Here is a solution to your question about Va as a function of Vref.

Below is the Millman's expressions for voltage at the opamp's positive and negative input.

\(e_+ = \frac{\frac{V_a}{R2}+\frac{Vref}{R1}}{\frac{1}{R1}+\frac{1}{R2}}\)

\(e_- = \frac{\frac{V_a}{R2}}{\frac{1}{R1} + \frac{1}{R2} + \frac{1}{RL}}\)

These two equations can be placed equal to each other due to the action of the opamp.

\(\frac{\frac{V_a}{R2}+\frac{Vref}{R1}}{\frac{1}{R1}+\frac{1}{R2}} = \frac{\frac{V_a}{R2}}{\frac{1}{R1} + \frac{1}{R2} + \frac{1}{RL}\)

This equation can be further simplified to obtain the equation:

\(\frac{V_a}{V_{ref}} = \frac{-((R1 + R2)RL + R1R2)}{R1R1}\)

hgmjr

 

So far, we've assumed that the opamp is ideal.

But, what if it isn't?

If the opamp gain is finite, equal to A, then what is the voltage between the inputs? Find an expression for (Vpos-Vneg).

 

The Electrician,

If the opamp gain is finite, equal to A, then what is the voltage between the inputs? Find an expression for (Vpos-Vneg).

Sorry I did not get back to this problem sooner. The solution is rather simple really. Using last equation submitted by hgmjr and using "R" to represent the expression of resistance values on the right side, we get Va/Vref =R. Now we know that A(Vpos-Vneg) = Va, because that is what an OP amp does. Simply substituting and arranging we easily get (Vpos-Vneg)= Vref*R/A . That's it. As you can see, when "A" approaches infinity, the difference between the OP amp terminals approaches zero.

Ratch

 

The Electrician,



Sorry I did not get back to this problem sooner. The solution is rather simple really. Using last equation submitted by hgmjr and using "R" to represent the expression of resistance values on the right side, we get Va/Vref =R. Now we know that A(Vpos-Vneg) = Va, because that is what an OP amp does. Simply substituting and arranging we easily get (Vpos-Vneg)= Vref*R/A . That's it. As you can see, when "A" approaches infinity, the difference between the OP amp terminals approaches zero.

Ratch

Wrong again Ratch. There are two flaws here. First, a ratio of voltages can not have units of resistance. Second, you are using a relation, which was derived assuming that the voltage difference is zero, to calculate the value of the voltage difference.

 

steveb,

Wrong again Ratch. There are two flaws here. First, a ratio of voltages can not have units of resistance. Second, you are using a relation, which was derived assuming that the voltage difference is zero, to calculate the value of the voltage difference.

You are right about the "second" flaw, but not not the first. A ratio of voltages is unitless, as is the "R" expression of the equation I referenced. The amplification factor "A" is also unitless. So there is no conflict there. The units are correct, but as you say, too many factors did not get into the equation when A is less than infinity.

Ratch

 

You are right about the "second" flaw, but not not the first. A ratio of voltages is unitless, as is the "R" expression of the equation I referenced. Ratch

Actually, my issue was not with the units in the equations (which I could see was not a problem), but with your verbal statement that R was a resistance. Perhaps the word "flaw" was too strong here, but you made a mistatement which could confuse someone trying to learn. I would have let that one go, but we are in the "homework help" section now.

 

steveb,

Actually, my issue was not with the units in the equations (which I could see was not a problem), but with your verbal statement that R was a resistance. Perhaps the word "flaw" was too strong here, but you made a mistatement which could confuse someone trying to learn. I would have let that one go, but we are in the "homework help" section now.

I don't see any confusion or misstatement. I specifically stated that "R" represented the expression of resistance values on the right side of the equation. That is different than saying that R represents a resistance.

By the way, I believe I have figured out V+-V- in terms of a finite A. I am checking it over before I present it.

Ratch

 

steveb,

I don't see any confusion or misstatement. I specifically stated that "R" represented the expression of resistance values on the right side of the equation. That is different than saying that R represents a resistance.

OK, if you want to defend that statement and don't think it can be misinterpreted, I won't argue about it.

 

Steve, I had the same initial impression as you. While "R" would normally be a good symbol for a ratio, in this case it was not, because of the Resistors in the expression. Mentioning "the expression of resistance values" only added to the confusion. I had to look at it a couple of times to realize that his "R" was not intended to represent resistance.

 

The Electrician,

So far, we've assumed that the opamp is ideal.

But, what if it isn't?

If the opamp gain is finite, equal to A, then what is the voltage between the inputs? Find an expression for (Vpos-Vneg).

Well, here it is by particular request. I used the resistor notation of AN1515, which is not what the OP used. Also Vpos and Vneg are the signal inputs, and Vp and Vn are the OP amp terminal inputs. So here it is for what it's worth.

Vp-Vn = -(-R4*Zl*R2*Vpos-R3*Zl*R2*Vpos+R1*Zl*R4*Vneg+Zl*R2*R4*Vneg+R1*R2*R4*Vneg)/(R3*Zl*R2*A+R3*R1*R2*A-R1*Zl*A*R4+Zl*R2*R4+Zl*R2*R3+R1*R2*R4+R1*R2*R3+R1*Zl*R4+R1*Zl*R3)

And, of course, when A approaches infinity, Vp-Vn approaches zero.

Now it would really be interesting if we try to compute something when the impedance between the OP input terminals is finite.

Ratch

 

The Electrician,



Well, here it is by particular request. I used the resistor notation of AN1515, which is not what the OP used. Also Vpos and Vneg are the signal inputs, and Vp and Vn are the OP amp terminal inputs. So here it is for what it's worth.

