Gain Bandwidth Product

Thread Starter


Joined Jun 18, 2015
Greetings! I have a question about the circuit below. I need help with the last question (d). I don't understand how to calculate the output voltage when the frequency starts changing. My answer does not match the solution. Namely, for part (i) Vin = 10sin(t) the output voltage amplitude is just 10 * 100 = 1000 (that answer is good). However, I'm confused on which formula to use for the next two inputs with higher input frequencies.
If I apply the GBP = (A)*(fc) my answer is wrong. Please help me understand this better.


Joined Nov 30, 2010
I've been thinking about this for 3 hours because I know how to get the answers, but I don't know how to put them in your terms. It's been way too long since I was in school.

What is the, "wrong" answer that you first derived?
What is, "t"?
What is, "fc"?
Isn't fc the -3db point on the frequency curve?
Why do you think Gain x Bandwidth = A x fc?

This line of questioning looks like my ignorance talking, but answering my questions might lead you to a better method.


Joined Feb 24, 2006
The GBW identifies a point on the frequency response where the gain is one and the bandwidth is GBW. This implies that for all frequencies that are less than or equal to GBW, the gain will be greater than or equal to one.

Thread Starter


Joined Jun 18, 2015
Ok, let's start with the transfer function and work from there. So, for each section of our filter we have the same 5MHz GBP of the op-amp itself. That means calculating each section's gain and dividing 5MHz by the gain gives that section's corner frequency, wc1, wc2, and wc3. In question (b) it asks for the lowest frequency pole (-3db point) which is wc3 = 250kHz. Does that determine the corner frequency of the entire filter/circuit, the fc or wc point (-3db)? Then, for question (c) I plugged in 100 for Acircuit, 1 for H(s), and corresponding wc1, wc2, wc3 in radians, and solved for s, the frequency where Vout = Vin. Another question which arose is: Why doesn't Vout = Vin occur at 5MHz? Isn't GBP the definition of where Vout = Vin? So now I'm confused with question (d) and with the formulas and definitions and need a nudge in the right direction.



Joined Feb 17, 2009
If OpAmp has GBP = 5MHz this means that if input frequency is 5MHz the OpAmp open loop gain is equal to 1V/V.
Vout = Vin only when closed loop gain is equal to 1V/V. So you need to find this frequency.
As for the -3db
For the first op amp -3db is equal to
Fc1 = 5Mhz/5 = 1MHz
Fc2 = 5MHz/1 = 5MHz
Fc3 = 5Mhz/20 = 250kHz So form this we can tell that -3db of the entire circuit is 250kHz


Joined Mar 14, 2008
Why would you think Vout=Vin would occur at 5MHz?
Where that point occurs depends upon the relative gain and rolloff of the three stages.

Note also that the frequency response of an inverting amplifier is different than for a non-inverting amplifier of the same closed-loop gain.
For an inverting amp the BW = GBW / (Av+1). This is very significant at low gains since, for example, the bandwidth of a inverting amp with a gain of 1 is 1/2 the bandwidth of a non-inverting follower amp.
And this means the bandwidth of the OP's third stage is (5MHz / 21) = 238.1kHz.
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Joined Oct 2, 2009
Everything you need to solve the problem you can learn from this:


Top pane shows the open-loop gain of your 5MHz GBW opamp vs freq.
At what freq does the gain become 1?
What is the gain at a low frequency? (not specified in your probem?)
Why is the corner at about 50Hz?

Bot pane shows the closed-loop gain vs frequency (with feedback) as a function of gain setting resistor R2. Where is the corner frequency at each gain? Why?


Joined Jun 17, 2014

A useful formula is:

w1 is the gain bandwidth product (like 1000000 for example),
G is the set gain of the stage (like 10 for example),
w is the operating frequency.

Recognizing that this is just an estimate as GBW products are anyway, there may be other ways to estimate. Take a good look at Mikes post for some good simulations.

Op amps act like filters and we all know that several filters in tandem change the gain at the operating frequency. What the above formula tells us is that for three stages in a row the gain bandwidth may be reduced by about 0.51 approximately over what it would be for one single stage, but here they just seem to want you to find the gain assuming that the slowest stage sets the total gain, not the true tandem combination.

Thread Starter


Joined Jun 18, 2015
Thank you for your support! So far, I've understood that the open-loop op-amp itself has an LPF characteristic and its unity gain occurs at BW = 5MHz. Now, for each stage we have its own LPF characteristic and BW because of closed-loop. Closed-loop decreases the gain, thereby increasing the BW. Also, the GBP of the circuit is 4.2 MHz according to the solution, although I got 3.2 MHz. The corner frequency of the entire circuit is 250kHz, Jony130 also confirmed on that. So, I'm still ambiguous on how to get a gain of 85 for w = 10^6 and 3E-5 for w = 10^9 as in the solutions?



Joined Oct 2, 2009
Does anyone have an idea on how we calculate gain of 85 and 3E-5?
I make the low freq gain to be -100. The frequency at which the gain is -50 (-6db) is ~375kHz. The frequency at which the gain is -70.7 (-3db) is ~225kHz.
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Thread Starter


Joined Jun 18, 2015
MikeML, what do you use as your GBP, 5MHz? I'm just confused on how to crank out the math for this?
I.e. given that we have 10^6 rad/s and 10^9 rad/s as input what value for GBP do we use to find the gain?