1
\$\begingroup\$

I have a varying source voltage for maximum of 15 V and am using an error amplifier to maintain it using an integrator. Reference voltage is of 5V. I want a time constant of 100 msec with a capacitor of 1 uF, so I'll need a resistance of 100 kOhms. What combination of resistors would equal that resistance? enter image description here

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Maybe it's just me, but I'm confused. Are you trying to create a variable output power supply? You write, "... a varying source voltage for maximum of 15 V..." What does that mean? Sounds kind of like a variable output to me. And since this is an "error amplifier" I have to imagine this is part of a closed loop -- the rest of which I cannot see nor does a single obvious one pop to mind where all other ideas are eliminated by what you wrote. So I guess I need to see and hear more about this system. And I certainly don't feel comfortable trying to answer your question at this time. \$\endgroup\$
    – jonk
    Sep 23 at 1:50
  • 1
    \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$
    – Community Bot
    Sep 23 at 4:49
  • \$\begingroup\$ The time constant would be infinite for the above circuit if the opamp is ideal. How would u get a finite time constant? \$\endgroup\$
    – prashanth
    Sep 23 at 5:36

1 Answer 1

0
\$\begingroup\$

Neglecting the problem about whether the time constant of 100 ms can be achieved with what you have shown and what you propose to add to the circuit (hint, the integrator you have drawn will simply ramp indefinitely), it's possible to answer the question of what combination of R1, R2 and R3 will yield an effective output resistance of 100 kΩ, and a voltage division of 3:1.

The effective output resistance of the R1/R2 voltage divider is their parallel combination. If this is not 100 kΩ, then R3 can be used to increase the output resistance.

The simplest solution is R1 = 300 kΩ, R2 = 150 kΩ, with R3 = 0.

There is an infinity of other solutions with a finite R3, as long as R3 < 100 kΩ

  • R1 = 3 x (100k-R3)
  • R2 = 1.5 x (100k-R3)

We normally describe a pure integrator as having a gain, rather than a time constant. With an input resistance of 100 kΩ, the gain is 10 V/s per V, or 10/s. When the integrator is used in a feedback loop, it's usually more useful to express the gain as the frequency at which it has unity gain, which is 10 rad/s, or roughly 1.6 Hz.

\$\endgroup\$

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.