Slew Rate MCQ Quiz in తెలుగు - Objective Question with Answer for Slew Rate - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Slew Rate:

• The slew rate is defined as the maximum rate of output voltage change per unit time.
• It is denoted by the letter S.
• The slew rate helps us to identify the amplitude and maximum input frequency suitable to an operational amplifier (OP amp) such that the output is not significantly distorted.
• The slew rate should be as high as possible to ensure the maximum undistorted output voltage swing.

The equation for the slew rate is given by,

\(S = \frac{dV_0}{dt}|_{maximum} Volts/ \mu sec\)

Concept:

Slew Rate:

• Slew rate (SR) is defined as the maximum rate of change in output voltage per unit of time.
• Slew rate indicates how rapidly the output of an op-amp can change in response to change in the input frequency.
• The slew rate of an op-amp is fixed therefore if the slope requirements of the output signal are greater than the slew rate, then distortion occurs.
   

\(SR = \frac{{d{v_o}}}{{dt}}\) V/μs

Explanation: 

Since slew rate defined as m the maximum rate of change of the output, from the given figure it can be seen that,

\(SR = \frac{{d{v_o}}}{{dt}}\)

\(SR = \frac{{3 - \left( { - 3} \right)}}{{\frac{{0.5}}{2}}}\)

\(SR = \frac{{12}}{{0.5}}\)

Slew rate = 24 V / μs

The slew rate of the op-amp = 1 volt/nsec

∴ For each nanosecond, the change in voltage = 1 V

Explanation:

(i) 0 < t < T/2, i.e. for a positive cycle of input, the output starts increasing from – 10 volts.

Since, S.R = 1 V/nsec

By the end of 20 nsec, the output reaches from -10 V to +10 V and after that it saturates.

(ii) T/2 < t < T

For a negative cycle of input, the output starts decreasing from +10 Volts.

Again, by the end of 60 nsec, the output will reach from +10 V to -10 V, and after that it saturates.

A similar trend will repeat for further cycles of the input cycle and the output waveform will be:

Note:

1) Shape of the output waveform is trapezoidal for T/2 > 20 nsec

2) For, perfect triangular shape of the output \(\frac{T}{2} = 20\;nsec\)  

Key Points

•  The slew rate is defined as the maximum rate of output voltage change per unit of time. (Denoted by letter S)
• The slew rate helps us to identify the amplitude and maximum input frequency suitable to an operational amplifier such that the output is not significantly distorted.
• It should be as high as possible to ensure the maximum undistorted output voltage swing.
• Formula:

                                 \(S=\frac{\mathrm{d} V_{0}}{\mathrm{d} x}\bigg\rvert_{maximum} \, \, Volts/ \mu S\)

Mathematical Analysis: 

The maximum rate of change of the output voltage in response to a step input voltage is defined as the slew rate of the op-Amp.

Mathematically, the slew rate is defined as:

\(S.R = {\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}}\)

When the input is a sinusoid given as:

V_i(t) = Vm sin2πf t

And the voltage gain of the op-amp is A_v, the output voltage will be:

V_o = A_V V_i

Vo = AV Vm sin2πf t

Now, the rate of change in output will be:

\(\frac{{d{V_0}}}{{dt}} =A_VV_m2\pi f~cos2\pi ft\)

The maximum rate of change (Slew rate) will be:

\({\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}}=A_VV_m2πf_{max}\)   ---(1)

Calculation:

Gain = 40 dB, i.e.

20 log₁₀ A_V = 40

A_V = 100

Putting on the respective values in Equation (1), we can write:

\(\frac{1}{10^{-6}}=100\times V_m\times 2π\times20k\)

\(V_m=\frac{1}{2\pi \times 2}=79.5~mV\)

∴ The input level must not exceed above 79.5 mV.

• The output of an Op-Amp can only change by a certain amount in a given time. This limit is called the slew rate of the Op-Amp.
• Op-Amp slew rate can limit the performance of a circuit if the slew rate requirement is exceeded.
• This is best explained with the help of the given waveform:

Explanation:

Slew Rate and its Significance:

The slew rate of an operational amplifier (op-amp) is defined as the maximum rate of change of output voltage per unit time. It is typically expressed in volts per microsecond (V/μs) or volts per millisecond (V/ms). The slew rate limits the ability of the op-amp to follow rapid changes in the input signal, causing distortion if the input signal frequency or amplitude exceeds certain limits.

In the given question, the op-amp has a slew rate of 62.8 V/ms and is connected in a voltage follower configuration. In this configuration, the output voltage ideally follows the input voltage exactly. However, when the input signal's rate of change exceeds the slew rate, the output cannot follow the input accurately, resulting in slew rate limited distortion.

Given Data:

• Slew rate of the op-amp (SR) = 62.8 V/ms
• Maximum amplitude of the sinusoidal input (V_m) = 10 V

Key Formula:

The maximum rate of change of a sinusoidal signal occurs at its peak (zero-crossing point), and it can be expressed as:

Rate of change of voltage = ω × V_m

where:

• ω = 2πf is the angular frequency of the signal
• f is the frequency of the sinusoidal signal
• V_m is the amplitude of the sinusoidal signal

The condition for avoiding slew rate limited distortion is:

ω × V_m ≤ SR

Substituting ω = 2πf, the condition becomes:

2πf × V_m ≤ SR

Rearranging for frequency (f):

f ≤ SR / (2π × V_m)

Solution:

Substitute the given values into the formula:

• SR = 62.8 V/ms = 62.8 × 10³ V/s (convert to V/s)
• V_m = 10 V
• 2π ≈ 6.28

Calculate the minimum frequency at which distortion sets in:

f = SR / (2π × V_m)

f = (62.8 × 10³) / (6.28 × 10)

f = 1 × 10³ Hz

f = 1 kHz

Therefore, the minimum frequency at which slew rate limited distortion sets in is 1 kHz. To convert this to MHz:

f = 1 kHz = 1 × 10⁻³ MHz = 1 MHz

Correct Answer: Option 1) 1 MHz

Important Information:

To further analyze the other options:

Option 2: 6.28 MHz

This option assumes a miscalculation in the formula. If the factor of 2π is misinterpreted, it might lead to this incorrect result. However, as shown in the detailed solution, the correct calculation yields 1 MHz, not 6.28 MHz.

