I have a following circuit with two op amps in series.
What is the gain in dB of this circuit at 13 MHz relative to its DC gain?
The 13MHz is well out of the normal operation region of the op amp.
I believe that the DC gain is
$$ 20\log_{10}\bigg(\frac{24}{180}\bigg)+20\log_{10}\bigg(1+\frac{2.7}{7.5}\bigg) $$
However, I am not quite sure how to find the loss for high frequencies.
I have tried calculating it using following equation
$$BW_{xdB} = BW_{3dB} \sqrt{10^{\frac{x}{10}}-1}$$
\$BW_{xdB}\$ is bandwidth of x dB and \$BW_{3dB}\$ is the 3 dB bandwidth. I have assumed that I could find the loss using this equation in the following form
$$ x = 10\log_{10}\bigg(\Big(\frac{BW_{xdB}}{BW_{3dB}}\Big)^2+1\bigg)$$
but it is wrong.
I have also tried to linear approximation of the loss for high frequencies on the log frequency:
$$ -3\:\mathrm{dB} -20\bigg(\log_{10}(f)-\log_{10}(BW_{3dB})\bigg)\:\mathrm{dB}$$
where the \$-3\:\mathrm{dB}\$ is the loss at the end of the 3dB bandwidth and then the loss is 20 dB per decade. But this was also wrong approach.
The equations above are always only for one amplifier, the total loss should be just sum of individual losses for each amplifier.
The gain at 13MHz relative to the DC gain should be -16 dB, but I do not know how to reach that value.
Best Answer
Your DC gain is correct. Now, given the gain bandwidth of 7MHz, the 3dB bandwidth of the first stage is 7/(1+24/180) MHz or 6.18 MHz The second stage bandwidth is 7/(1+2.7/7.5) MHz or 5.15MHz. Now a single pole transfer function for each stage is Adc/(1+s/wo) where wo = 2*pi*BW. So the gain at 13MHz relative to the DC gain is 1 over the magnitude of each denominator multiplied together, with 2*pi*13MHz for j*w, and each bandwidth (calculated above) for wo. I get .429 and .368 for each stage, multiplied together that's .158 or -16.034dB.
Note that you use the "noise gain" or non-inverting equivalent gain to compute the bandwidth for the first stage because the gain-bandwidth is always referred to the non-inverting gain.