Op-amp calculating bandwidth frequency

Thread Starter

Ross Satchell

Joined Jan 2, 2017
I'm working on a homework problem and I'm stumped.
The question is:
A measurement of the open loop gain of an internally compensated op-amp at very low frequencies shows it to be 98dB; at 100kHz this shows it is 40dB. Estimate values for Ao, fb, ft:
where: Ao is open loop gain; fb is the break frequency; ft is the unity gain frequency.

From what I have learned:
The open loop gain is 98dB.

Now I figured I would use the point slope formula to get the break frequency.
(y2-y1) / (x2-x1) = m
rearranging gives me:
x1 = x2 - (y2-y1)/m

where m = -20dB/dec, y2 = 98dB, y1 = 40dB, x2 = 100kHz
Now I know I have to be careful because the x-axis is logarithmic, so x2 would be 10^5 (100kHz)

The second part of the equation => (y2-y1)/m gives units of dB/(dB/decade) => decades.
So the first part of the equation (x2) has to be in decades as well.
So then x2 = 100kHz = 5 decades

So then the equation should be:
x1 = 5 - (98-40)/(-20)
x1 = 5 - (-2.9) = 7.9 decades which would be 10^(7.9) which is definitely WRONG as it should be somewhere down near 10Hz (10^1).

I can't figure out where I'm going wrong, I spoke to the instructor yesterday and he said using the point slope approach was correct. So I'm doing something wrong but I can't figure it out.



Joined Feb 17, 2009
But the exact solution you will get only if you use this equation:

\(\large A = \frac{ A_{O}}{\sqrt{1 + (\frac{F}{F_b})^2}}\)

Thread Starter

Ross Satchell

Joined Jan 2, 2017
Thanks for your reply Jony130.

Interestingly I got the same answer using both the point slope formula and the equation you provided.

The class lectures had indicated that the gain starts to roll off fairly rapidly around 10Hz, so when I got a value an order of magnitude larger I assumed it was wrong.