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

[aesoph13442]

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.

 

[davenn]

"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.

this is correct

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

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

 

[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)?

 

[NascentOxygen]

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.

 

[davenn]

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

 

[LvW]

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.