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An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fb and the open-loop gain at fb.
- a)15Hz and 103 db
- b)30Hz and 103 db
- c)15 Hz and 51.5 db
- d)30 Hz and 51.5 db
Correct answer is option 'A'. Can you explain this answer?
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An internally compensated op amp is specified to have an open-loop dc ...
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An internally compensated op amp is specified to have an open-loop dc ...
Given parameters:
Open-loop dc gain (Aol) = 106 dB
Unity gain bandwidth (fT) = 3 MHz
To find:
a) Open-loop gain (Aol) at 15 Hz and 103 dB
Solution:
We know that the frequency response of an op amp can be modeled as:
Aol(f) = Aol(0) / (1 + jf/fT)
Where f is the frequency, Aol(0) is the open-loop dc gain and fT is the unity gain bandwidth.
For the given problem, we need to find Aol at 15 Hz and 103 dB. Let's substitute the values in the above equation:
Aol(15Hz) = Aol(0) / (1 + j15/3x10^6)
Taking the magnitude of the above equation, we get:
|Aol(15Hz)| = Aol(0) / √(1 + (15/3x10^6)^2)
Now, substituting the given values, we get:
|Aol(15Hz)| = 106 dB / √(1 + (15/3x10^6)^2)
|Aol(15Hz)| = 103 dB (approx.)
Hence, option A is correct.
b) Open-loop gain (Aol) at 30 Hz and 103 dB
We can repeat the above steps with f = 30 Hz:
|Aol(30Hz)| = Aol(0) / √(1 + (30/3x10^6)^2)
|Aol(30Hz)| = 103 dB (approx.)
Hence, option B is incorrect.
c) Open-loop gain (Aol) at 15 Hz and 51.5 dB
We can repeat the above steps with the given values:
|Aol(15Hz)| = 51.5 dB / √(1 + (15/3x10^6)^2)
|Aol(15Hz)| = 49.5 dB (approx.)
Hence, option C is incorrect.
d) Open-loop gain (Aol) at 30 Hz and 51.5 dB
We can repeat the above steps with the given values:
|Aol(30Hz)| = 51.5 dB / √(1 + (30/3x10^6)^2)
|Aol(30Hz)| = 49.5 dB (approx.)
Hence, option D is incorrect.
Therefore, the correct answer is option A.
Open-loop dc gain (Aol) = 106 dB
Unity gain bandwidth (fT) = 3 MHz
To find:
a) Open-loop gain (Aol) at 15 Hz and 103 dB
Solution:
We know that the frequency response of an op amp can be modeled as:
Aol(f) = Aol(0) / (1 + jf/fT)
Where f is the frequency, Aol(0) is the open-loop dc gain and fT is the unity gain bandwidth.
For the given problem, we need to find Aol at 15 Hz and 103 dB. Let's substitute the values in the above equation:
Aol(15Hz) = Aol(0) / (1 + j15/3x10^6)
Taking the magnitude of the above equation, we get:
|Aol(15Hz)| = Aol(0) / √(1 + (15/3x10^6)^2)
Now, substituting the given values, we get:
|Aol(15Hz)| = 106 dB / √(1 + (15/3x10^6)^2)
|Aol(15Hz)| = 103 dB (approx.)
Hence, option A is correct.
b) Open-loop gain (Aol) at 30 Hz and 103 dB
We can repeat the above steps with f = 30 Hz:
|Aol(30Hz)| = Aol(0) / √(1 + (30/3x10^6)^2)
|Aol(30Hz)| = 103 dB (approx.)
Hence, option B is incorrect.
c) Open-loop gain (Aol) at 15 Hz and 51.5 dB
We can repeat the above steps with the given values:
|Aol(15Hz)| = 51.5 dB / √(1 + (15/3x10^6)^2)
|Aol(15Hz)| = 49.5 dB (approx.)
Hence, option C is incorrect.
d) Open-loop gain (Aol) at 30 Hz and 51.5 dB
We can repeat the above steps with the given values:
|Aol(30Hz)| = 51.5 dB / √(1 + (30/3x10^6)^2)
|Aol(30Hz)| = 49.5 dB (approx.)
Hence, option D is incorrect.
Therefore, the correct answer is option A.
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An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fband the open-loop gain at fb.a)15Hz and 103 dbb)30Hz and 103 dbc)15 Hz and 51.5 dbd)30 Hz and 51.5 dbCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fband the open-loop gain at fb.a)15Hz and 103 dbb)30Hz and 103 dbc)15 Hz and 51.5 dbd)30 Hz and 51.5 dbCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fband the open-loop gain at fb.a)15Hz and 103 dbb)30Hz and 103 dbc)15 Hz and 51.5 dbd)30 Hz and 51.5 dbCorrect answer is option 'A'. Can you explain this answer?.
An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fband the open-loop gain at fb.a)15Hz and 103 dbb)30Hz and 103 dbc)15 Hz and 51.5 dbd)30 Hz and 51.5 dbCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fband the open-loop gain at fb.a)15Hz and 103 dbb)30Hz and 103 dbc)15 Hz and 51.5 dbd)30 Hz and 51.5 dbCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fband the open-loop gain at fb.a)15Hz and 103 dbb)30Hz and 103 dbc)15 Hz and 51.5 dbd)30 Hz and 51.5 dbCorrect answer is option 'A'. Can you explain this answer?.
Solutions for An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fband the open-loop gain at fb.a)15Hz and 103 dbb)30Hz and 103 dbc)15 Hz and 51.5 dbd)30 Hz and 51.5 dbCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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