An Audio pre-amplifier needs to reproduce signals as high as 20 kHz. The maximum output swing is 10 V peak. The minimum acceptable slew rate for the op-amp used is:
1.257 V/μs
The question asks for the minimum acceptable slew rate for an operational amplifier (op-amp) used in an audio pre-amplifier circuit. We are given the maximum signal frequency and the maximum output voltage swing required from the op-amp.
Slew rate (SR) is a critical parameter for op-amps, especially when dealing with high-frequency signals or large output voltage swings. It represents the maximum rate of change of the output voltage per unit of time. If the required rate of change of the output signal is faster than the op-amp's slew rate, the op-amp will not be able to reproduce the signal accurately, leading to distortion.
For a sinusoidal signal, the maximum rate of change occurs when the signal crosses the zero-voltage point. The output voltage of a sinusoidal signal can be described as:
\(v_{out}(t) = V_{peak} \sin(2 \pi f t)\)
The rate of change of the output voltage is the derivative with respect to time:
\(\frac{dv_{out}(t)}{dt} = \frac{d}{dt} (V_{peak} \sin(2 \pi f t))\)
\(\frac{dv_{out}(t)}{dt} = V_{peak} (2 \pi f) \cos(2 \pi f t)\)
The maximum rate of change occurs when \(|\cos(2 \pi f t)| = 1\), which is \(2 \pi f V_{peak}\).
Therefore, the minimum slew rate required for an op-amp to faithfully reproduce a sinusoidal signal of frequency \(f\) and peak voltage \(V_{peak}\) without slew-induced distortion is given by the formula:
\(SR \ge 2 \pi f V_{peak}\)
We are given the following information for the audio pre-amplifier:
First, let's convert the frequency to Hz:
\(f = 20 \text{ kHz} = 20 \times 10^3 \text{ Hz} = 20,000 \text{ Hz}\)
Now, we can use the formula to calculate the minimum required slew rate (SR):
\(SR \ge 2 \pi f V_{peak}\)
\(SR \ge 2 \pi (20,000 \text{ Hz}) (10 \text{ V})\)
\(SR \ge 400,000 \pi \text{ V/s}\)
To compare this value with the given options, we need to convert V/s to V/μs. We know that 1 μs = 10\(^{-6}\) s. Therefore, 1 V/s = 10\(^{-6}\) V/μs.
\(SR \ge 400,000 \pi \times 10^{-6} \text{ V/μs}\)
\(SR \ge 0.4 \pi \text{ V/μs}\)
Let's calculate the numerical value:
\(SR \ge 0.4 \times 3.14159 \text{ V/μs}\)
\(SR \ge 1.256636 \text{ V/μs}\)
Rounding this value to three decimal places, we get approximately 1.257 V/μs.
The minimum acceptable slew rate for the op-amp used in this audio pre-amplifier application must be at least 1.257 V/μs to accurately reproduce the 20 kHz signal with a 10 V peak swing.
Let's compare our calculated minimum slew rate with the provided options:
Our calculated minimum slew rate is approximately 1.257 V/μs. This matches option 2.
For an audio pre-amplifier designed to handle signals up to 20 kHz with a maximum output peak voltage of 10 V, the operational amplifier must have a slew rate of at least 1.257 V/μs to avoid slew-rate-induced distortion. Selecting an op-amp with a higher slew rate would provide better performance margins, but 1.257 V/μs is the theoretical minimum.
| Parameter | Value |
|---|---|
| Maximum Frequency (\(f\)) | 20 kHz (20,000 Hz) |
| Maximum Output Peak Voltage (\(V_{peak}\)) | 10 V |
| Formula for Minimum Slew Rate (SR) | \(2 \pi f V_{peak}\) |
| Calculated Minimum SR (V/s) | \(2 \pi \times 20,000 \times 10 = 400,000 \pi \text{ V/s}\) |
| Calculated Minimum SR (V/μs) | \(400,000 \pi \times 10^{-6} \approx 1.257 \text{ V/μs}\) |
| Concept | Explanation | Formula |
|---|---|---|
| Slew Rate (SR) | Maximum rate of change of op-amp output voltage. | \(SR = \frac{\Delta V_{out}}{\Delta t}\) (units typically V/μs) |
| Minimum SR for Sinusoidal Signal | Required SR to reproduce a sine wave without slew-rate limiting distortion. | \(SR_{min} \ge 2 \pi f V_{peak}\) |
| f | Signal frequency (in Hz) | - |
| \(V_{peak}\) | Peak output voltage (in Volts) | - |
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