0
\$\begingroup\$

I have been experimenting with an opamp as a noninverting amplifier for a Piezoelectric vibration sensor 0-1Khz expected. I am only interested in the positive portion of the signal as an average hence the big Cap on the output signal. My problem is bias voltage injection. When the gain network is tied to the ground the circuit works as expected. but I have since tried to inject a 0.5v bias voltage via a divider network but I'm not getting the response expected.

I suspect it's the divider network because it's in the same order of magnitude as the gain network but I don't have larger values to test at the moment so I hope someone can shed some light on the issue. I feel like I'm missing something obvious. The circuit is supplied from a bench 3.0V supply so it's trustworthy. enter image description here

The Opamp used has a 1.3mv offset and with an x101 gain, I expect to see 131mV on the output with no input signal and am currently getting 129mv so it seems accurate, when I apply a steady-state 7.5mv I get 760mV out which is also expected. but when I inject the biased voltage the output is stuck at a steady state of 35mV.

EDIT*https://www.ti.com/lit/ds/symlink/lmv321-n.pdf?ts=1626621566114&ref_url=https%253A%252F%252Fwww.ti.com%252Fstore%252Fti%252Fen%252Fp%252Fproduct%252F%253Fp%253DLMV321M5X%252FNOPB

So as an update thanks to the comments from members I have found a way that seems to work to provide Bias on a Noninverting amplifier. The Bias can be driven from the same supply rail on the condition that the output is buffered. in addition, the Piezo Sensor Negative ( if that's the correct term) has to be connected to the virtual ground of the same bias feedback path and not ground. Still need to test it on real circuit but simulation seems to be accurate enter image description here

\$\endgroup\$
2
1
\$\begingroup\$

but I have since tried to inject a 0.5v bias voltage via a divider network but I'm not getting the response expected.

That's because you haven't accounted for -Vin being a virtual earth and it connects to the voltage divider (2 k and 10k) via the input resistor R3. R3 in effect, is partially grounding that node between the 2k and 10k (that otherwise would produce 0.5 volts): -

enter image description here

You need to factor it in. Maybe get rid of the 2 k and reduce the 10k to 5k.

However, that then upsets your piezo amplification gain (as it also does in your current circuit) so, another method should be found that both gives you the correct offset and doesn't alter the gain.

I don't have larger values to test at the moment so I hope someone can shed some light on the issue.

No, larger values are even worse. If you want to try something change 2k to 200R and 10k to 1k. If you can live with the smaller error then that's a viable solution.

I feel like I'm missing something obvious.

Hopefully, not any more.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you Andy. ill try that to confirm, I was trying to stay away from the lower values to reduce leakage current the voltage devider, might need to find a different way to inject Bias voltage them. \$\endgroup\$
    – Rustie0125
    Jul 18 at 15:47
1
\$\begingroup\$

An op amp works by driving its output to a voltage which makes the two inputs be at almost the same voltage as each other.

With a positive bias voltage at the left hand side of R3 the op amp is trying to drive its output negative with respect to ground in a desperate and failing attempt to make the inputs have the same voltage level.

The output can't actually go negative because there is no negative supply voltage and so it limits at the negative saturation limit. I haven't looked at the data sheet but the negative saturation limit is probably a few 10s of mV above ground when the op amp's negative supply rail is 0V.

Slowly reduce the 2k resistor and when it reaches the optimum value you will see the op amp's output just start to rise (leave saturation).

\$\endgroup\$
6
  • \$\begingroup\$ Hi James, that is my understanding of an opamp aswell, Iv been referencing a white paper regarding the matter from ocw.mit.edu/courses/media-arts-and-sciences/… but re-reading this document and the comments above its clear that the Negative rail is missing. which leads me to more questions. I don't want to run split rail supplies. \$\endgroup\$
    – Rustie0125
    Jul 18 at 16:10
  • \$\begingroup\$ @Rustie0125 That's a very good document, I printed a copy of that one off for my own learning file several years ago. A smaller bias voltage (less than 0.5V) will remove most of the offset voltage but (and I haven't looked at the data sheet) if the op amp saturates at a few 10s of mV above 0V then you won't remove it all without a negative rail. Also, when an input signal is received and the output rises it will alter the bias voltage slightly. Keep the bias resistor values low to minimise this or buffer the potential divider. If you insert a buffer you'll probably need a negative rail. \$\endgroup\$
    – James
    Jul 18 at 16:22
  • 1
    \$\begingroup\$ @Rustie0125 One approach to this would be to apply a fixed input voltage of say 5mV and then adjust the bias voltage until the correct output voltage is achieved (remove the offset). This then lets you remove the offset whilst avoiding saturation problems but, if I'm correct about the saturation limit, they'll still be a saturation problem at very low output voltages. \$\endgroup\$
    – James
    Jul 18 at 16:36
  • 1
    \$\begingroup\$ @Rustie0125 One problem with this approach is that when you alter the resistance in the unbuffered bias divider it alters the bias divider's Thevenin equivalent resistance which is in series with R3 which will then alter the gain and the expected output voltage. Buffering the bias divider removes this problem by isolating the bias divider from R3. \$\endgroup\$
    – James
    Jul 18 at 17:48
  • \$\begingroup\$ Hi James, I just redrew the circuit in simulation software and suggested what you said, or at least my understanding of it. inserted a buffer amplifier and injected the output on the center of the gain network. This does not work as expected as the output goes static negative. if I remove the Gnd of R3 and Inject the 0.5v instead fo the ground I get the same result. \$\endgroup\$
    – Rustie0125
    Jul 18 at 18:08

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.