Vp-Vn = -(-R4*Zl*R2*Vpos-R3*Zl*R2*Vpos+R1*Zl*R4*Vneg+Zl*R2*R4*Vneg+R1*R2*R4*Vneg)/(R3*Zl*R2*A+R3*R1*R2*A-R1*Zl*A*R4+Zl*R2*R4+Zl*R2*R3+R1*R2*R4+R1*R2*R3+R1*Zl*R4+R1*Zl*R3)

And, of course, when A approaches infinity, Vp-Vn approaches zero.

Congratulations! You got it correct and typed it here without a single typo.

Now it would really be interesting if we try to compute something when the impedance between the OP input terminals is finite.

Ratch

If another impedance, ZB, is connected between the + and - terminals of the opamp, the expression for Vp-Vn is given by the expression shown in the attachment.

The second expression is just the somewhat simplified version.

It's amazing how rapidly the expressions grow with just the addition of one more component.

Professor Middlebrook (http://ardem.com/whatis_D_OA.asp) notes that analysis of relatively simple circuits can generate complicated expressions which he calls "high-entropy" expressions.

Also see:
www.aus.edu/engr/ele/documents/AnalogCircuitAnalysisUsingaReuseMethodology.pdf

 

Vp-Vn = -(-R4*Zl*R2*Vpos-R3*Zl*R2*Vpos+R1*Zl*R4*Vneg+Zl*R2*R4*Vneg+R1*R2*R4*Vneg)/(R3*Zl*R2*A+R3*R1*R2*A-R1*Zl*A*R4+Zl*R2*R4+Zl*R2*R3+R1*R2*R4+R1*R2*R3+R1*Zl*R4+R1*Zl*R3)

This result helps demonstrate the instability issue.

If the denominator of the above expression is set to zero, we get the following condition for instability.

\( Z_L^\infty={{R_1\; R_2\;R_4+(A+1)\;R_1\;R_2\;R_3} \over{A\;R_1\;R_4-A\;R_2\;R_3-R_2\;R_4-R_2\;R_3-R_1\;R_4-R_1\;R_3}}\)

As long as the above value of \( Z_L^\infty\) is very large or negative, and the real ZL is a reasonable value, there is no stability issue.

When A is very large and R1=R3 and R2=R4, then we see that ZL must be infinity for marginal stability, so there is no problem if a finite load is used. However, it is possible for mismatched values such that R1*R4 > R2*R3 to create instability for finite values of ZL. The good news is that as A gets smaller, those negative terms in the denominator become more significant and drive the instability point to higher load impedance. This means that at high frequency, when the gain naturally drops, there is no fundamental problem with this circuit.

Note that the above is just a quick way to make sure that the circuit is not ill-conceived, which it isn't. A proper stability analysis would use a more complex opamp model, and a real circuit would likely use some stabilization capacitors too.

This may appear to be of just academic interest but consider the use of 5 % resistors and imagine that R1*R4 is 10% higher than R2*R3, by chance. This would mean that a load impedance of RL=10*R1 would create instability.

 

The Electrician,

Thank you for your compliment. The attachment and link appear to be very interesting. I will certainly look them over. Prof Middlebrook certainly seems to be a very interesting fellow.

Ratch

 

Note that the above is just a quick way to make sure that the circuit is not ill-conceived, which it isn't. A proper stability analysis would use a more complex opamp model, and a real circuit would likely use some stabilization capacitors too.

This may appear to be of just academic interest but consider the use of 5 % resistors and imagine that R1*R4 is 10% higher than R2*R3, by chance. This would mean that a load impedance of RL=10*R1 would create instability.

Did you read the AN-1515 app note? I was thinking of emailing Pease about his appendix D. It's practically unreadable because he kept to the two column format and just embedded mathematical expressions within his text.

Go back to the original circuit at the beginning of this thread. Delete RL (make it infinite) and ground Vref. Then ask yourself what output voltage Va would make the + and - inputs the same. Any Va would do the trick; Va is indeterminate. Now, if Vref is something other than zero, there is no Va that will make the + and - inputs the same; we must have a load impedance. And, as you point out, if we have resistor mismatches, the impedance at the + input may be negative even with a finite RL.

For this circuit, a person can just look at it and draw some useful conclusions about stability. Those 24 GIC topologies we discussed in another thread take things up another level.

An interesting paper on this topic was published in the IEEE Transactions on Circuits and Systems, titled "How to Identify Unstable DC Operating Points", October 1992, pp. 820-832. If anybody wants a copy, PM me a usable email address.

 

Did you read the AN-1515 app note? I was thinking of emailing Pease about his appendix D. It's practically unreadable because he kept to the two column format and just embedded mathematical expressions within his text.

Yes, I see your point. It's more than just the formating. He has equations in which he adds voltages and the rate of change of voltages together. Probably these are typos, otherwise I dont get it.

I often have trouble following his analyses because he takes some non-traditional approaches and seems to avoid frequency based methods. However, he always gives good insight into the important issues.

Personally, in this case, I would just take his point that there is an output capacitance as part of the output impedance, if the OPAMP has a finite gain-bandwidth product. Then, I would just derive the output impedance in the frequency domain. The capacitance value will then become very obvious. Of course, so far we have only considered infinite gain-bandwidth so the equations will not yet show the capacitance. Under our assumptions, the output impedance is resistive if the resistors are mismatched and it is infinite if they are precisely matched.

 

If you look at Pease's latest column, he makes an issue of working in the time domain exclusively.

 

I read that Pease just got laid off.

 

If true, Bob will survive the layoff and prosper wherever he lands. I hold both he and Forrest Mims in very high esteem.

hgmjr

 

I read that Pease just got laid off.

that's too bad. though you have to admit that Bob is more of an entertainer than an engineer, as of late.

But he has a good brand name and his let-go by National suggests that things are really bad down there.