Analyzing Gain Curve for 741 Op-Amp

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Homework Help Overview

The discussion revolves around analyzing the gain curve of a 741 operational amplifier, focusing on the estimation of gain loss in decibels per octave and the implications for bandwidth in relation to the gain-bandwidth product.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss estimating gain loss in dB per octave and question the accuracy of their approximations. There is exploration of the relationship between dB/decade and dB/octave, with some participants expressing uncertainty about the calculations involved. Others suggest using log-log graph paper to improve resolution for analysis.

Discussion Status

The discussion is active, with participants sharing insights about the gain-bandwidth product and its relevance to the problem. Some guidance has been offered regarding the interpretation of the graph and calculations, but there remains a lack of consensus on certain numerical values and methods.

Contextual Notes

Participants note the limitations of the provided graph's resolution and the need for clarity on terms like noise gain and closed-loop gain. The original poster expresses uncertainty about the calculations and the implications of their findings.

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Homework Statement



I have attached the graph, and questions.


Homework Equations





The Attempt at a Solution



For part a, do I just have to estimate how many db's are lost during one octave? One octave is halving of frequency, so could part a be approximated to be about 8 db/octave?
For part b, I'm not sure what to do, so any help is appreciated. Thanks
 

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maherelharake said:

Homework Statement



I have attached the graph, and questions.


Homework Equations





The Attempt at a Solution



For part a, do I just have to estimate how many db's are lost during one octave? One octave is halving of frequency, so could part a be approximated to be about 8 db/octave?
For part b, I'm not sure what to do, so any help is appreciated. Thanks

The plot you are given does not have very good resolution. I'd recommend that you google log-log graph paper, and re-draw the plot on that paper. That will give you the resolution to be able to answer the questions.
 
Are you suggesting I print it out and redraw it on a new sheet? Unfortunately, I don't think my drawing ability will be as good as the given, computer-generated plot. Is there no way to use the one I uploaded? Thanks.
 
maherelharake said:
Are you suggesting I print it out and redraw it on a new sheet? Unfortunately, I don't think my drawing ability will be as good as the given, computer-generated plot. Is there no way to use the one I uploaded? Thanks.

Yes, that is my suggestion for you to solve this problem set, and to learn how to solve this type of problem in the future. Once you solve it with full resolution log-log graph paper, you will understand how to solve this question intuitively without graph paper.

It would be good if you could scan and post your solution to this problem. That would help to teach other students who are encountering similar problems.
 
I dunno. The graph provided seems to give a pretty good indication that there's a drop of 100 dB over 4 decades. dB/decade ≈ 3.322 db/octave. The "3.322" comes about as the ratio of the logs of the bases -- log(10)/log(2).
 
I can definitely try to post some solutions when I get it all sorted out.

As far as how you got that number 3.322, I'm still a bit unsure about how it was attained. I agree with the 100 dB over 4 decades, so why wouldn't it be 25 instead of 3.322?
Thanks.
 
3.322 is not the answer. It's the scaling factor that relates dB/decade to dB/octave.

dB's are logs of ratios. Over a decade, the frequency changes by a factor of 10. Over an octave it changes by a factor of 2. log(10)/log(2) = 3.322. You can take the logs in any base you wish, and the ratio of them will be the same: ln(10)/ln(2) = 3.322.
 
Ohh. So since the dB/decade is about 25, I would have to multiply that by 3.322 to get the ultimate gain fall-off in terms ofdB/octave?
 
Try dividing. A decade contains several octaves. dB/decade ≈ 3.322 db/octave so that db/octave ≈ (db/decade)/3.332.
 
  • #10
Ahh right. So about 7.5 db/octave. My initial rough estimate was fairly close. Great.

Any hints for part b?
Thanks
 
  • #11
The Gain-Bandwidth product should come in handy for part b. You can read its value right off your plot, and use it to determine the closed-loop bandwidth for the required gain.

Have a look http://www.analog.com/static/imported-files/tutorials/MT-033.pdf" .
 
Last edited by a moderator:
  • #12
It says the closed-loop bandwidth is equal to gain bandwidth product divided by noise gain. Would the gain-bandwidth product just be 100? I'm not sure how to get the noise gain in this case though.
 
  • #13
The gain-bandwidth-product is equal to the unity-gain-bandwidth, which is equal to the frequency where the open loop gain is unity (1 x freq = freq). Your graph crosses the zero dB line at 105 Hz. So that should be your UGBW.

The NG, or "noise gain" is just nomenclature for the closed loop gain.

Now, you want a closed loop gain of 300. So 300*BW = UGBW. BW = 333.3 Hz.

Seems small, but heck, the op-amp appears to crap out entirely at only 100 kHz, and has a critical frequency for open loop gain at less than 10Hz.
 
  • #14
Oh ok. Gotcha. Thanks for all your help.
 

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