I'm reading the TI-precision lab of opamp. And I'm confused with the calculation of the BW_CL (bandwidth of th close-loop).

I'm prettry sure the BW_CL= GBPW/GAIN_CL. This is very clear refer to enter link description here

What I cannot understand is why can we find/use the close-loop gain on the bode plot of "Open-loop gain vs frequency" ?

As I see I get the bandwidth when open-loop gain = 40dB !?!?

enter image description here

  • \$\begingroup\$ In this context, the closed-loop gain is just a number. If you can find that number on the y-axis of your open-loop gain plot, then you've found it on that plot. \$\endgroup\$
    – The Photon
    Commented Nov 5, 2020 at 3:16
  • \$\begingroup\$ @The Photon Sorry but would you explain a bit more? I can understand it's just a number.. but there should be some connnection;otherwise why we can find the coresponding frequency.. I think I might be stuck in sth.. \$\endgroup\$
    – FNJU
    Commented Nov 5, 2020 at 3:39
  • \$\begingroup\$ The connection is shown in the image you posted. You draw a horizontal line from that number and where it intersects the open-loop gain tells you what the bandwidth will be in the closed-loop circuit. \$\endgroup\$
    – The Photon
    Commented Nov 5, 2020 at 3:42
  • \$\begingroup\$ @The Photon Thanks but still cannot get it... I'm still stuck in: BW_CL= GBPW/GAIN_CL, it's right, and I draw a horizontal line of the close loop gain, it's also fine. but this line intersects the open-loop gain curve, and the frequecy I get is when open-loop is equal to this gain value??? \$\endgroup\$
    – FNJU
    Commented Nov 5, 2020 at 6:07
  • \$\begingroup\$ Just like it shows in the picture you posted \$\endgroup\$
    – The Photon
    Commented Nov 5, 2020 at 6:09

3 Answers 3


What I cannot understand is why can we find/use the close-loop gain on the bode plot of "Open-loop gain vs frequency" ?

Example: Non-inverting amplifier with closed-loop gain:

Acl=Aol/(1+kAol)=1/[(1/Aol)+k] (with open-loop gain Aol and feedback factor k.)

For low frequencies (Aol>>k and 1/Aol<<k) we have Acl=1/k

We can insert the closed-loop gain 1/k in the Bode plot (red horizonal line in the shown diagram) - as long as the approximation Acl=1/k holds. There is a certain frequency wo where we have 1/k=Aol (crossing of both graphs) - and for larger frequencies w>>wo the horizontal line (1/k) is above the Aol curve - now we have 1/k>>Aol or k<<Aol.

As a consequence, the feedback factor k can be neglected in the Acl expression and the closed-loop gain now can be approximated by Acl=Aol.

Both regions are separated by the frequency wo where we have Aol=1/k (loop gain k*Aol=1).Hence, wo is the frequency that determines the bandwidth for the closed-loop gain Acl.

Note that the region between the 1/k line and the Aol response gives you the loop gain in dB - and at w=wo we have unity loop gain. In the above diagram the loop gain for small frequencies is 120dB-40dB=80dB and rolls off with 20dB/dec until it reaches 0dB at w=wo (2Pi*220kHz).

Comment (with respect to the contribution from James): The inverse feedback factor 1/k is sometimes called "noise gain".

  • \$\begingroup\$ Graeat! I almost understand it. And so based on Alo=1/k. I can get: at the wo, the Acl = 1/2*1/k ,so can we figure out it's -3dB for Acl? \$\endgroup\$
    – FNJU
    Commented Nov 6, 2020 at 10:13
  • \$\begingroup\$ I mean math calculation based on above, many thank! @LvW \$\endgroup\$
    – FNJU
    Commented Nov 6, 2020 at 10:15
  • \$\begingroup\$ FNJU - analyze the denominator of the Acl expression (1+kAol). At wo we have Aol=1/k (unity loop gain or 0 dB) and the magnitude of the denominator is SQRT(2) - which means: -3dB. \$\endgroup\$
    – LvW
    Commented Nov 6, 2020 at 11:55

The DC low pass filter with a breakpoint of 20 Hz and a gain of 120 dB is no accident. This is performed by an small integrating capacitor in the intermediate stages of the Op Amp.

