I would be very happy if anyone can help me to find these questions regarding the opamps.

In the first picture the parameters are given as Aol = VOA/VM and 1/B = Vout/Vm. ( 1'st picture is from TI)
In the second picture the parameters are given as Aol = Vo/Vfb and 1/B= 1/Vfb ( 2nd picture is from TI)
So my questions are as follows :

  1. Why is the 1/B is different from each other although the circuits are similar ?
  2. When i try to solve for AolB for the first picture, i found AolB = VOA/VOUT. This result doesn't make sense to me. Do you have any suggestions ?
  3. The Vout in the 1'st picture corresponds to node betweem L1 and C1 in the 2'nd picture so why they seperated the Voa and Vout as two different output in the 1'st picture ? they seems same to me. enter image description here

enter image description here

  • \$\begingroup\$ Second figure: Is "beta" identical to a voltage? \$\endgroup\$
    – LvW
    Nov 6, 2023 at 9:30
  • \$\begingroup\$ @LvW , as far as i can see beta is equal to Vfb in the second picture. \$\endgroup\$
    – Mhan
    Nov 6, 2023 at 9:43
  • 1
    \$\begingroup\$ Hello! Can you please update your question with the sources for those pictures, or if they are your own work clearly state that they are. Thank you. \$\endgroup\$
    – MiNiMe
    Nov 6, 2023 at 10:06
  • \$\begingroup\$ Hi dear @MiNiMe , sure ı can provide as follow : The 1'st ficture is from picture.iczhiku.com/resource/eetop/SHKlYzWoRirirbNX.pdf The 2'nd picture is from Texas opamp videos - stability analysis - spice simulation \$\endgroup\$
    – Mhan
    Nov 6, 2023 at 10:15
  • 1
    \$\begingroup\$ Good - and I should've been more clear about this - edit this into your question. Also provide that the first picture is from TI as well, the link itself doesn't tell. \$\endgroup\$
    – MiNiMe
    Nov 6, 2023 at 10:23

1 Answer 1


1/beta = 1/Vfb is an incorrect statement and doesn't make sense. It is correct to say that 1/beta = Vo/Vm.

The third statement in the second picture is also incorrect, it should read Aol*beta = Vo/Vin

In both of those pictures the loop is being opened at the output and a very large inductor is being inserted in the break in the loop. The very large inductor is a short to dc so maintaining DC conditions around the opened loop but the large inductor is an open circuit to ac. This means that a signal can be injected into one side of the break (into the feedback beta circuit) and the output (phase and gain) of the complete loop can be measured at the output of the loop, at the other side of the inductor. So we see that VOA/Vout is actually the loop gain which is equal to beta * Aforward. (strictly speaking the loop gain is equal to -beta * Aforward, the minus sign being included in the loop gain expression because of the inversion caused by the op amp's inverting input).

Now there is sometimes some extra confusion. This is because "open loop gain" (Aol) can be defined in two ways - firstly it can refer to the gain around the complete loop -beta*Aforward but this in electronic circles is usually referred to as the "loop gain", secondly it can refer to just the forward gain which is also usually the case when talking about amplifiers. In my formula I have referred to the forward gain as Aforward to avoid any ambiguity caused by the two different definitions of open loop gain.

Reserving the term "open loop gain" for just the forward part of the loop and then using the term "loop gain" for the whole loop removes this ambiguity.

The purpose of opening the loop, injecting a signal into one side of the break and measuring the gain and phase of the signal being returned at the other side of the break after it has travelled around the loop and doing this over a wide frequency range enables us to analyse the data using a Nyquist plot, a Bode plot or a Nichols chart thereby giving us a technique for assessing the stability of the circuit by looking at gain and phase margins and then modifying the circuit (compensation) to achieve the stability margins we require.

  • \$\begingroup\$ thanks alot for your explanations i rewrite everyting as follows. Is there any mistake in the below: Actually the both VOA and Vout are output depending on the situation of inductor (short or open). when L is short, Vout and when L is open VOA. This is what ı see from the 1'st picture. Considering the the beta is related to how much portion of the output is fed back to input, when the inductor is short 1/beta = VOUT/ Vm. When inductor is open Aol = VOA/Vm At last, the loop gain becomes Aol*beta = VOA/Vin These can be also applicable on the 2'nd picture as well right ? @user350400 \$\endgroup\$
    – Mhan
    Nov 6, 2023 at 15:09
  • \$\begingroup\$ @Mhan For DC it's as though the inductor is a piece of wire where as for AC (the injected test signal) the inductor acts as an open circuit (a break in the loop). For the first picture the loop gain = VM/Vout *VOA/VM = VOA/Vout (where Vout is actually equal to the injected test signal Vin) = beta * Aol (where Aol is the forward gain). For the second picture, the loop gain = Vo/Vfb * Vfb/Vin = Vo/Vin which again, as you say = beta * Aol(where Aol is the forward gain). As I mentioned previously there should be a minus sign before the loop gain expression resulting from op amp input inversion \$\endgroup\$
    – user350400
    Nov 6, 2023 at 18:28
  • \$\begingroup\$ Sir, since ı am new to this subject, ı am confused at another point. How can Vout be equal to Vin? when L is short (at DC) which means Vout of the opamp is fedback to inverting input and capacitor is open at the same time. at that point how Vin is equal to Vout ? @user350400 \$\endgroup\$
    – Mhan
    Nov 6, 2023 at 20:18
  • \$\begingroup\$ @Mhan With reference to the first picture - The inductor is a short to DC but an open circuit to AC. A capacitor is the opposite, an open to DC but a short to AC. This means that Vin can can wiggle up and down (AC) and Vout follows it. The effect is to send this AC signal to the feedback beta circuit and around the loop to VOA where it can be measured. VOA is isolated with regard to the AC input signal at Vout because the large inductor is open to AC. However the inductor is a short to DC so DC conditions around the loop are maintained and Vin is isolated from this DC by the input capacitor. \$\endgroup\$
    – user350400
    Nov 6, 2023 at 20:47
  • \$\begingroup\$ Okay Sir. so Vout and Vin appear at the same node as you said they follow each other. Then we can say that 1/beta= Vin/Vm at AC (Vin is being fed into beta circuit) or it can be written as 1/beta=Vout/Vm at DC (Vin is isolated so Vout is being fed into circuit). Since ratios are important here, both of them represents the same thing right ? @user350400 \$\endgroup\$
    – Mhan
    Nov 7, 2023 at 7:25

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