0
\$\begingroup\$

I am trying to make an inverting Op-amp circuit in LTSpice where I can invert an analog signal without any gain on my signal, I want the signal to stay the same but inverted. I have followed the basic inverting Op-amp circuits that I have seen in electronics text books. Inverting Op-amp example

I have followed it through using the OP27 op-amp provided in LTSpice and got results looking like this: enter image description here

I don't know how to explain what happened here but it happens with the other models of op-amps as well on LTSpice except for the ideal op-amp. Can someone explain what happened to the signal? Is it a problem with LTSpice? or is there something missing in my circuit that could solve it?

\$\endgroup\$
  • 1
    \$\begingroup\$ Change your supply from +5V to +15V and - 15V. \$\endgroup\$ – SteveSh Jan 6 at 1:23
  • 1
    \$\begingroup\$ GAin is -R2/R1 = -10. For the output to go to below ground the opamp MUST have a negative supply rail. \$\endgroup\$ – Russell McMahon Jan 6 at 2:12
5
\$\begingroup\$

OP27 is not a rail-to-rail input op-amp. Here's the relevant section of the datasheet:

enter image description here

Notice the input range is specified with +/- 15 V supplies. This means the input voltage must be (for a typical chip) at least 2.7 V above the negative rail and 2.7 V below the positive rail for the op-amp to work as expected. This is obviously impossible to achieve with a 5 V single supply.

You'll need to increase your supply voltages, and be sure to bias your input within the recommended operating range. As mentioned in comments, you will also need to have a negative supply if you want the output voltage to be able to reach below ground.

Or choose a more modern op-amp designed to work with lower supply voltage.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Thanks very much. I put the +15V and -15V to supply the op-amp and it works as intended now. I'll be sure to pay attention to this detail in the future. \$\endgroup\$ – ForeverLearningJP Jan 6 at 21:21
0
\$\begingroup\$

One problem is that since R2 is 10X the value of R1, the gain of your amplifier is also 10X. Another problem is that you are only using a single suipply. You need to bias the input of the op amp at one-half the supply voltage to keep it in its linear region. Otherwise use a negative supply.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ I think the top circuit is just an example used to get the topology. In the simulation the feedback resistors have equal values. \$\endgroup\$ – Bruce Abbott Jan 6 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.