a. How do i find the correct bias point ?
b. The RC circuit is an alternative test ? Can you elaborate on that issue ?

 

a) Apply a dc voltage source at the inv. opamp input and make a dc sweep +- 5mV (use split supply, of course.). Use a sufficient resolution (1uV).
The transfer curve for Vout crosses the zero volt line at a voltage Voff. That is the offset voltage of the opamp model. Apply this voltage Voff at the inv. terminal.
Then, connect an ac source at the non-inv. terminal and perform an ac analysis.
Display magnitude and phase.
b) Alternative (without offset compensation): Use a large feedback resistor between output and inv. input (1megohm). Use a large capacitor (100uF) between inv. terminal and ground.
Thus you have 100% dc feedback (no offset problems) and the proper ac behavior about a lower cut-off frequency Fc=1/RC which is rathe small.

 

a) when i apply the dc voltage, what value do i give ?
What is the meaning of Voff in our issue and why is it so important ?

BTW - thanks so much for your kind help. it helps me a lot and it's very much appreciated.

 

Perhaps you did overlook the values I have given in my former post (first line):Approximately a range between minus and plus 5 mV.

Voff (offset voltage) is a measure of unsymmetry for the whole amplifier.
Because of the large dc gain (up to 1E5) an unsymmetry of 1mV (not unusual) would cause a dc output of 1E-3*1E5=100 volt (theortically!) - in fact, latch-up at power supply rail.
Therefore, for open-loop analysis an offset correction is necessary.

More than that, try to become familiar with negative feedback and its consequences on stability margin, gain, input/output resistances, linearity, bandwidth, slew rate. It`s really important.

 

Great, thanks.
I am actually learning about this stability issue. If you can help with something i cant grasp i would really appreciate it.
On page 22, in the gain compensation section, i didn't understand the second paragraph at all (the four lines under equation 27). I've tried Google it but nothing is useful, can you please help me with that ?

 

Great, thanks.

On page 22, in the gain compensation section, i didn't understand the second paragraph at all (the four lines under equation 27). I've tried Google it but nothing is useful, can you please help me with that ?

Please, reconsider your question.
I don`t know where you have problems.
The gain compensation section is on page 18 and on page 22 I can find equ. (35) and (36) only.

Edit: OK - I think I have identified your problem. Eq. (27) is on page 18 - and it shows some loop gain curves.
Instead of giving some explanation to the figure (showing loop gain versus frequency) I ask you at first:
* Do you know what "loop gain" is?
* Do you know the stability conditions for circuits with feedback (Barkhausen, Nyquist, Leonhard)?

This knowledge is necessary to understand the meaning of the whole chapter.

 

Eladta, I recommend to forget (at least) the whole chapter 6 of the Application note, because Fig. 17 and the corresponding explanations contain errors.
Thus, I understand that you have problems.
Look at Eq. (23) and it seems that K is the open-loop dc gain of the opamp.
However, this is in contrast to the explanations to Fig. 17 where K is changed from K/2 to K/10. In fact, I don`t know the meaning of K.
More than that, the whole Eq.(23) is wrong because in the denominator a frequency is added with a time (s+tau).
As mentioned before, the meaning of the curves in Fig. 17 is easy to explain if you know about the background of the stability criterion, do you?

 

I do, the basic criteria to my knowledge is to avoid open loop gain magnitude of 1 and phase of 180 degrees. Is there something else i'm missing ? I will explain what i don't understand on figure 17.
1. They are saying that the original loop gain curve is unstable, is it because the slop from 1/tao2 is very high (40dB/dec) ?
2. why does the compensated curve produce stability ? aside from the gain the curve slopes are the same !

 

I do, the basic criteria to my knowledge is to avoid open loop gain magnitude of 1 and phase of 180 degrees. Is there something else i'm missing ? I will explain what i don't understand on figure 17.
1. They are saying that the original loop gain curve is unstable, is it because the slop from 1/tao2 is very high (40dB/dec) ?
2. why does the compensated curve produce stability ? aside from the gain the curve slopes are the same !

Hi eladta, i didn`t want to "examine" you - I only wanted to know where to start with explanations.
1.) The stability criterion requires that the loop phase (that is the phase response of the LOOP GAIN LG=Aol*beta) must be already beyond -180 deg if the loop gain (magnitude) is LG=0 dB.
2.) That means (for typical opamps) that the SLOPE of the loop gain must be considerably smaller than -40 dB/dec at the zero-crossing frequency.
3.) For 100% feedback (feedback factor beta=1) the loop gain response is identical to the open loop gain Aol.
4.) To evaluate the slope you must use the slope of the real curve and NOT the slope of the asymptotic lines (which are shown in the figure only).
Thus, the lower curve (asymptotes) may indicate instability - however, as the slope of the real curve will be only marginally smaller, the stability will be, most probably, not sufficient.
5.) But, in general, it is correct that larger gain gives better stability properties. The most critical case is always beta=1 (100% feedback, closed-loop gain Acl=+1)
______________
Further questions?