Option 3: 10 MHz

This option is incorrect because it overestimates the frequency at which distortion sets in. The slew rate of the op-amp limits the maximum rate of change of the output voltage, and the correct calculation shows that distortion begins at 1 MHz, not 10 MHz.

Option 4: 62.8 MHz

This option is also incorrect and represents a significant overestimation. This value might come from a misunderstanding of the slew rate value itself or an error in applying the formula. The slew rate of 62.8 V/ms corresponds to a much lower frequency limit of 1 MHz for the given input amplitude of 10 V.

Conclusion:

The slew rate is a critical parameter in op-amp applications, especially in high-frequency or high-amplitude signal processing. For the given op-amp with a slew rate of 62.8 V/ms and an input amplitude of 10 V, the minimum frequency at which slew rate limited distortion sets in is 1 MHz. This frequency is calculated using the formula f = SR / (2π × V_m), ensuring accurate analysis of the op-amp's performance limits.

• Voltage follower is an Op-amp circuit whose output voltage follows the input voltage. That is, output voltage is equivalent to the input voltage.
• Op-amp circuit does not provide any amplification. The high input impedance of voltage follower is the reason for it being used in several circuits.
• A voltage follower can be used as a buffer because it draws very little current due to the high input impedance of the amplifier, and hence it eliminates the loading effects while still maintaining the same voltage at the output.

Input, V_s = V_m sin ωt

Given that: V_m = 10V and slew rate = 62.8 V/msec

In a voltage follower, output voltage, V₀ = V_s (input voltage)

∴ V₀ = V_m sin ωt

Slew rate = dV₀ / dt = ω V_m cos wt

Maximum of dV₀ / dt = ω V_m >₌ s [gain]

⇒ 2πfV_m ≥ s

⇒ f ≥ s / 2 × π × V_m ≥ 62.8 / ( 10^-6 × 6.28 × 10 ) ≥ 10⁶ Hz

Therefore, f_min = 1 MHz

The large signal bandwidth of an operational amplifier is limited by its slew-rate of op-amp because

slew-rate is the maximum rate at which the op-amp can respond to a large change in the input

signal.

For a band-limited sinusoidal signal with bandwidth f_m Hz.

V_in = V_m sin 2πf_mt

And for an op-amp, the output voltage is:

V₀ = A_v V_in

\(S.R = {\left| {\frac{{d{V_0}}}{{dt}}} \right|_{maximum}}\) 

∴ V₀ = A_v V_m sin (2π f_m t)

\(\frac{{d{V_0}}}{{dt}} = {A_v}{V_m}\cos (2\pi {f_m}t) \cdot 2\pi {f_m}\) 

\(\therefore S.R = {\left| {\frac{{d{V_0}}}{{dt}}} \right|_{max}}\) 

S.R. = A_v V_m 2π f_m

∴ S.R. ∝ f_m

i.e. the slew rate is proportional to the signal bandwidth.

Important Point:

CMRR (common-mode rejection ratio)

\({A_{cm}} = \frac{{{V_{o\;cm}}}}{{{V_{cm}}}}\) = common-mode gain

\({A_{dm}} = \frac{{{V_{odm}}}}{{{V_p} - {V_n}}} = \frac{{{V_{odm}}}}{{{V_d}}}\) = differential mode gain.

\(\therefore CMRR = 20{\log _{10}}\left| {\frac{{{A_{dm}}}}{{{A_{cm}}}}} \right|dB\) 

CMRR is the ratio of differential-mode gain to the common-mode gain of op-amp.

Gain-Bandwidth product (GBP):

The Gain-bandwidth product of op-amp is the product of gain and bandwidth, and it must be equal

to a constant value.

i.e. if the gain is decreased by a factor A than bandwidth will increase by the same factor A.

ω = 10⁶ rad/s

\(\Rightarrow f = \frac{{{{10}^6}}}{{2\pi \;}}Hz\)

V_i = 1 V

|V₀| = KV_i = KV

Voltage supply limit (saturation):

\(V_s^ - < {V_0} < V_s^ +\)

K ≤ 15

Frequency response limit

A₀f₀ = f_u = (1 + A)f = 10⁷

\( \left( {1 + K} \right)\frac{{{{10}^6}}}{{2\pi }} \le {10^7}\)

K ≤ 2π – 1 = 62

Maximum output current limit:

\({I_L} = \frac{{{V_0}}}{{{R_L}}} \le {I_{SC}}\)

\( \frac{K}{{500}} \le 10 \times {10^{ - 3}} \)

\(K \le 5\)

Slew rate:

\(\frac{{d{V_0}}}{{dt}} = A\;{V_{in}}\;\omega \le {S_0}\)

K × 1 × 10⁶ ≤ 4 × 10⁶

K ≤ 4.

The maximum value of K = 4