Upvoted Answer
In an ideal op amp the open-loop gain is 103. The op amp is used in a ...
Given Information:
Open-loop gain (A) = 103
Input terminals (v1, v2) = unknown
Output terminal (v3) = 2V

Finding Differential and Common-Mode Signals:
To find the differential and common-mode signals, we first need to assume an output voltage at v3. Let's assume that the output voltage is equal to the voltage at v2, which is 0V. Therefore, the input voltage difference (vd) would be:

vd = v1 - v2 = v1 - 0V = v1

Now, let's find the common-mode voltage (vcm) by taking the average of input voltages:

vcm = (v1 + v2)/2 = v1/2

Given that v2 is 0V, we can rewrite the above equation as:

vcm = v1/2

Now, let's substitute the given values to find the differential and common-mode signals:

vd = v1 = v3/A = 2/103 = 0.0194V or 19.4mV

vcm = v1/2 = 0.0194/2 = 0.0097V or 9.7mV

Since v1 is positive, vcm is negative. Therefore, the correct answer is option 'B' (vd = 2mV and vcm = -1mV).
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In an ideal op amp the open-loop gain is 103. The op amp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured as v2= 0V and v3= 2V where it is assumed that v1and v2are input terminals and v3is the output terminal. The value of the differential (vd) and common-mode (vcm)signal isa)Vd= 2 mV and vcm= 1 mvb)Vd= 2 mV and vcm= -1 mVc)Vd= 2 mV and vcm= 2mVd)Vd= 2 mV and vcm= -2mVCorrect answer is option 'B'. Can you explain this answer?
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In an ideal op amp the open-loop gain is 103. The op amp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured as v2= 0V and v3= 2V where it is assumed that v1and v2are input terminals and v3is the output terminal. The value of the differential (vd) and common-mode (vcm)signal isa)Vd= 2 mV and vcm= 1 mvb)Vd= 2 mV and vcm= -1 mVc)Vd= 2 mV and vcm= 2mVd)Vd= 2 mV and vcm= -2mVCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about In an ideal op amp the open-loop gain is 103. The op amp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured as v2= 0V and v3= 2V where it is assumed that v1and v2are input terminals and v3is the output terminal. The value of the differential (vd) and common-mode (vcm)signal isa)Vd= 2 mV and vcm= 1 mvb)Vd= 2 mV and vcm= -1 mVc)Vd= 2 mV and vcm= 2mVd)Vd= 2 mV and vcm= -2mVCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an ideal op amp the open-loop gain is 103. The op amp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured as v2= 0V and v3= 2V where it is assumed that v1and v2are input terminals and v3is the output terminal. The value of the differential (vd) and common-mode (vcm)signal isa)Vd= 2 mV and vcm= 1 mvb)Vd= 2 mV and vcm= -1 mVc)Vd= 2 mV and vcm= 2mVd)Vd= 2 mV and vcm= -2mVCorrect answer is option 'B'. Can you explain this answer?.
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