741C Inverting Amplifier Distortion Analysis
Problem Setup
We need to find the maximum peak-to-peak input signal ($V_{in, pp}$) for an inverting amplifier using a $741C$ op-amp. The amplifier has a voltage gain ($|A_v|$) of $10$ and must maintain a flat frequency response up to $40 kHz$. Distortion is the main concern.
Op-Amp Limitations
The $741C$ op-amp has limitations that can cause distortion, primarily:
- Gain-Bandwidth Product (GBWP): Limits the bandwidth for a given gain. For a $741C$, GBWP is typically $1 MHz$. The closed-loop bandwidth is $BW \approx GBWP / |A_v| = 1 MHz / 10 = 100 kHz$. Since $100 kHz > 40 kHz$, the GBWP does not limit the required frequency response.
- Slew Rate (SR): Limits the maximum rate of change of the output voltage. For a $741C$, SR is typically $0.5 V/\mu s$. This often limits the maximum signal amplitude at higher frequencies.
- Output Voltage Swing: The output voltage cannot exceed the power supply rails minus some overhead voltage. This limits the peak output voltage.
At $40 kHz$, the slew rate is the most likely limiting factor.
Slew Rate Calculation
The maximum rate of change of the output voltage ($V_{out}$) for a sinusoidal signal is given by:
$ \left| \frac{dV_{out}}{dt} \right|_{max} = V_{p,out} \omega $Where $V_{p,out}$ is the peak output voltage and $\omega$ is the angular frequency ($ \omega = 2 \pi f $).
The slew rate limits this rate:
$ V_{p,out} \omega \le SR $For the given frequency $f = 40 kHz$:
$ \omega = 2 \pi (40 \times 10^3) \text{ rad/s} = 80,000 \pi \text{ rad/s} $Using the typical SR for a $741C$ ($SR = 0.5 V/\mu s = 0.5 \times 10^6 V/s$):
$ V_{p,out} (80,000 \pi) \le 0.5 \times 10^6 V/s $ $ V_{p,out} \le \frac{0.5 \times 10^6}{80,000 \pi} \approx \frac{50}{8 \pi} \approx 1.989 V $Input Signal Calculation
The peak input voltage ($V_{p,in}$) is related to the peak output voltage by the amplifier's gain:
$ V_{p,out} = |A_v| V_{p,in} $Therefore, the maximum peak input voltage is:
$ V_{p,in} = \frac{V_{p,out}}{|A_v|} \approx \frac{1.989 V}{10} = 0.1989 V $The question asks for the maximum peak-to-peak input signal ($V_{in, pp}$):
$ V_{in, pp} = 2 \times V_{p,in} \approx 2 \times 0.1989 V = 0.3978 V $This value is approximately $0.398 V$. The output voltage swing limitation ($|A_v| \times 0.398 V \approx 3.98 V$ peak) is well within typical op-amp output limits (e.g., $\pm 13V$ for $\pm 15V$ supplies), confirming slew rate is the limiting factor.
Conclusion
The maximum peak-to-peak input signal that can be applied without causing slew-rate distortion at $40 kHz$ is approximately $0.398 V$.
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