# Bandwidth of inverting and non-inverting op-amps

If the same op-amp is used in both inverting and non-inverting modes (with same closed loop gain using appropriate resistors), will the closed loop bandwidth of the op-amp in both cases be the same?

Now, if I assume unity gain frequency = 10 MHz, is the bandwidth for both 5 MHz? If I am correct, then why is the GBWP of the closed loop inverting op-amp less than that of the non-inverting counterpart?

• Do both circuits have the same gain? There's your answer. – Matt Young Nov 13 '15 at 13:06
• Yea they do have a gain of 2 ,So?? – Ashik Anuvar Nov 13 '15 at 13:08
• @RespawnedFluff i have updated my try as well.I am getting different bandwidth which GBP and unity gain different for inverting case (assuming single pole opamp ) – Ashik Anuvar Nov 13 '15 at 14:15
• You are correct. They are different: books.google.com/books?id=j5upBgAAQBAJ&pg=PA29 – Fizz Nov 13 '15 at 14:32
• And the same thing is said/derived in ti.com/lit/an/sloa035d/sloa035d.pdf (p.6-7) – Fizz Nov 13 '15 at 14:50

There is a simple answer: The bandwidth for the closed-loop gain is determined by the frequency where the LOOP GAIN is 0 dB. In your example circuits the loop gain is not the same - hence, the bandwidth will not be the same. The circuit with the largest loop gain (non-inverter) has the largest bandwidth.

Explanation why the Loop Gain (LG) determines bandwidth:

The denominator of the closed-loop gain formula is

$D(s) = 1 - LG$

From this, we can derive that "something" happens when $LG=1$ (0 dB). At the corresponding frequency $\omega_{o}$ we have a real pole (think of the behaviour of a first-order lowpass). And this pole gives the frequency where the 3dB-bandwidth is defined.

I should add that this is a simplified explanation; a detailed explanation involves the open-loop gain Aol and its frequency response:

$A_{CL} = \dfrac{H_{FW} \cdot A_{OL}}{1 - Hr \cdot A_{OL}}$

with $LG=Hr * A_{OL}$ and forward factor $H_{FW}$.

We can see that for low frequencies (large $LG$) and negative feedback factor ($Hr$ negative) the "1" can be neglected and the gain is

$A_{CL}= \dfrac{H_{FW}}{Hr}$ = constant.

However, for large frequencies ($A_{OL}$ and $LG$ smaller) we cannot neglect the "1". When we reach the frequency $\omega_{o}$ where $|LG|=1$ the "1" starts to dominate for larger frequencies and we can neglect the loop gain LG.

In this case the numerator $H_{FW} \cdot A_{OL}$ determines mostly the frequency response ($A_{CL}= H_{FW} \cdot A_{OL}$, approximately a first order lowpass).

Hence, the transition from the first region to the second region is at the cut-off frequency wo.

For inverter: $H_{FW}=\dfrac{-R2}{R1+R2}$

For non-inverter: $H_{FW}=1$.

• I agree that your explanation should be the reason.So does that mean we can shift the response of the opamp to a different unity gain frequency ? Also, why is bandwidth dependent on loop gain rather than closed loop gain ? – Ashik Anuvar Nov 13 '15 at 14:24
• See my extended answer. – LvW Nov 13 '15 at 14:37
• I was tempted to -1 this for lack of math markup, but I just went and added it anyway. – Adam Lawrence Nov 13 '15 at 15:30

You're are basically correct. The derivation of the formulas can be found in multiple places e.g. 1 or 2 and the books cited therein, so I'm not going to do it here; you've also done it correctly.

In a nutshell, if $f_T$ denotes the Unity Gain Frequency for the open loop opamp and $f_B$ denotes the same for a given circuit with circuit gain $G_0$, then

• for the non-inverting opamp circuit, the equation is simply $G_0 f_B = f_T$.
• for the inverting circuit, the equation is however $G_0 f_B = - f_T (1-\beta)$, where $\beta = \frac{R_1}{R_1 + R_2}$ using your notations.

What this means is that for the inverting opamp circuit the worst case is going to be $\beta=1/2$, $G_0=-1$, when you'll only get half the bandwidth of the non-inverting circuit!

