has so many flaws that define these things.

Those flaws make the chip a perfect example to teach people about real opamps.

 

...................
And don't get all cranked up about my age comment above. That really is my version of humor. I'm 64.833, and I think Wally is near that. If he isn't, he should be.

Thanks. Is that because I'm so cantankerous? Actually I've got you beat by more than a decade.

 

Those flaws make the chip a perfect example to teach people about real opamps.

I thought so.

 

Try this:

Connect the non-inverting input to ground and put a 1Meg resistor between the output and the inverting input. Measure the output voltage, it should be mostly due to the voltage being developed across the resistor because of the input bias current.

Next connect an identical resistor (as matched in value as you can find) from the non-inverting input to ground. Measure the output voltage, it should be mostly due to input offset current and input offset voltage, plus whatever (probably small) difference there is between the resistors. The two inputs should have roughly equal voltages because both inputs have the same resistance attached to them.

 

Try this:

Connect the non-inverting input to ground and put a 1Meg resistor between the output and the inverting input. Measure the output voltage, it should be mostly due to the voltage being developed across the resistor because of the input bias current.

Next connect an identical resistor (as matched in value as you can find) from the non-inverting input to ground. Measure the output voltage, it should be mostly due to input offset current and input offset voltage, plus whatever (probably small) difference there is between the resistors. The two inputs should have roughly equal voltages because both inputs have the same resistance attached to them.

1 M output to inverting. Non-inverting to ground, 25 nV out !!!. Wow, 25 nA bias current???
1 M non-inverting input to ground, 0.3 mV out!! I can't measure anything much smaller than that. I readjusted offset. No significant change in result.
Not bad for an LM741.
Maybe that explains why I wasn't seeing much of a difference in my tests. My LM741 wasn't bad enough.

 

1 M output to inverting. Non-inverting to ground, 25 nV out !!!. Wow, 25 nA bias current???
1 M non-inverting input to ground, 0.3 mV out!! I can't measure anything much smaller than that. I readjusted offset. No significant change in result.
Not bad for an LM741.
Maybe that explains why I wasn't seeing much of a difference in my tests. My LM741 wasn't bad enough.

Conclusion ...
Bipolar op amps in high gain circuits ... yes, a good idea to use the resistor. (input bias in the nA.)
FET op amps with bias in the pA, is it often necessary? Maybe if input resistance is in the Meg ohms?

 

So how did you measure this 25 nV?

 

I'm 64.833,

I'm 65.263
Never thought in those terms before.

You are responsible for the bulk of the effort to learn the concepts and we are willing to assist. I told you the what the topic is that you need to research.

I was just browsing through and thought, "It would take a whole book to explain all of this." That's why the books were written. This is just the place for examining one point at a time. If you want the whole book, read it. We aren't going to type it here for you.

 

Here is how I approach the problem.

Begin with the basic four resistor difference amplifier circuit:



Set R1 = R3, R2 = R4.
Voltage gain = R2/R1

Non-inverting amplifier:


Inverting amplifier circuit:


Replace the parallel combination of R3||R4 with one resistor:

 

I vaguely remember some books suggesting a resistor be added to an inverting amplifier design, but I never understood why. I remember it suggesting a value for R1 to be equal to the parallel resistance of the other two resistors. In retirement now I have the time to pursue answers to questions I never found an answer to. So what is R1 supposed to accomplish? I built a circuit with and without R1 and see no difference.
I used an LM741 with an offset pot if it makes a difference.
Is it not apparent in my circuit?
Is there an equivalent resistor in a non-inverting amplifier?

It's for input compensation....not necessary with most op amps...or circuits.

 

@hp1729 since you are being such a problem child about this, I thought I would share a saying a math teacher of mine had, "contribute to a solution and not to a problem".

Do things that benefit you when you design a circuit instead of ignoring good engineering. Input bias current is real, op-amp manufactures work hard to make it as small as possible, but it is there. All you have to do is match the impedance of both the positive and the negative inputs and you have minimized the issue. You might try spending some quality time reading and trying to understand op-amp data sheets.

There are cases where omitting the non-inverting input to ground resistor is desired, but I don't think you are ready for that.

 

So how did you measure this 25 nV?

That dawned on me after I had a nap. Repeating exercise. Wrong meter scale 2 V, not 200 mV. Reading is 25 mV. Through 1 M ohm, is 25 nA. Right?

(edited to add ...)
I tried four of my "cheaper than dirt LM741's" and get similar readings. What am I doing wrong?
Do you find this method of measuring bias acceptable?

Ground NI input. 1 Meg resistor IN to out and measure output voltage to calculate bias current?

