Upper 3db Frequency of Op Amp Help

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Discussion Overview

The discussion revolves around calculating the upper 3 dB frequency of an operational amplifier (op-amp) in the context of a homework problem. Participants explore the implications of gain and slew rate on frequency response, addressing both theoretical and practical aspects of the problem.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculated a gain of 11 and an approximate frequency of 80 kHz but expressed uncertainty about the method for part (b).
  • Another participant questioned the definition of "upper 3 dB frequency," seeking clarification on whether it refers to a gain reduction of 3 dB from the maximum gain or a specific frequency value.
  • A participant confirmed that the amplitude is 1 V and suggested that the upper 3 dB frequency is when the gain is 3 dB, though they expressed uncertainty.
  • One participant proposed a method to relate the slope of the sine wave to the slew rate of the op-amp, asking for further elaboration on the calculations.
  • Another participant attempted to apply the formula for frequency based on slew rate but questioned the validity of their approach.
  • A later reply acknowledged a correction regarding the slope and provided a method to calculate the frequency based on a gain drop of 3 dB.
  • One participant calculated a frequency of approximately 850 kHz based on a logarithmic relationship but questioned whether their approach was correct.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of "upper 3 dB frequency" or the correct method to calculate it. Multiple competing views and methods remain, with some participants expressing uncertainty about their calculations.

Contextual Notes

There are unresolved assumptions regarding the definitions of amplitude measurements (peak, peak-to-peak, or RMS) and the specific conditions under which the gain is considered to drop by 3 dB.

mmmboh
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So I did the first part, and got the gain as 11.
I also did part c, and got approximately 80khz,
but I'm not sure what to do for b).
I know that at the 3db frequency, the voltage is at 1/20.5 of it's amplitude.
So can I just use the equation f=(slew rate)/(2pi*A), and set A to 1V/20.5, and solve for f? In which case f=112.5khz
If not, what should I do?

Thanks!
 
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Questions:

(b) What does "upper 3 dB frequency" mean? Does it mean the frequency at which the gain rolls off 3 dB from the value of 11 or does it mean the frequency at which the gain is 3 dB. (Does "upper" refer to gain or frequency in the question?)

(c) Does the input voltage amplitude of 1 V represent a peak to peak, peak or RMS measurement?
 
The amplitude is 1 V. And I believe for b) it's when the gain is 3db, although your guess is as good as mine. This is due tomorrow so I don't have time to ask the teacher.
 
I get a different answer for (c) than you do. Here's how I thought about it.
What is the slope at 180 deg? (or the absolute value of the slope) That value must relate to the slew rate of the opamp of 0.5 V/uS. How many microseconds long must one cycle be so that the slope of the sine wave when it crosses the x-axis is equal to 0.5 V/uS?

If you show me your work I'll make some suggestions.
 
Well I said f=(slew rate)/(2pi*A) = (0.5/us)/2pi=80khz. Why doesn't think work?
Any thoughts on b?

Thanks.
 
My mistake, you are correct about the slope.

(b) The voltage gain drops by 2 dB per 1 dB change in frequency. An output gain of 3 dB should be at a frequency 1.5 dB less than 1.2 MHz. Can you figure that out?
 
Hm...1.5=10*log(1.2MHz/f), f=0.85 MHz?...Or should the f be in the numerator of the log since its upper 3db frequency? or is this just wrong..
 
That's what I get.
1.5 dB = 1.413
1.2 MHz / 1.413 = 850,000 Hz
 

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