By swamping (shunting) the gain of the amplifiers well below what they are capable of, means that the frequency limitations of each transistor stage are replaced with an ideal amplifier and an integrator. First order filters are much easier to deal with in closed loop designs than higher order feedback with issues with instability, oscillations, ringing, overshoot.

So essentially the bandwidth limitations of each transistor stage are ignored. Thus a high order, high phase-shift effect at unity gain is converted from an nth order filter to a first order low pass filter with -90 phase shift up to unity gain where it might be starting to see the residual phase shift and gain of the internal transistors so -100' phase shift with 180 deg inverting feedback means you would expect a 180-100 = 80 deg. phase margin at unity gain. I recall 60 deg. is often considered adequate for Phase margin in a closed loop design, unless you really need zero overshoot on a step pulse.

Thus the reason we use the Bode plot to read phase margin, is because it is necessary to have less than 180 phase shift for intended negative feedback gain control with impedance ratios and always converting to 1st order integrator works easiest. But for "video amps" that need high gain, this method is not used.

Now the Bode GBW is just a linear 1st order product of the integrator of 20dB/decade or 10x per decade = Gain * BW= GBW


As I see I get the bandwidth when open-loop gain = 40dB !?!?

Yes. The Open Loop gain, Aol at 220 kHz is 100 or 40 dB and if you choose a gain using Rf/Rin=100 then the -3dB BW is also 220 kHz so the actual gain is 37 dB , the half power point.

What is nice about Log-Log graphs here is that a straight line can represent of Gain vs BW instead of an inverse hyperbolic line on a linear scale. Thus a product or a ratio displays this easily like a slide rule used to for computing ratios or products of 2 numbers, Gain and Bandwidth. Since the LPF attenuates at 20 dB/decade for a 1st order filter, the result is this linear slope on a log-log scale.

So the Bode Plot shows your Op Amp linear R ratio for fixed gain up to the point where the open loop gain intersects with the planned fixed gain.

  • \$\begingroup\$ stewart I've edited my question trying to make my confusion more clearly.. \$\endgroup\$
    – FNJU
    Commented Nov 5, 2020 at 6:24
  • \$\begingroup\$ FNJU I edited my answer to help resolve your confusion Closed loop gain uses feedback to amplify the error then the only current is supposed to be just going thru the resistors whose ratio determines the gain. I didn't know what you didn't know at first. \$\endgroup\$ Commented Nov 5, 2020 at 7:35

Actually we don't use closed loop gain to determine bandwidth using GBW. What you should use is the Noise Gain.

Bandwidth = GBW/Noise Gain

Where Noise Gain is defined as 1/Beta Beta being the feedback fraction which is equal to R1/(R1 + R2) for both the inverting and non-inverting amplifier configurations. (R2 being the feedback resistor).

For a purely resistive feedback network, the Noise Gain plot is a horizontal line on a Open Loop Gain vs Frequency plot and where it crosses the Open Loop Gain plot the loop gain is equal to unity. At this point the closed loop gain is down 3dB.

Lets consider a non-inverting amplifier with equal resistance values for R1 & R2 and it having a GBW of 1MHz.

Its closed loop gain is equal to its Noise gain which is equal to 2 and therefore its bandwidth is 500kHz

Now lets consider an inverting amplifier also with equal resistance values for R1 & R2 it also having a GBW of 1MHz.

Its closed loop gain is equal to 1 but its noise gain is equal to 2 therefore it also has a bandwidth of 500kHz.

So it is apparent when comparing the inverting and non-inverting configurations of amplifier with equivalent closed loop gains that the non-inverting amplifier will have the higher bandwidth.

In fact if both configurations of amplifier are configured for a closed loop gain of 2 then the non-inverting amplifier has a bandwidth of 500kHz but the inverting amplifier, which now has a noise gain of 3, only has a bandwidth of 333kHz.


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