The text says that "the amplifier requires compensation because its basic open loop gain is still higher at frequency where internal phase shift are reaching 180 degrees..This turn negative feedback to positive at higher frequencies which causes oscillation."

My question is that why they tell to roll off the open loop gain…rather than it should be closed loop gain because oscillation develops when negative feedback is turning to positive at higher frequencies, so feedback anyhow is necessary to cause oscillation..and when we talk about feedback we mean about close loop gain…then from where open loop gains comes into picture…??


why they tell to roll off the open loop gain

Saying it this way is just a convenience but the practicality is that there will be negative feedback applied that both shapes the amplifier gain to what you want AND corrects for instability.

First of all, familiarize yourself with the important features of the open-loop gain. This is for an OPA192: -

enter image description here

This op-amp is pretty stable when you connect negative feedback and you can see that in the open-loop response because, for instability to occur, the phase angle would have be passing through 0 degrees when the gain was at some positive dB value. Clearly the OPA192 is going to be stable when resistive negative feedback elements are used.

Now let's consider the OPA338 op-amp. The graph below shows the OPA338 (unity gain unstable) and the OPA337 (unity gain stable): -

enter image description here

My question is that why they tell to roll off the open loop gain

If your basic amplifier is unstable, the open-loop gain can be "tweaked" using a combination of both attenuation and feedback components to make an unstable amplifier stable at the gain you require when you eventually apply the required negative feedback.

However, the bottom line is that you might apply feedback components to correct the open-loop response that are barely indistinguishable from the components you apply to give you the desired gain.

It's a little "pretty" and "theoretical" to say you should "roll-off the open loop gain" but it's a useful way look at things. practicality means that the feedback you need for the desired performance merges with other components to prevent instability.

  • \$\begingroup\$ Just one comment to Andy aka`s nice drawings (for a better understanding): The arrows (stable/unstable) are relevant for 100% feedback only (feedback factor k=1). In this case, the LOOP GAIN (only this gain matters for stability) is identical to the open-loop gain Aol of the opamp which is shown in the drawings. For a closed-loop gain of Acl=6dB the OPA 338 would be at its stability limit - and for larger gains Acl both alternatives are stable. \$\endgroup\$ – LvW Feb 3 '16 at 12:46
  • \$\begingroup\$ Forgot to mention that the data sheet for OPA338 requires a gain>5 for stable operation (with sufficient phase margin). \$\endgroup\$ – LvW Feb 3 '16 at 14:20
  • \$\begingroup\$ Your explanation especially last three paragraphs are making sense to me...but take it this way...During oscillations a controlled closed loop gain at negative feedback changes into an uncontrolled closed loop gain at positive feedback.....so the whole deal about compensation is to control this uncontrolled closed loop gain below unity. \$\endgroup\$ – partykid Feb 4 '16 at 6:44
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    \$\begingroup\$ @sixcab pure laziness and not thinking. It should start close to 180 degrees at low frequency but other times the same graph might start at 0 degrees. I just get used to converting in my head what they mean when different numbers are shown. Yes, it can be confusing to folk when first getting into this sort of study. One plausible answer - they are making a non-inverting gain measurement with the inverting input grounded so to speak. \$\endgroup\$ – Andy aka Feb 17 '18 at 18:38
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    \$\begingroup\$ Yes it would be the same. \$\endgroup\$ – Andy aka Feb 18 '18 at 10:23

The closed loop gain consists of two cascaded gains
a) the open loop gain
b) the feedback gain

For an amplifier to be stable when fed back, the closed loop gain must be stable.

When we apply feedback to a high gain amplifier, we are creating a system where the output/input signal gain is the inverse of the feedback gain. For instance, if we want an overall signal gain of 10, we might use a 9k series feedback resistor and a 1k shunt feedback, to give a feedback gain of 0.1, wrapped round a high gain op amp.

Therefore if we want an amplifier with a given signal gain, we do not have any scope to change the feedback gain, it has already been defined.

If it unstable, we can therefore only change the open loop gain.

To summarise - if all you want is a stable amplifier, you can alter the feedback and/or the open loop until it's stable. If you want a stable amplifier with a specific signal gain, you may have to alter the open loop response as well.