And to actually apply these equations to your example[s]:

• for the non-inverting one: $f_B = f_T / 2 = 5\text{Mhz}$.
• for the inverting one: $\beta = R_1/(R_1+R_2) = 10/30 = 1/3$, so $$f_B = - \frac{f_T}{G_0}({1-\beta}) = -\frac{10}{-2}\frac{2}{3} = 3.33\text{MHz}$$

Here's an even quicker way to remember/solve this right, based on J.H. Krenz's textbook. The equality $f_B = \beta f_T$ is valid for both inverting and non-inverting opamp circuits, and $\beta$ (which is called the feedback fraction) has the same formula as above for both circuits, i.e. $\beta = \frac{R_1}{R_1 + R_2}$ where $R_2$ is the resistor in the feedback loop. However, to get a gain of 2 for the inverting amp you need a beta of 1/3 as above, while for the non-inverting circuit (of gain 2) beta will be 1/2.

• I think understand when G0*f_b=-ft*(1-B), because I run the calculations of the OP. But the result is that I don't get why is used the equality f_cl* A_cl=f0*A0*k. From the OP you have [A0/(1+A0*B)]*k=A_cl*k, while instead I thought that [A0/(1+A0*B)]*k=A_cl, so k was part of the closed loop gain, not a constant reducing the open loop A0. Is it a rule to consider always the Acl=A0/(1+A0*B) ? Is not very intuitive to me – thexeno Oct 23 '18 at 20:55

Okay, so I am pretty late to the party, but I just thought I would provide future ponderers with a single equation to solve this for either inverting or non-inverting amplifiers of this nature:

$f_B = \frac{f_t}{1+\frac{R_2}{R_1}}$

To show this with the above circuits:

Non-inverting circuit: $f_b = \frac{10\text{MHz}}{1+\frac{10}{10}} = 5\text{MHz}$

Inverting circuit: $f_b = \frac{10\text{MHz}}{1+\frac{20}{10}} = 3.33\text{MHz}$

EDIT:

The above example assumes an ideal op-amp. If you wish to find the true bandwidth of the circuit taking into consideration the effect of finite open loop gain and frequency dependence, you must consider the gain of the circuit at the -3dB point.

$G_{\text{dB}}-3\text{dB}=20\text{log}(A)$ then: $A=10^{\frac{G_{\text{dB}}-3\text{dB}}{20}}$

Thus: $f_b=\frac{f_t}{A}$

So for the non-inverting circuit: $G_{\text{dB}}=20\text{log}(2)=6\text{dB}$ Then $A=10^{\frac{3}{20}}=1.41$

Finally: $f_b=\frac{10\text{MHz}}{1.41}=7.08\text{MHz}$

Then the same is also true for the inverting circuit. Therefore, to answer your original question...yes, the two circuits will indeed have the same bandwidth. However, the bandwidth is not 5MHz, it is 7.08MHz. I hope this helps.

Yes, barring limitations in the external components. To get a rough idea of minimum bandwidth, divide the opamp's gain-bandwidth-product by the absolute value of the closed loop gain. That is the same whether inverting or non-inverting. Therefore in your example, assuming the opamp has a minimum GBP of 10 MHz, then both the circuits have a minimum bandwidth of 5 MHz.

However, you also have to think about the external components. There will always be some parasitic capacitance. To get the value computed above, the R-C low pass filters formed by any resistance and some parasitic capacitance must have a rolloff comfortably above the bandwidth you want.

To be pessimistic, assume 20 pF caps are added to ground and maybe 10 pF across components wherever they would reduce the bandwidth. For example, assume 10 pF across R2 in the second example. 10 pF and 20 kΩ have a 800 kHz roloff, so 5 MHz is way past reasonable expectations. We can work this backward and find the resistance that has 5 MHz rolloff with 10 pF, which is 3.2 kΩ. Since you'd actually be 3 dB additional down for every filter at the rolloff frequency, you want that to be at least a octave, preferably 2-3 octaves, past the frequency of interest. In this case, 1 kΩ would be a good choice for R2, with the other resistance scaled accordingly.

High bandwidth requires low impedances and costs current.

• I have updated the problem. I get different bandwidth!! – Ashik Anuvar Nov 13 '15 at 14:11
• Correct, dexter, we have different bandwidth values (see my detailed answer). – LvW Nov 13 '15 at 14:17