Inverting opamp amplifier: frequency response not compatible with LTspice simulation

Question

I have a simple inverting opamp amplifier and I want to study the frequency response of the circuit and finally calculate the frequency band (or cutoff frequency). Unfortunately the model created in LTspice gives very different output from the data I collected doing the circuit manually in laboratory.

What I get is a very different cutoff frequency, but the amplification is okay. I'm gonna link to you the pictures of our frequency response with our data, and then also the circuit we use in LTspice and the output. As you can see in the LTspice simulation cutoff frequency is something like 1MHz but in our data we estimated it to be 200kHz more or less.

Answer 1

Looks like you are using a behavioural opamp model. If so, you need to adjust the gain-bandwidth (GBW) of the model to match the GBW of the TL082 which is typically 3 MHz. Depending on the breadboard system you are using, the contact capacitance can alter the operation of the actual circuit. In the simulation below, the UniversalOpamp model GBW is set to 3 MHz and 3 pF was added to the feedback capacitor which mimics the breadboard capacitance which gives a -3 dB frequency around 200 kHz.

It is better to use the SPICE model for the actual part. You can download the model from the manufacturer which is on Texas Instruments web site for the TL082. This is a PSpice model which is normally compatible with LTspice.

Answer 2

The model that you use can make a big difference in the frequency response. I don't see any number for the op-amp in the schematic, which suggests you may be using a generic op-amp model. You should be using the model for whatever real op-amp you are using in the lab.

The construction method also makes a difference, for instance if you are building the circuit on a breadboard there will be capacitance between the connector strips that will affect the response.

Answer 3

Well, notice that your transfer function is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\displaystyle\text{V}_\text{o}\left(\text{s}\right)}{\displaystyle\text{V}_\text{i}\left(\text{s}\right)}=-\frac{\displaystyle\text{R}_\text{f}}{\displaystyle\text{R}_1\left(1+\text{C}_\text{c}\text{R}_\text{ser}\text{s}\right)+\text{R}_\text{ser}}\tag1$$

So, for the amplitude plot (in dB) you get:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|_{\space\text{dB}}=20\log_{10}\left(\frac{\displaystyle\text{R}_\text{f}}{\displaystyle\sqrt{\left(\text{R}_1+\text{R}_\text{ser}\right)^2+\left(\text{C}_\text{c}\text{R}_1\text{R}_\text{ser}\omega\right)^2}}\right)\tag2$$

Using your values, we get:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|_{\space\text{dB}}=20\log_{10}\left(\frac{\displaystyle55580}{\displaystyle\sqrt{\left(5590+50\right)^2+\left(10^{-10}\cdot5590\cdot50\omega\right)^2}}\right)\tag3$$