When the phase shift reaches 180 degrees the negative feedback becomes positive feedback. changing the closed loop gain cannot change this.(If you lower the closed loop gain, you will still have positive feedback,though the entire system will have a smaller output.) However, if the open loop gain is lowered so that at the frequency which causes the 180 degree shift the gain is less than one, the positive feedback is attenuated rather than amplified, and does not destabilize. (and at negative feedback frequencies, we still have enough gain for the system to behave)

  • \$\begingroup\$ How can open loop gain(gain when no feedback is applied) is lowered....I mean that it is a static quantity which is composed during the time of manufacture of the opamp and remain fixed.. Please justify? \$\endgroup\$ – partykid Feb 4 '16 at 6:11
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    \$\begingroup\$ There are two general ways to stabilize an amp. 1) internally: a capacitor is added between amplification stages in some op-amps this has already been done( i.e. 741) and others allow you to add a capacitor between to pins. The general purpose of the capacitor is to roll off the gain so that before the phase reaches 180,the returned signal (A*Beta) is attenuated. \$\endgroup\$ – Mordechai Salomon Feb 7 '16 at 11:57

The bandwidth of linearity and stability of a closed loop amplifier is different from the one of the open loop amplifier (op amp) used to build the closed loop system. The open loop gain varies with the frequency omega, A(omega), and depends on the internal structure of the op amp while the gain of the closed loop system is: G(omega) = A(omega)/(1+A(omega)*beta(omega))

where beta(omega) is the frequency response of the feedback network (the two-port network you can identify as the one having its input connected to the output of the op amp, and its output connected to the feedback node of the op amp).

The A and beta needs to be evaluated in terms the Barkhausen's stability criterion which states that, in order to avoid auto oscillations, it must be that: |Abeta| < 1 and phase(Abeta) < 45 degrees

Knowing this, one can calculate the frequency response beta(omega), given the feedback network values, and can read the A(omega) from the data sheet of the op amp.

By plotting A(omega) superposed to 1/beta(omega), one can observe what happens at the critical point where the two lines cross each other: at that critical point A = 1/beta, or conversely A*beta = 1.

enter image description here

The common method is to use ROC = Rate Of Closure analysis in order to study the stability around the critical point. Since the phase response of any transfer function is correlated with the slope of its frequency response, it is sufficient to look at the slope of the A(omega) and 1/beta(omega) curves where they cross each other (in a logarithmic scale plot these are the so called Bode diagrams).

The open loop A(omega) usually have a dominating pole (slope = -20dB/dec) and A(omega) crosses the frequency axis at a frequency equal to fGBP (GBP = Gain Bandwidth Product, this piece of data can be read from the op amp's data sheet), mainly due to the input capacitance of the op amp.

The feedback network of a linear closed loop amplifier is usually a voltage divider made by two resistors. This kind of network has no poles, nor zeros.

In order to have the phase margin of A*beta close to 45 degrees, it is necessary to have at least one supplementary pole in the beta(omega) diagram: this corresponds to have one zero and one pole in the 1/beta(omega) diagram, making the latter one becoming flat for high frequencies.

If we can make the the 1/beta(omega) diagram intercepting the A(omega) in a way that 1/beta(omega) is flat when A(omega) is sloping down at -20dB/dec then this situation grants the phase margin of A*beta is close to 45 degrees: the system is hence stable.

This can be done by putting a shunt capacitor Cf in parallel to the feedback resistor (the one connected to the output of the op amp in the feedback network): this solution is called the Miller's compensation. That capacitor makes the pole we need in beta(omega) in order to introduce the supplementary pole we need: f_pole = 1/(2*piRf(Ci + Cf))

where Rf is the feedback resistor and Ci is the input capacitance of the op amp (read it from the data sheet).

The approximate triangle formed by the interception of the two curves with the horizontal axis, in logarithmic scale, gives the intercept frequency f_int to be the average of the two other vertices at the base: log(f_int) = (log(f_pole) + log(f_GBP))/2

The latter formula can be substituted for the expression of f_pole and expanded at the first order with the Taylor series. From that expansion, inverting the formula, the proper value for the feedback capacitor can be obtained as:

Cf = (1 + sqrt(1 + 8*piRfCi*f_GBP))/(4*piRff_GBP)

A larger capacitor is still good but unnecessary: values larger than this leads to an unnecessary overcompensation.

The resulting bandwidth of the closed loop system is f_int < f_GBP, in order to grant stability. This can be read conversely: for a given desired bandwidth of the closed loop system, the bandwidth of the open loop op amp must be higher, according to what previously said. From the point of view of control theory, the control system (in this case the op amp) must always be faster than the controlled system (here being the feedback network).

This last sentences are the final answer to your question, whatever is before is the explanation of "why" it is like this.

Best regards,


  • \$\begingroup\$ Thank Erik for your such an elaborative explanation.... But I was seeking quite simple explanation as my query was also to simple.....What I just meant to say that was......since oscillation develop when negative feedback change to positive..that means that feedback is necessary for oscillations..and if feedback is involved then the talk is about closed loop gain and if talk is about closed loop gain then why in textbooks its written to roll off the open loop gain..rather than it should be written to roll of the close loop gain for compensation.. \$\endgroup\$ – partykid Feb 4 '16 at 6:26

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