 

@hp1729 since you are being such a problem child about this, I thought I would share a saying a math teacher of mine had, "contribute to a solution and not to a problem".

Do things that benefit you when you design a circuit instead of ignoring good engineering. Input bias current is real, op-amp manufactures work hard to make it as small as possible, but it is there. All you have to do is match the impedance of both the positive and the negative inputs and you have minimized the issue. You might try spending some quality time reading and trying to understand op-amp data sheets.

There are cases where omitting the non-inverting input to ground resistor is desired, but I don't think you are ready for that.

If I thought I were "ready for it" I would not be here asking questions.
I don't deny the existence of bias current. I am just trying to get a grasp on when that resistor can be ignored. The data sheet does not tell me that. when would you choose not to use it?

 

Here is how I approach the problem.

Begin with the basic four resistor difference amplifier circuit:

View attachment 101048

Set R1 = R3, R2 = R4.
Voltage gain = R2/R1

Non-inverting amplifier:
View attachment 101051

Inverting amplifier circuit:
View attachment 101052

Replace the parallel combination of R3||R4 with one resistor:

View attachment 101053

How does "using the parallel equivalent" deal with a specific op amp's bias current? Using that method I found little difference in my LM741 circuit with or without the resistor at gains below 100. Is my LM741 really that good? I have trouble setting up a reliable circuit with a gain of 1,000.
Still working on that.
Using the other suggested method (using the 1 M ohm resistor ...) means I always have R1 of 1 M Ohm.
Not a parallel equivalent. ????

 

If I thought I were "ready for it" I would not be here asking questions.
I don't deny the existence of bias current. I am just trying to get a grasp on when that resistor can be ignored. The data sheet does not tell me that. when would you choose not to use it?

The data sheet doesn't tell you when to ignore it because the data sheet tells you about the characteristics of that part. The data sheet can't read your mind to know how you are going to use that part or what things about your application are important to you and how the characteristics of that part are going to impact the things about your application that are important to you. That's YOUR job. They give you the data and those textbooks that you disdain show you how those issues affect circuit behavior and how to compensate for them.

As to when not to use the compensation, that's simple -- ignore the compensation when doing so will have negligible (or at least tolerable) negative impact on your circuit's performance. To answer that question you have to understand what them phenomenon is that you are considering compensating for, but also what the tolerable impact on your particular circuit in your particular application is.

 

How does "using the parallel equivalent" deal with a specific op amp's bias current?

You will never understand this until you understand what the bias current is and how it affects circuit operation. Understand that before you try to understand the magic behind a particular compensation technique.

 

You also need to recognize that the output offset is a combination of the input voltage offset, and the input bias current through the input resistance.
To determine which is which you first need to measure the voltage offset with a low input resistance.
Then you increase the resistance to see the effect of the bias current on the offset. It may either add or subtract from the voltage offset, depending upon their relative polarities.

 

You will never understand this until you understand what the bias current is and how it affects circuit operation. Understand that before you try to understand the magic behind a particular compensation technique.

Yes, that is what I am trying to do.
I tried to repeat the exercise with an LF351, bias current in the pA. My meter won't go down far enough to get a measurement. So I am seeing something related to bias but I find it difficult to believe my $0.25 LM741's are lower than the data sheet specs with a bias of 25 nA..

 

Reading the story on superposition theory I see why we get to the proper value being equal to the parallel resistances. Understand it, no, but accept it, yes.
We do not need to know the bias current, only compensate for it in bipolar transistor input op amps.
Do we need to do this for FET or MOSFET input op amps?
I am still lacking an example of what the result of this resistor is. How would you demonstrate to a class room what effect this has?
Using the worst op amp I can find it doesn't seem to make much difference.

 

You also need to recognize that the output offset is a combination of the input voltage offset, and the input bias current through the input resistance.
To determine which is which you first need to measure the voltage offset with a low input resistance.
Then you increase the resistance to see the effect of the bias current on the offset. It may either add or subtract from the voltage offset, depending upon their relative polarities.

That makes sense.
So by adjusting the offset am I compensating, at least in part, for bias currents?
Is that why my LM741 appear to be so good?
My offset adjustment is not perfect. I get down to tweaking over a fraction of a degree on a 25-turn pot, and accept it is only "as close as possible".
Going back to play with it. I have not tried readjusting offset when monitoring the output when checking that bias check output.

(EDITED TO ADD ...)
No adjusting offset doesn't make much difference.
Grounded both inputs. Left the 1 M ohm resistor between 2 and 6. Adjusted offset as best possible. Removed the jumper between pins 2 and 3 so only 3 goes to ground. Output was the usual 25 mV. Adjusting offset pot a full turn either way made no change in output.