Clarifying the Confusion: Op Amp Gain in Non-Inverting Amplifier Circuits

Click For Summary

Discussion Overview

The discussion revolves around the gain characteristics of operational amplifiers (op amps) in non-inverting amplifier circuits, specifically focusing on the relationship between open-loop gain, closed-loop gain, and the impact of resistor ratios on voltage gain. Participants explore theoretical models, practical implications, and the significance of different op amp designs.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant references a professor's model indicating that the transfer function involves multiplying the open-loop gain by the closed-loop gain, but expresses uncertainty about this approach.
  • Another participant supports the idea that the voltage gain depends solely on the resistor ratio, citing a textbook that states if R1 and R2 increase proportionally, their ratio remains constant.
  • Some participants question the necessity of different op amp models if gain is primarily determined by external components, prompting discussions about op amp characteristics like frequency response and input current draw.
  • One participant argues that while the open-loop gain is theoretically very high, practical applications rarely require such values, and different op amps serve various needs such as noise performance and precision.
  • A later reply challenges the initial claim about multiplying gains, providing a formula for closed-loop gain that suggests a different understanding of the relationship between open-loop gain and feedback factors.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between open-loop gain and closed-loop gain, with some supporting the notion that the gain is primarily determined by resistor ratios, while others contest this and provide alternative formulas. The discussion remains unresolved with multiple competing perspectives on the topic.

Contextual Notes

Participants highlight assumptions regarding the conditions under which the resistor ratio can approximate gain and the implications of high open-loop gain in practical scenarios. There are also references to specific operational characteristics of different op amp models that may influence circuit performance.

aesoph13442
Messages
11
Reaction score
0
According to my professor's model answer for an assignment, when we derive the transfer function of an op amp circuit, we have a "K" value, which is the open loop gain of the op amp (I am not 100% certain), multiplied to the op amp circuit (1+R1/R2 for non-inverting amplifier for example). I need to multiply the "open loop gain" with "close loop gain" in order to get the transfer function for a specific circuit. However, when I read the non-inverting amplifier section in "fundamental of microelectronics" by Ravazi, his explanation is in the following: "Interestingly, the voltage gain depends on only the ratio of the resistors; if R1 and R2
increase by 20%, R1/R2 remains constant." It's on page 358 for 2nd edition. I just wonder which version is correct.
 
Engineering news on Phys.org
  • Like
Likes   Reactions: berkeman and aesoph13442
this is correct

watch this video
http://www.bing.com/videos/search?q...E950D0C45487CFCBDAEFE950D0C45487CFC&FORM=VIRE

Dave (not me) explains opamps very well :)
Dave

Thank you Dave! I have one more question. If the gain only depends on other components rather than the op amp itself, what is the point of having different models of op amp? I know some op amps are designed to remain a "normal" behavior in higher frequencies or have a larger bandwidth. But besides those reasons, what is the point of using a different model of op amp? Let's say I am plotting the bode plot of some transfer functions, some of them start at 0dB and some start at 60dB or a different value. The starting point depends on the constant coefficient of the transfer function. Does the op amp determine the value of the constant? Is it the open loop gain (gain of the op amp)?
 
Last edited:
aesoph13442 said:
If the gain only depends on other components rather than the op amp itself, what is the point of having different models of op amp?
You can rely on the resistor ratio to be an approximation to the gain of the op-amp-with-feedback only if the op-amp's open-loop gain far exceeds that. And as you say, if you want high closed-loop gain at higher frequencies then the op-amp itself must be able to deliver even higher gain at those higher frequencies. A good high frequency response equates to a snappy behavior with squarewaves.

Some op-amps are designed to draw very very low currents at the inputs so these will take almost no power from the signal source.
 
aesoph13442 said:
If the gain only depends on other components rather than the op amp itself, what is the point of having different models of op amp?

The open loop gain of any/all op-amps is very high, in theory, infinity, practically tho, 10's or 100's of 1000's, which for normal use is never needed.
10's or 100's or occasionally a few 1000 gain is the most ever used ( depending on requirements)

Why are there so many ?
Many reasons ... many of the earlier versions are still around they may have lower gain before noise becomes a problem
other newer ones have extremely low noise, high current handling output, higher working voltage etc

looking through different datasheets, you will be able to see the beneficial features some have over others
for some basic circuits where input bias currents etc are not critical to the op-amp or circuit operation beans that cheaper more basic op-amps can be used

Other circuits may call for high precision very low bias current leakage and so there are op-amps made for that purposeDave
 
aesoph13442 said:
I need to multiply the "open loop gain" with "close loop gain" in order to get the transfer function for a specific circuit. .
That`s definitely wrong. The following formula says everything (Aol=open-loop gain, k=feedback factor):

Closed-loop gain (non-inverting amplifier with k=R2/(R1+R2))
Acl
=Aol/(1+Aol*k)=1/(1/Aol+k).
For 1/Aol<<k we get Acl=1/k.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 33 ·
2
Replies
33
Views
6K
Replies
12
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
14
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K