# 3dB Frequency of non-inverting amplifier

The question is

The op-amp is described to have that transfer function and I am supposed to find the 3dB frequency of the entire amplifier. Here is what I did

I used the fact that 3dB frequency occurs at a gain reduced by a factor of 1/sqrt(2). I'm getting a wrong answer.

• Fo = Wo/2pi = 100Hz and GBP =Fo*Ao = 10MHz, so F(-3db) = 10MHz/(1+ 50k/10k) = 10MHz/6 = 1.666MHz – G36 Mar 25 '18 at 18:05
• How come we are using open loop gain in this closed loop circuit for the calculation? Also what does Wo stand for? – AlfroJang80 Mar 25 '18 at 18:16
• en.wikipedia.org/wiki/Gain%E2%80%93bandwidth_product – G36 Mar 25 '18 at 18:19
• Ah. I think I understand now. GBP is consistent between open loop gain and closed loop gain with negative feedback. – AlfroJang80 Mar 25 '18 at 19:20

The story is in a short form written in the comment of user @G36. I am afraid that understanding it needs a moderate amount of experience. That explanation uses the concept of gain-bandwidth product as a shortcut. You should learn it, too, if you are going to make estimations fast when the used opamp has typical dominant pole frequency response like yours.

A hard work approach:

You have driven out of the road when deriving a formula for the transfer function of the amplifier circuit. At least the beginning equation of your third image needs square root over the whole right half. There can be more errors.

Here's the correct result and how to use it:

There is found that the closed loop DC voltage gain is 6. The denominator of the closed loop transfer function should have absolute value sqrt(2) at -3dB frequency. The denominator has real part about =1, so the whole denominator should be = 1+j. That happens when f=1667kHz.

Here is a zoomed-in plot.

Notice the gain-error is huge: 45%, at F3dB.