0
\$\begingroup\$

I'm making a simple amp > voltage doubler rectifier circuit to drive an lm3916 meter circuit and when simming to choose between JFET, BJT or OpAmp (just for my learning, I'm looking to a Jfet in the end I think) I have this weird behaviour I don't understand.

The basic circuit works fine Basic circuit

I go to add a blocking cap for real world use

enter image description here

And the input goes low to -7v or so, rendering the output unuseable. N001 is OpAmp Non-Inv input

But add a resistor to the input stage

enter image description here

And it all works again.. What is causing this? I don't know why the coupling cap causes the circuit to 'not work' without an input impedance of some sort, separate to the OpAmps near infinite impedance. I simmed many different values of R3 and as long as there's 'some resistance' it works, but given what I know about OpAmps, and the fact that it works without a coupling cap I have no idea what causes the fail when I add the cap..

Any advice well appreciated!

\$\endgroup\$
3
  • \$\begingroup\$ All Op Amps have specs for Vcm, I_in that must be met. AC coupling won't work as it is not self biased \$\endgroup\$ – Tony Stewart EE75 Feb 22 at 0:06
  • \$\begingroup\$ There is nothing to define the DC at that input so it will settle wherever parasitic components allow. Consider what the input bias current of the opamp does... \$\endgroup\$ – user_1818839 Feb 22 at 13:19
  • \$\begingroup\$ Ok, thanks very much, I think I get it more or less. I can visualise, it, with a positive bias in a given circuit you are clearly defining, eg. 4.5v at the given input. For this circuit it should be 0v but without defining that it is just floating and free to settle somewhere the circuit allows. I'm not 100% clear on the mechanism that allows the leakage to pull the input to one of it's rails but maybe as thats because I'm not really clear on exactly how the leakage manifests in real world circuits yet! Thanks again. \$\endgroup\$ – OwenM Feb 22 at 21:54
4
\$\begingroup\$

You need to some DC path to allow the op-amp's bias or leakage current to go to ground. Without it the input capacitor charges to one of the rail voltages.

The current is small but it's there.

\$\endgroup\$
7
  • \$\begingroup\$ Thanks very much, I think I get 90% of it! I'm not certain on how the input when floating can move to one of it's rails, but as mentioned above, I'm only aware 'in principle' of opamp leakage, and not how it effect real circuits, yet, so I'm sure the remaining 10% will come in time! Thanks. \$\endgroup\$ – OwenM Feb 22 at 21:56
  • \$\begingroup\$ It would be rather like an RC circuit charging up. If you have a capacitor connected to ground and you feed it a little current it will charge up or if you drain a little current it will discharge. The op-amps aren't perfect. If you have a look at the datasheet you should find some information on the input bias currents or similar. \$\endgroup\$ – Transistor Feb 22 at 22:00
  • \$\begingroup\$ Hmm, cool, thanks! You've just led me toward a little lightbulb moment that the leakage in question here is simpler than I had assumed! \$\endgroup\$ – OwenM Feb 22 at 22:14
  • \$\begingroup\$ Good. In general look for a DC path for each input. This can be to ground or through the feedback circuit to the op-amp output. If you have a look for op-amp integrator you will find that this is a big issue for them. The other thing that the bias current does is it can introduce an input offset on one input that gets amplified on the output. For this situation you might see that the non-inverting input that you expect would be connected to ground has a resistor in that connection and it is set to the combined parallel connection of the input and feedback resistors on the inverting input. \$\endgroup\$ – Transistor Feb 22 at 22:19
  • \$\begingroup\$ Will do, I THINK I can see how that could be a big issue in the basic integrator circuit, with the feedback incorporating a capacitor rather than a resistor any leakage on the Inverting Input is stuck floating in a similar way to my middle circuit, unless we build in a reference for the Inverting Input elsewhere? Yup, I get the principle of adding the resistor on the non-inverting input to compensate, or balance, I think. So far I have just made guitar pedal circuits with opamps as a buffer, where you just remove the desired bias after the op amp! So new territory for me, but getting there! \$\endgroup\$ – OwenM Feb 22 at 22:49

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.