EDIT: I have detected some more errors connected with Fig. 17. It`s completely useless! For example, the 14 dB are garbage.

 

EDIT: I have detected some more errors connected with Fig. 17. It`s completely useless! For example, the 14 dB are garbage.

14dB is the ratio of K/2 to K/10. Makes sense to me.

EDIT: The text above Fig. 17 says

The original loop-gain curve for a closed-loop gain of one is shown in Figure 17,
and it is or comes very close to being unstable. If the closed-loop noninverting
gain is changed to 9, then K changes from K/2 to K/10. The loop-gain intercept
on the Bode plot (see Figure 17) moves down 14 dB, and the circuit is stabilized.

First, I think K/Z should read K/2. This implies a noninverting closed loop gain of 2, or an inverting closed loop gain of 1. Then it mentions changing the noninverting closed-loop gain to 9, making the loop gain=K/10.
There was apparently some editing that went on during the drafting of this note that left this section with inconsistencies.

 

Understood. Although they are saying that the lower curve is stable, you are saying it isn't (with all do respect keep in mind that this tutorial is of Texas Instrument) ?
Another thing, say i wanna use the AD8639 OP AMP, on page 8 we can see the open loop response. we can see that in 0 dB crossover the phase is ~ 50 degrees, that is phase margin of 180 - 50 = 130 degrees. sounds very respecable. but you can see that the slop of the phase is pretty high (about 130 degrees/less than decade,) so is the op amp stable or not ? In any case don't we need to examine the loop gain (A * beta) instead of the open loop ?

 

14dB is the ratio of K/2 to K/10. Makes sense to me.

Ron, can you explain figure 17 perhaps ? and the four lines above it ?

 

Ron, can you explain figure 17 perhaps ? and the four lines above it ?

See my edit in post #30.

I agree that it is not clear from the plot that the circuit will be stable with the lower curve. I like to see the second pole corner be below 0dB. The phase shift when the 2nd pole is at 0dB is 135°, assuming tau1 >> tau2. So, with the 2nd pole being above 0dB, there can still be some phase margin.

 

14dB is the ratio of K/2 to K/10. Makes sense to me.
EDIT: The text above Fig. 17 says First, I think K/Z should read K/2. This implies a noninverting closed loop gain of 2, or an inverting closed loop gain of 1. Then it mentions changing the noninverting closed-loop gain to 9, making the loop gain=K/10.

Hi Ron, of course one could guess what they mean in the TI note - however, do you really think it makes sense?

They speak about a "closed-loop gain of one" and later about a "non-inverting gain" of 9. And your guess is that they compare - in one sentence - an inverting gain of "1" with a non.inv. gain of 9 ?
May be - but for my opinion, very confusing.
And the whole "calculation" is based on a (factor?) K, which is not defined for my opinion. Do YOU know what K is? Which dimension? Perhaps the DC open-loop gain?
Shall I mention further "inconsistencies" in Fig. 17?
* The frequency axis shows "f" - however some corners are given with 1/tau.
As you know, the classical convention is 1/tau=w=2*Pi*f.
* And what is shown on the gain axis: An expression equivalent to K*beta
which is fixed value. Correct: A*beta.

Ron, did you inspect Fig. 15 and Fig. 16? You can find similar errors (with your words: inconsistencies).
They speak about open-loop gain - and what can you read at the vertical axis: A*beta (which, probably, is the gain of the open loop, i.e. loop gain.)
I am afraid here the same error has happened as in some other articles (even some textbooks): They confuse loop gain with open-loop gain of an opamp!

More than that, the head line is "gain compensation".
I must state that I have seen a lot of books and papers about opamps and the methods to improve stability. But it is the first time to read in this funny paper that applying higher gains is a method of "compensation".
For my opinion: Simply false! This is by far not a kind of compensation. Rather, it is the classical phenomenon which is known from each system with feedback: Higher closed-loop gain means lower loop gain and improved stability. That`s all.

In summary, I do not like notes that confuse the reader instead of defining parameters and explaining relations and formulas. And I don`t like to make guessing in this context and to discuss all the errors and inconsistencies.

Therefore, I repeat my my recommendation to the OP:
Try a google search for some other papers which explain some basics about opamps and the associated stability properties in a correct manner (and forget Fig. 17 and the whole note).
And when you have specific questions - do not hesitate to ask again.

