# Relation between crossover freqquency of loop gain and closed loop's corner frequency when feedback is not unity

In this Control Loop Cookbook from TI, in the red box the author said that the crossover frequency (indicated in figure 2) of the open-loop is same as 3dB down frequency or corner frequency of the closed loop.

However, I don't think this statement is correct with the specific case in figure 2. The statement is only correct if feedback factor Kfb is unity. But from the closed-loop gain, we can infer that Kfb is not 1 (Kfb = 0.5).

What do you think about this? Is the article wrong? In general case, the crossover frequency of open-loop fc(actually I think call it "loop gain" is more accurate) makes the closed loop corner frequency at Kfb*fc.

• It doesn't make much sense to me either. Mar 7, 2018 at 13:08
• So do you agree with my conclusion? With that system above, Kfb =0.5 and the crossover frequency of loop gain fc, the closed-loop gain 3dB corner frequency should be Kfb*fc or 0.5fc not fc as the article. Mar 7, 2018 at 13:16
• I don’t think the figure 2 is that great either because they show the closed loop gain remaining higher at frequencies higher than the open loop gain can sustain. Mar 7, 2018 at 13:34

It is not a mistake, the author is correct.

..at $$\\ f_c\$$ (where $$\\ T=1\$$, with associated $$\ 90^{\circ}\$$. phase lag), the closed loop gain $$\\ G(s)\$$ is 3dB down..

Let $$\\ T=1\angle90^{\circ}=\cos(90^{\circ})+j\sin(90^{\circ})=j\$$

The closed-loop transfer function is:

$$\\ G(s) = \frac{1}{K_{FB}} \cdot \frac{T}{1+T}\$$

Substituting $$\\ T=j\$$ and solving for the gain:

$$\\ |G(s)|= \frac{1}{K_{FB}} \cdot |\frac{j}{1+j}|=\frac{1}{K_{FB}} \cdot |\frac{1}{2}+\frac{1}{2}j|=\frac{1}{K_{FB}} \cdot \sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2}=\frac{1}{K_{FB}} \cdot \frac{1}{\sqrt{2}}\$$

$$\\ 20 \cdot \log_{10}(\frac{1}{\sqrt{2}})=-3 \$$ [dB]

As the author stated; when the open loop gain $$\\ T(s)\$$ equals one, with associated $$\ 90^{\circ}\$$ phase lag, the closed loop gain $$\\ G(s)\$$ is down 3dB, or $$\\ \frac{1}{\sqrt{2}}\$$

• I can't follow your maths but with a closed loop gain of 2 the closed loop gain is down 3dB when the open loop gain is also equal to 2 (not 1). Jan 21, 2020 at 10:32
• G(s) = T/(1+(0.5T)) Jan 21, 2020 at 10:40
• No, not true, the closed loop gain is down 3db when the open loop gain T(s), as defined by the author, is equal to 1 with an angle of 90deg, or in other words when T(s) =j.. Jan 21, 2020 at 11:18
• Ah! I see where the confusion is. "As defined by the author" is the crux. What the author calls the "open loop" gain is usually referred to as the "loop" gain. By conventional definition the "open loop" gain is the forward loop gain or Vout/Vdiff for an op amp and the "loop" gain is the gain right around the loop. I agree that where the loop gain in the usual sense, (open loop gain in the author's sense), is unity, the closed loop gain is down -3dB whatever the value of the low frequency closed loop gain. Jan 21, 2020 at 16:43
• Yes exactly, as I wrote; when $\ T=1\angle 90^{\circ}$ then $\ G(s)=\frac{1}{\sqrt{2}}G_0$.. $\ T$ is what the author calles the open loop gain, and I agree that I would have called this the loop gain. So the statements the author make are all completely true and accurate, and the math holds true, the confusion comes from two facts; the author used the term "open loop gain" instead of "loop gain" and also the author uses the term "gain" when he means "transfer function" the gain is not the transfer function it self but the magnitude of the transfer function. Jan 21, 2020 at 16:49

You are absolutely right. The loop gain (open loop gain times the feedback fraction) is unity at the frequency where the closed loop response is down 3dB at -45 degrees. It is only for unity closed loop gain (feedback fraction = 1) where the closed loop gain is down 3dB when the open loop gain is unity. For a closed loop gain of 2 (feedback fraction = 1/2), the closed loop gain is down 3dB when the open loop gain = 2 (loop gain = 2*0.5 = 1).