0
\$\begingroup\$

I have a weird but annoying problem. I have an inverting low pass filter as in the figure below:

schematic

simulate this circuit – Schematic created using CircuitLab

I want this circuit to act as a RL circuit and in simulations it does. So when giving 5V as an input I would expect to have -7.12V at the output, except that I don't. The output voltage decreases in the beginning as it should but once of a sudden it stops at -4.3V or so and decreases slowly but when I say slowly I mean 0.001V every second so not even remarkable. When I remove the capacitor everything is okay, I get my -7.12V so I suppose the problem has something to do with the capacitor.

Anyone experienced a similar problem or knows what's going on?

Thanks!

\$\endgroup\$
  • \$\begingroup\$ how do the supply voltages of the TL081 look like? \$\endgroup\$ – Nils Pipenbrinck Jun 4 '16 at 13:22
  • \$\begingroup\$ They're +/-15V if that's what you mean. Nothing unusual about that. By the way, I am not using TL081 but LM741, I've \$\endgroup\$ – MarkoP Jun 4 '16 at 13:37
  • \$\begingroup\$ The turnover frequency of your filter is 15mHz. It won't pass anything except DC in practice and even then the time constant is enormous, ten minute or so, and therefore its settling time as well. Is that really the frequency you're aiming at? \$\endgroup\$ – user207421 Jun 4 '16 at 19:09
  • \$\begingroup\$ Yes it is. The circuit should emulate a magnet with a really small R and L, few mOhms and mHs. \$\endgroup\$ – MarkoP Jun 5 '16 at 8:59
  • \$\begingroup\$ @MarkoP I am guessing you are using a polarized capacitor (e.g. tantalum or electrolytic) which is probably rectifying. If you are, use a non polarized cap. \$\endgroup\$ – user5108_Dan Jun 6 '16 at 12:33
1
\$\begingroup\$

I think you have two problems here:

  • Your choice of the OpAmp:

The 741 is ancient and has very bad specs compared to modern parts. Most notably the input impedance is 300k worst case. That's in the same ballpark of the output impedance of your voltage divider at the negative input terminal. This, along with the input bias current will cause a huge voltage error to develop.

The remedy: Lower R1 and R2 by a factor of 10 (or so) and adjust C1 accordingly. Or use a better opamp.

  • The leakage current of your capacitor:

If you've used an electrolytic capacitor for C1 there will always be some leakage current. You can think of this leakage as a resistor in parallel to C1. How high this leakage is depends on the capacitor, but it can be surprisingly low.

Now take a look at your circuit: R2 is of a relative high value, and the output voltage is directly defined by it: What happens if you put a second (leakage simulating) resistor in parallel with it: The total resistance from output to inverting input will drop and so will the gain. You'll get a lower voltage at the output. Just as you have measured.

Again the remedy is to change parts: Use a capacitor with a very low guaranteed leakage or use a capacitor type (film) that has negligible leakage and use a capacitance multiplier circuit to scale it up to the value you want.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the reply! I can surely chnage the OpAmp but what would you recommend me? Changing R1 and R2 would be a bit trickier because then my C1 will increase and it can sometime be 2mF and I dont' really want to something that big on my board. I am using electrolytic capacitors and they can get bit. Even now sometimes I need to mount 2-3 capacitors to obtain a value of C1 I want. Problem is that this is part of a bigger circuit, something like bellow: imgur.com/zb8tHHb In reality there are more R2 resistors and a switch between the feedback path and the inverting OpAmp. \$\endgroup\$ – MarkoP Jun 5 '16 at 8:57
  • \$\begingroup\$ Also, could it be that I need to discharge my capacitor? \$\endgroup\$ – MarkoP Jun 5 '16 at 9:11

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.