 

Here comes another TI document - much, much better and correct!!!
Read, in particular, starting with page 31.

http://www.google.de/url?sa=t&rct=j...bcScdLQKEiWTPlcDe80gWjg&bvm=bv.48705608,d.Yms

In this document, the quantities open-loop gain, loop gain and closed-loop gain are visualized in a clear and understandable manner.
Some explanations to Fig. 34:
The line at 40 dB is the value of "1/beta" (1/feedback factor). This can be explained using the loop gain expression

Loop gain LG=Aol*beta=Aol/(1/beta) >> Aol in dB minus 1/beta in dB.
Aol=opamps open-loop gain .

Thus, if you plot the difference Aol(dB)-1/beta (dB) the remaining function is LG(dB).

Note that the non-inv. closed-loop gain is (nearly) identical to 1/beta (therefore, 40 dB=closed-loop gain in the figure).
Any questions?

 

14dB is the ratio of K/2 to K/10. Makes sense to me.

EDIT: The text above Fig. 17 says First, I think K/Z should read K/2. This implies a noninverting closed loop gain of 2, or an inverting closed loop gain of 1. Then it mentions changing the noninverting closed-loop gain to 9, making the loop gain=K/10.
There was apparently some editing that went on during the drafting of this note that left this section with inconsistencies.

Help me clarify the terms, when you say "closed loop gain" you mean the "loop gain" (A*beta according to these article) ? I guess not because if so, I've realized that they both the same.
Bottom line is why does K/2 implies a noninverting closed loop gain of 2, or an inverting closed loop gain of 1 ?

 

Loop gain LG=Aol*beta=Aol/(1/beta) >> Aol in dB minus 1/beta in dB.
Aol=opamps open-loop gain .

Any questions?

Great LvW, that helps a lot !!
What i didn't understand is the "rate of closer" term they are using.
They are stating (WITHOUT explanation) that the intersection point between the closed and open loop
curves is important because the angle between the two curves - or, more precisely, the “rate of
closure” since the curves aren’t actually straight lines - determines whether the closed loop
amplifier, being designed will be stable.
Principle: If the rate of closure between the open and closed loop sections of the Bode plot is greater than 40 db per decade the system is likely to be unstable.

Can you explain the term “rate of closure” and the principle of 40 dB ?

 

Never mind LvW, I got it. Thanks for the article.
Can you please write the equation of the closed loop gain ?
This is the only ambiguous term for me

EDIT: another basic question: unity gain meaning Aol=1 or beta=1 ?

 

Can you please write the equation of the closed loop gain ?
This is the only ambiguous term for me
EDIT: another basic question: unity gain meaning Aol=1 or beta=1 ?

1.) Closed-loop gain
Acl=Aol/(1-LG)=1/[(1/Aol) + beta]
with open-loop gain Aol, loop gain LG=-beta*Aol,
feedback factor beta=R1/(R1+R2)

For 1/Aol<<beta we have the classic formula for non-inverting gain
Acl~1/beta.
Note that for inverting operation we have an additional factor (input
damping) k=-R2/(R1+R2) resulting in Acl=-R2/R1
____________________________________________________________

2.) unity gain means Acl=1/beta=+1 >>> beta=1 (100% feedback).
In this case: Aol=LG (loop gain identical to open-loop gain of the opamp)

3.) Rate of closure always means: When two lines are crossing forming an angle, this angle can be expressed by another quantity in dB/dec. In most cases one of the lines (A) is horizontal. In this case the rate of closure (ROC) is identical to the slope of the other line. But in some cases (e.g. frequency dependent feedback) none of both crossing lines is horizontal. I this case, the ROC is the difference in dB between slope A(db) and slope B(dB).

 

Eladta, perhaps it helps to explain WHY the rate of closure (ROC) plays such an important role for stability properties.
For all feedback circuits without zeros in the right half of the s-plane (RHP) - in particular for operational amplifiers - there is a fixed relation between the slope of the magnitude response and the phase response. And it is the phase that matters!
However, because of this known relation, it is not necessary to observe the phase response - we can translate it into the magnitude response.

Examples: -20 dB/dec >> -90 deg ; -40 dB/dec >> -180 deg.

Thus, you get asymptotes in the phase plot which allow a rough estimate of the real phase angles - for example: -30 dB/dec >> -(120...140) deg (roughly).

Now, because the stability criterion for amplifiers requires for the LOOP GAIN (LG) a minimum phase margin of app. 30...40 deg (better: 60 deg) this requirement - if transferred to the LG magnitude response - means: Rate of closure (ROC) for the loop gain at the point LG=0 (crossing point) must NOT be -40 dB/dec but app. -30 dB/dec or less. (In theory, a ROC of -39 dB/dec would be stable, however, the amplifier could not be used because of a catastrophic step response with many ringing periods).
Hope this helps and improves your understanding.

EDIT: Example for ROC:

Aol decreases with -20 dB/dec (normal behaviour),
1/beta increases with +20 dB/dec (differentiator response),
ROC at LG=0 (crossing point): ROC=20-(-20)=40 dB/dec >> instability.