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In a digital voltmeter, the oscillator frequency is 400 kHz and the ramp voltage falls from 8V to 0V in 20m sec. The number of pulse counted by the counter is
Correct answer is '8000'. Can you explain this answer?
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In a digital voltmeter, the oscillator frequency is 400 kHz and the ra...
Frequency = 400 kHz.
fclk = 400kHz
t = 20m sec
⇒ Clock pulses, n = t fclk = 20m sec × 400 kHz = 8000
fclk = 400kHz
t = 20m sec
⇒ Clock pulses, n = t fclk = 20m sec × 400 kHz = 8000
Most Upvoted Answer
In a digital voltmeter, the oscillator frequency is 400 kHz and the ra...
Explanation:
A digital voltmeter utilizes an oscillator to generate a pulse train, which is then used to charge a capacitor with a known value. The voltage across the capacitor is compared with the input voltage, and the number of pulses required to charge the capacitor to the input voltage is counted.
Given:
- Oscillator frequency = 400 kHz
- Ramp voltage falls from 8V to 0V in 20m sec
Solution:
The time period of the oscillator is given by:
T = 1/f = 1/400 kHz = 2.5 µs
The ramp voltage is falling from 8V to 0V in 20 ms or 20,000 µs. Therefore, the ramp voltage is falling by 8V/20,000 µs = 0.4 mV/µs.
The slope of the ramp voltage is given by:
Slope = (Change in Voltage)/(Change in Time) = 0.4 mV/µs = 400 V/s
The digital voltmeter generates a pulse train at a frequency of 400 kHz, which means that there are 400,000 pulses per second.
The time it takes for one pulse to occur is given by:
Time per Pulse = 1/f = 1/400 kHz = 2.5 µs
The number of pulses required to charge the capacitor to the input voltage is given by:
Number of Pulses = (Input Voltage)/(Step Voltage)
where Step Voltage is the voltage corresponding to one pulse.
Step Voltage = Slope x Time per Pulse
Substituting the values:
Step Voltage = 400 V/s x 2.5 µs = 1 mV
Number of Pulses = 8V/1 mV = 8000
Therefore, the number of pulse counted by the counter is 8000.
A digital voltmeter utilizes an oscillator to generate a pulse train, which is then used to charge a capacitor with a known value. The voltage across the capacitor is compared with the input voltage, and the number of pulses required to charge the capacitor to the input voltage is counted.
Given:
- Oscillator frequency = 400 kHz
- Ramp voltage falls from 8V to 0V in 20m sec
Solution:
The time period of the oscillator is given by:
T = 1/f = 1/400 kHz = 2.5 µs
The ramp voltage is falling from 8V to 0V in 20 ms or 20,000 µs. Therefore, the ramp voltage is falling by 8V/20,000 µs = 0.4 mV/µs.
The slope of the ramp voltage is given by:
Slope = (Change in Voltage)/(Change in Time) = 0.4 mV/µs = 400 V/s
The digital voltmeter generates a pulse train at a frequency of 400 kHz, which means that there are 400,000 pulses per second.
The time it takes for one pulse to occur is given by:
Time per Pulse = 1/f = 1/400 kHz = 2.5 µs
The number of pulses required to charge the capacitor to the input voltage is given by:
Number of Pulses = (Input Voltage)/(Step Voltage)
where Step Voltage is the voltage corresponding to one pulse.
Step Voltage = Slope x Time per Pulse
Substituting the values:
Step Voltage = 400 V/s x 2.5 µs = 1 mV
Number of Pulses = 8V/1 mV = 8000
Therefore, the number of pulse counted by the counter is 8000.
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In a digital voltmeter, the oscillator frequency is 400 kHz and the ramp voltage falls from 8V to 0V in 20m sec. The number of pulse counted by the counter isCorrect answer is '8000'. Can you explain this answer?
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