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A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will be
- a)0.847
- b)0.628
- c)0.734
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A load of 1 KΩ is connected to a diode detector which is shunted by a...
We know,
Given, R = 1kΩ = 1000Ω
C = 10000pF = 10−8F
ωm = 2π × 104rad/sec
On putting these values in (1) we will get,
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Community Answer
A load of 1 KΩ is connected to a diode detector which is shunted by a...
Calculation:
1. The maximum permissible depth of modulation can be calculated using the formula:
\(m_{max} = \frac{1}{1 + k^2}\)
2. Given that the load resistance (R) is 1 KΩ, the forward resistance of the diode (r) is 1 Ω, and the modulating signal frequency (fm) is 10 kHz.
3. The reactance of the shunt capacitor (Xc) can be calculated using the formula:
\(Xc = \frac{1}{2πfC}\)
4. Substituting the values of frequency (10 kHz) and capacitance (10 pF) into the formula, we get Xc = 1591.55 Ω.
5. The depth of modulation can be calculated using the formula:
\(m = \frac{Xc}{Xc + r + R}\)
6. Substituting the values of Xc, r, and R into the formula, we get m = 0.847.
Result:
Therefore, the maximum permissible depth of modulation to avoid diagonal clipping with a modulating signal frequency of 10 kHz is 0.847. Hence, the correct answer is option A.
1. The maximum permissible depth of modulation can be calculated using the formula:
\(m_{max} = \frac{1}{1 + k^2}\)
2. Given that the load resistance (R) is 1 KΩ, the forward resistance of the diode (r) is 1 Ω, and the modulating signal frequency (fm) is 10 kHz.
3. The reactance of the shunt capacitor (Xc) can be calculated using the formula:
\(Xc = \frac{1}{2πfC}\)
4. Substituting the values of frequency (10 kHz) and capacitance (10 pF) into the formula, we get Xc = 1591.55 Ω.
5. The depth of modulation can be calculated using the formula:
\(m = \frac{Xc}{Xc + r + R}\)
6. Substituting the values of Xc, r, and R into the formula, we get m = 0.847.
Result:
Therefore, the maximum permissible depth of modulation to avoid diagonal clipping with a modulating signal frequency of 10 kHz is 0.847. Hence, the correct answer is option A.
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A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
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