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Original circuit

From the circuit above, this is what I have to do: " Connect the circuit with R3 equal to 470 kΩ. Set the input frequency to 100 Hz. Adjust the signal level to get an output of 5 Vp-p. Record gain and upper cut off frequency. Multiply gain and bandwidth."

What I did was this:

My circuit

I used a function generator and made the frequency to 100Hz. As for amplitude, I chose 1.25 Vp so that in PR1 it shows 5vp-p (I drew a red arrow and labelled it as 1 to make it easier to locate). This is what it meant by "Adjust the signal level to get an output of 5Vp-p" right?

Next, in order to get the gain. I placed a probe (PR2) at the output and made PR1 as the reference. Then in the list of values, there's a V(gain_AC) which is equal to 5.66 (Again, I drew a red arrow and labelled it as 2). Is this the gain of the circuit already?

Lastly, to get the cut off frequency I did an ac sweep and located where it drops by 3db (I already changed the vertical scale to decibels). The result is the following:

AC sweep

Is the value that I highlighted which is 2.1348k the upper cut-off frequency I'm looking for? And, this is also the bandwidth right?

To sum it up, I basically have three questions:

  1. Was I able to adjust the signal level properly to get an output of 5 Vp-p?
  2. Is the V(gain_AC) shown by the probe, the actual gain of the circuit?
  3. In the AC sweep I did, is the value I highlighted the upper cut-off frequency and the bandwidth?

If you need more information, please tell me. Thank you so much for your help!

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Was I able to adjust the signal level properly to get an output of 5 Vp-p?

No.
You have probed circuit output properly. It shows a voltage amplitude of 28.2V peak-to-peak. However, this measurement is not to be trusted. Had you actually looked at the waveform, it would not be sinusoidal. It would look more like a square wave: when you apply a sine wave at amplifier input, you should see a sinusoidal wave shape at amplifier output.
A too-large input signal asks of the amplifier to generate an output amplitude that exceeds its DC supply voltages. Since your DC supply range is 30V, that 28.2V p-p output signal is limited to a little less than this range (output can only approach, not exceed DC supply).

Is the V(gain_AC) shown by the probe, the actual gain of the circuit?

No
As stated above, a sine wave applied should yield a sinusoidal wave at amplifier output. This amplifier is driven with an input amplitude that is too high. A lower input amplitude should be set on the function generator. Set it low enough so that output amplitude eventually drops to 5V p-p.
As a check, if you reduce input amplitude in half, output amplitude should also reduce by half - this check indicates that the amplifier is running linearly.

In the AC sweep I did, is the value I highlighted the upper cut-off frequency and the bandwidth?

Your method seems correct. Once you get the amplifier running linearly, do this again - you'll likely get a different result.

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  • \$\begingroup\$ Thank you for this! I already adjusted it. But just to be sure, the V(gain_AC) after it was adjusted, is the gain right? \$\endgroup\$ – Rose Sep 19 at 15:50
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    \$\begingroup\$ V(gain_AC) should end up having a magnitude a little under 470, not 5.66. The reported gain of 5.66 is so low because the calculation does not recognize that the output wave shape is not sinusoidal, but is limited within the DC supply voltages of +15V and -15V. Trust these box numbers warily: supplement them with your own calculation using other methods. \$\endgroup\$ – glen_geek Sep 19 at 16:36

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