# Bode diagram calculating for random frequency

The max Gain of an opamp is 100dB (=100000). I have a cutting frequency at 10 Hz and then the slope of my bode diagram decrease by -20dB/dec . I have unity gain at 1MHz.

For 100 Hz its increasing by 1 decade so the Gain is (100dB-20dB)= 80dB=10000. But if i want to calculate for a random value like 20kHz how should i calculate the gain?

• For frequency f the approximate gain g is g = 100 dB − 20 ∙ log(f ∕ 10 Hz), or, more precisely, this gives the asymptote for the gain... – aschipfl May 15 '19 at 11:47

## 2 Answers

The exact amplitude or gain formula for this kind of first-order low-pass filter characteristics is:

a = a_0 ∕ (1 + f ∕ f_0)


where a_0 = 100 000 is the linear gain (expressed in dB it is A_0 = 100 dB), f_0 = 10 Hz is the corner frequency, f is the variable frequency and a is the resulting gain. To express it in Decibels (dB), use this:

A = 20 ∙ log(a)


The asymptote for the gain A, which you are looking for, is:

a' = a_0 ∕ (f ∕ f_0)


or, expressed in dB:

A' = 20 ∙ log(a') = 100 dB − log(f ∕ f_0)


This lets you estimate the gain for an arbitrary frequency f that is at least about ten times greater than f_0 or more. For a frequency f that is at most less than about a tenth of f_0 the gain is approximately A_0. In case f is f_0, the gain is about A_0 − 3 dB.

A roll-off of 20 dB per decade means that the gain reduces by ten for an increase in frequency of ten. This means that at 100 Hz the gain is 10,000 (as you said) and at 200 Hz it is 5,000 or 74 dB i.e. 6 dB lower than 80 dB.

At 10 kHz the gain in decibels is 40 dB or 100 and, at 20 kHz the gain is half of this at 50 (or 34 dB).