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I have an inverter opamp with the next data :

  • \$A=-25\$
  • \$A_{ol}=20\frac{mV}{V}\$
  • Unity Gain Bandwith\$=10^{6}\$

Then asks for the cutt off frequency, so the open loop its passed to db
\$20\,\frac{V}{mV}=20log\frac{20}{0.001}=86.02059\,db\$

\$86.020=20log(\frac{Vo}{Vi})\$

\$\frac{86.020}{20}=log(\frac{Vo}{Vi})\$

\$4.301=log(\frac{Vo}{Vi})\$

So the \$A_{ol}=10^{4.301}\$

Usint the GBWP \$10^{4.301}f_{3db}=10^{6}\$

\$f_{3db}=\frac{10^{6}}{10^{4.301}}=50.0034\,Hz\$
Later its asked the values of output voltage if the frequencies are \$0.25f_{3db}\$ and \$5f_{3db}\$
so calculating the gains
\$0.25f_{3db}=(0.25)50=12.5\$

\$Af_{3db}=GBWP\$

\$A=\frac{GBWP}{f_{3db}}=\frac{10^{6}}{12.5}=80000\$

\$V_{o}=80000V_{in}\$

\$5f_{3db}=5(50)=250\$
and
\$A=\frac{GBWP}{f_{3db}}=\frac{10^{6}}{250}=4000\$

\$V_{o}=4000V_{in}\$
Is this the way to go? the results make me sense since its stated that before the cuttoff frequency the gains are very high but the bandwith its too narrow and over the cuttoff the gain is less but the band its wider. But then why Im supplyed with the closed loop gain?

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    \$\begingroup\$ "Then asks for the cutoff frequency". Are you sure it's not asking for the closed loop cutoff frequency rather than what you have taken it to mean - the frequency where the open loop gain starts to roll off which is a much lower frequency. \$\endgroup\$
    – James
    Apr 22, 2021 at 11:05
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    \$\begingroup\$ Perhaps this is the hint, as it is redacted I cant really tell. \$\endgroup\$
    – avelardo
    Apr 22, 2021 at 19:11
  • \$\begingroup\$ The part about it says What are the 3𝑑𝐵 frequencies for both the opamp and the inverter amplifier?/¿Cuáles son las frecuencias de 3𝑑𝐵 tanto del op-amp como del amplificador inversor? So I wass thinking it is saying the cuttoff frequency. \$\endgroup\$
    – avelardo
    Apr 22, 2021 at 19:28

1 Answer 1

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You calculated \$A_{ol}\$ well. You also know that the slope is 20 dB/dec, so with a \$G_{BW}\$ of 1 MHz, the open loop amplification will be:

   10 (20 dB) @ 100 kHz
  100 (40 dB) @  10 kHz
 1000 (60 dB) @   1 kHz
10000 (80 dB) @ 100  Hz

At 10 Hz it should be 100000 (100 dB), but Aol = 86 dB = 20000 (You wrote mV/V first time, instead of V/mV). That means the gain is double that at 100 Hz, which means the frequency is half of 100 Hz, or 50 Hz. Or, in simpler terms:

$$\dfrac{f_{max}}{A_{ol}}=\dfrac{1\;\mathrm{MHz}}{20000}=50\;\mathrm{Hz}\tag{1}$$

Which means your result is right (save roundings), just that there are fewer steps. Then you ask for:

the values of output voltage if the frequencies are \$0.25f_{3\,dB}\$ and \$5f_{3\,dB}\$

but you don't specify for what input. Given your calculations afterwards, I'll assume you didn't mean the value of the output voltage, but the gain. Then, for the first one it's \$A_{ol}\$, because it can't amplify more than 86 dB, and for the second one it's:

$$20\log_{10}{\Biggl(A_{ol}\dfrac{1}{5}\Biggr)}=20\log_{10}{\Biggl(\dfrac{20000}{5}\Biggr)}\approx 72\;\mathrm{dB}\quad(4000)\tag{2}$$

So, you calculated it correctly, just that you went a longer way after it.

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    \$\begingroup\$ It was ny bad the mV/V, indeed; I dont specify the input because it isnt given none, so I was thinking the same, it only can be represented, not a value specified. Thanks the real opamp gives me a little "noise" \$\endgroup\$
    – avelardo
    Apr 22, 2021 at 19:23

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