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Question

A conventional oscilloscope has a bandwidth of 100 MHz. The rise of the CRO is approximately ______

The correct answer is

3.5 ns

Oscilloscope Rise Time Calculation

An oscilloscope is a vital electronic test instrument that graphically displays varying signal voltages, usually as a two-dimensional plot of one or more signals as a function of time. Two important specifications for an oscilloscope are its bandwidth and its rise time.

  • Bandwidth (BW): The bandwidth of an oscilloscope refers to the frequency range over which it can accurately measure a signal. It is typically defined as the frequency at which the input signal is attenuated by 3 dB (approximately 70.7% of its original amplitude). A higher bandwidth means the oscilloscope can capture faster signals more accurately.
  • Rise Time (\(t_r\)): The rise time of an oscilloscope (or any electronic system) is the time it takes for a signal to change from a specified low value to a specified high value, typically from 10% to 90% of its final amplitude. It indicates how quickly the instrument can respond to a sudden change in input voltage.

Bandwidth-Rise Time Relationship

For a conventional oscilloscope, there is an inverse relationship between its bandwidth and its rise time. This relationship is often approximated by the formula:

$$t_r \approx \frac{0.35}{\text{BW}}$$

Where:

  • \(t_r\) is the rise time (in seconds)
  • BW is the bandwidth (in Hertz)
  • The constant 0.35 is an empirical factor commonly used for systems with a single dominant pole response, which is a good approximation for many general-purpose oscilloscopes.

CRO Rise Time Calculation for 100 MHz Bandwidth

Given the bandwidth of the conventional oscilloscope is 100 MHz. We need to calculate its approximate rise time.

  • Given Bandwidth (BW) = 100 MHz
  • Convert Bandwidth to Hertz:
  • $$ \text{BW} = 100 \times 10^6 \text{ Hz} $$
  • Now, use the formula to find the rise time \(t_r\):
  • $$t_r = \frac{0.35}{\text{BW}}$$
  • Substitute the value of Bandwidth:
  • $$t_r = \frac{0.35}{100 \times 10^6 \text{ Hz}}$$
  • Perform the division:
  • $$t_r = 0.0035 \times 10^{-6} \text{ s}$$
  • To express this in nanoseconds (ns), recall that \(1 \text{ ns} = 10^{-9} \text{ s}\).
  • $$t_r = 3.5 \times 10^{-3} \times 10^{-6} \text{ s}$$
  • $$t_r = 3.5 \times 10^{-9} \text{ s}$$
  • Therefore, the rise time is:
  • $$t_r = 3.5 \text{ ns}$$

This calculation shows that for a 100 MHz oscilloscope, the approximate rise time is 3.5 ns. This means the CRO can respond to a fast-changing signal and reach 90% of its final value within 3.5 nanoseconds, which is characteristic of its ability to display high-frequency signals.

The final answer is 3.5 ns.

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Important Questions from Cathode Ray Oscilloscope

  1. The function of a trigger level knob on a CRO is:

  2. Aquadag coating is most commonly used in CROs to:

  3. CRO stands for:

  4. Calculate the maximum velocity of the beam of electrons in a CRT having a cathode and anode voltage of 182 V. Assume that the electrons leave the cathode with zero velocity. (Charge of electron = 1.6 × 10-19 C and mass of electron = 9.1 × 10-31 kg)

  5. Which of the following expression is the correct formulae for the deflection sensitivity ‘S’ of a CRT, if

    D = deflection on the fluorescent screen

    L = distance from the center of the deflection plates to the screen

    Ld = effective length of the deflection plates

    d = distances between the deflection plates

    Ed = Potential between deflecting plates

    Ea = accelerating voltage
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