0
\$\begingroup\$

I am trying to simulate in LTSPICE a circuit containing two op amps cascaded in series (so that individual gains will be multiplied) but for some reason the gain of the first stage is higher than the last stage, but when I remove the power pins of the last stage, the circuit behaves normally. What is happening?

I am trying to do a frequency sweep and have tried two different op amps (TL072, NE5532) but they both behave abnormally. I don't know what i am doing wrong.

I got my op amp spice models from TI.

enter image description here

\$\endgroup\$

1 Answer 1

3
\$\begingroup\$

There is no path to ground for the input bias current in the non-inverting input of the second op-amp.

Add a 51k resistor between the non-inverting input of the second op-amp and ground.

\$\endgroup\$
4
  • \$\begingroup\$ It worked! Thanks!.I cant believe I haven't tried that yet. Can you please explain why that's necessary even though the two op amps are tied to the same supply rails? \$\endgroup\$
    – Dave L
    Apr 21, 2021 at 5:01
  • \$\begingroup\$ There is a small dc current called the input bias current into (or possibly out of) both inputs on op-amps. A circuit designer must always allow for this bias current to flow in order for the input stage of the op amp to function correctly. You had only a capacitor connected to the non-inverting input and because the input bias current is a dc signal it could not pass through the capacitor (capacitors block dc) and so the op amp's input stage could not function correctly. Adding the resistor enabled the input bias current to flow which enables the correct operation of the op amp's input stage. \$\endgroup\$
    – user173271
    Apr 21, 2021 at 5:15
  • 1
    \$\begingroup\$ @DaveL Give this a read: analog.com/media/en/technical-documentation/application-notes/… \$\endgroup\$
    – Ste Kulov
    Apr 21, 2021 at 5:16
  • \$\begingroup\$ Thanks, I always forget about those bias currents. \$\endgroup\$
    – Dave L
    Apr 21, 2021 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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