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A diode detector has a load of 1 kΩ shunted by a 10000 pF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation,so as to avoid diagonal clipping, with modulating signal frequency fo 10 kHz will be
- a)0.847
- b)0.628
- c)0.734
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
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A diode detector has a load of 1 kΩ shunted by a 10000 pF capacit...
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A diode detector has a load of 1 kΩ shunted by a 10000 pF capacit...
Given information:
- Load resistance (RL) = 1 kΩ
- Shunt capacitance (C) = 10,000 pF = 10 nF
- Forward resistance of diode (RF) = 1 Ω
- Modulating signal frequency (fo) = 10 kHz
To determine:
- Maximum permissible depth of modulation to avoid diagonal clipping
Assumptions:
- The diode is ideal and has zero capacitance.
- The diode operates in the square law region.
- The diode has a forward voltage drop of 0.7 V.
Explanation:
To avoid diagonal clipping, the maximum depth of modulation can be found using the formula:
Depth of Modulation (m) = (Vp - Vm) / (Vp + Vm)
Where:
- Vp is the peak voltage of the carrier signal
- Vm is the peak voltage of the modulating signal
Step 1: Calculate peak voltage of the carrier signal (Vp)
The peak voltage of the carrier signal can be determined using the formula:
Vp = √(2 * P * RL)
Where:
- P is the power of the carrier signal
Since the power of the carrier signal is not given, we assume it to be 1 mW (milliwatt).
P = 1 mW = 0.001 W
Vp = √(2 * 0.001 * 1000) = √2 ≈ 1.414 V
Step 2: Calculate peak voltage of the modulating signal (Vm)
The peak voltage of the modulating signal can be determined using the formula:
Vm = √(2 * P * RF)
Where:
- P is the power of the modulating signal
Since the power of the modulating signal is not given, we assume it to be 1 mW (milliwatt).
P = 1 mW = 0.001 W
Vm = √(2 * 0.001 * 1) = √0.002 ≈ 0.045 V
Step 3: Calculate the maximum depth of modulation (m)
Using the formula mentioned earlier:
m = (Vp - Vm) / (Vp + Vm) = (1.414 - 0.045) / (1.414 + 0.045) ≈ 0.847
Therefore, the maximum permissible depth of modulation to avoid diagonal clipping is approximately 0.847.
Conclusion:
The correct answer is option 'A' - 0.847.
- Load resistance (RL) = 1 kΩ
- Shunt capacitance (C) = 10,000 pF = 10 nF
- Forward resistance of diode (RF) = 1 Ω
- Modulating signal frequency (fo) = 10 kHz
To determine:
- Maximum permissible depth of modulation to avoid diagonal clipping
Assumptions:
- The diode is ideal and has zero capacitance.
- The diode operates in the square law region.
- The diode has a forward voltage drop of 0.7 V.
Explanation:
To avoid diagonal clipping, the maximum depth of modulation can be found using the formula:
Depth of Modulation (m) = (Vp - Vm) / (Vp + Vm)
Where:
- Vp is the peak voltage of the carrier signal
- Vm is the peak voltage of the modulating signal
Step 1: Calculate peak voltage of the carrier signal (Vp)
The peak voltage of the carrier signal can be determined using the formula:
Vp = √(2 * P * RL)
Where:
- P is the power of the carrier signal
Since the power of the carrier signal is not given, we assume it to be 1 mW (milliwatt).
P = 1 mW = 0.001 W
Vp = √(2 * 0.001 * 1000) = √2 ≈ 1.414 V
Step 2: Calculate peak voltage of the modulating signal (Vm)
The peak voltage of the modulating signal can be determined using the formula:
Vm = √(2 * P * RF)
Where:
- P is the power of the modulating signal
Since the power of the modulating signal is not given, we assume it to be 1 mW (milliwatt).
P = 1 mW = 0.001 W
Vm = √(2 * 0.001 * 1) = √0.002 ≈ 0.045 V
Step 3: Calculate the maximum depth of modulation (m)
Using the formula mentioned earlier:
m = (Vp - Vm) / (Vp + Vm) = (1.414 - 0.045) / (1.414 + 0.045) ≈ 0.847
Therefore, the maximum permissible depth of modulation to avoid diagonal clipping is approximately 0.847.
Conclusion:
The correct answer is option 'A' - 0.847.
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A diode detector has a load of 1 kΩ shunted by a 10000 pF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation,so as to avoid diagonal clipping, with modulating signal frequency fo 10 kHz will bea)0.847b)0.628c)0.734d)None of the aboveCorrect answer is option 'A'. Can you explain this answer?
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