# Question on the close loop gain and bandwidth of this OpAmp configuration

The OpAmp configuration below has two switch:

The close-loop gain and bandwidth on switch 1 is:

: A_v(CL) =1 and bandwidth of 1MHz

• I don't understand why at switch1 the gain is 1, I thought it will be zero since its common-mode signal.

• And if ever the gain is really 1,why would the upper bandwidth be 1MHz? I learned before that: $$f_{2(CL)}=\frac{f_{unity}}{A_v(CL)+1}$$ So the bandwidth must be:

$$f_{2(CL)}=\frac{1MHz}{1+1}=500KHz$$

Why at switch 1 the closed-loop gain is 1 and not 0 and why the bandwidth is 1MHz and not 500KHz(if ever the gain is really 1)?

• 1. Since at Connection 1: V+ and V- are shorted to Vin due to virtual ground of opamp in negative feedback causing potential difference across R1 =0, hence, no current flows. Therefore Vrf = 0 also that means Vout=Vin
– DivB
Commented Jul 9, 2020 at 10:39
• 2. Bandwidth is calculated using open-loop frequency response. That is why Unity Gain Bandwidth = 1MHz which you are getting.
– DivB
Commented Jul 9, 2020 at 10:46
• Since Vrf is zero, using KVL, Vout = Vrf + V-. Therefore Vout = Vin(Since current does not flow through Rf, the potential acoss its end-points would remain same). Yes it would have been zero, if feedback was not used. Since (V+) - (V-) would not be exactly zero. There would be some potential difference. Very Small.
– DivB
Commented Jul 9, 2020 at 10:51
• The current flowing through the resistors would be negligible. Hence, we neglect their drop and essentially, you can approximate with the voltage KVL.
– DivB
Commented Jul 9, 2020 at 10:53
• Yes, as you said.
– DivB
Commented Jul 9, 2020 at 11:04

The closed-loop gain of such a configuration (switch in position "1") can found using the classical formula :

Acl=Hf*Ao/(1+HrAo).

(Ao: Open.loop gain; Hf:forward factor; Hr:return (feedback) factor), with

Hf = 1 - Rf/(Rf+R1)=R1/(Rf+R1) and Hr=R1/(Rf+R1).

Hence, we have (for Ao infinite):

Acl=Hf/Hr=1.

Because the feedback factor is Hr=0.5 the loop gain is Ao/2 - and that is the reason for the closed-loop bandwidth to be smaller (500 kHz) than the unity gain frequency (1 MHz). Note that the closed-loop bandwidth is the frequency where the loop gain is unity.

This bandwidth reduction is the advantage that is connected with this circuit configuration because the stability properties (phase margin) can be selected without changing the closed-loop gain.

• I also tried to simulate it, it was down - 3dB at around 500KHz Commented Jul 10, 2020 at 10:40
• OK - this confirms the theoretical considerations...fine.
– LvW
Commented Jul 10, 2020 at 10:52
• Can I get a reference on your formula(site, pdf or book)? The closed gain formula on my book and in some sites online are generalised I think as there is no forward factor described Commented Jul 16, 2020 at 11:56
• The formula simply results from the superposition theorem - applied to the diff. input of the opamp. That means: Find the diff. voltage vd between the opamps input nodes (caused by Vin resp. Vout) and set Vout=vd*Ao. Thats all.
– LvW
Commented Jul 16, 2020 at 12:21

Just think about it and draw on the voltages: -

• A is the input voltage $$\V_{IN}\$$

• B also equals $$\V_{IN}\$$

• C is made to equal $$\V_{IN}\$$ due to op-amp feedback

• D has to be also $$\V_{IN}\$$ because no current flows through $$\R_1\$$ hence, the gain is unity.

In effect, $$\R_1\$$ could be removed. And, this means the unity gain BW is 1 MHz.

• The advantage of this special form of a unity-gain amplifier is as follows: With Rf and R1 we have the freedom to vary the loop gain (and hence, the closed-loop bandwidth) without touching the closed-loop gain.This offers the chance to use a high-frequency opamp that is NOT unity-gain compensated. And the price for this advantage: The closed-loop bandwidth is smaller than in case of the classical unity-gain configuration. In the case under discussion the reduction factor is 0.5!
– LvW
Commented Jul 9, 2020 at 14:37