I'm working on a project based on this publication. I've built the circuit but something is not working.

I've tried checking the voltage with a multimeter (I do not have an oscilloscope at hand) and there is voltage before the band-pass filter (1-1.1 V). As expected, there is no DC voltage after the band-pass filter (black arrow).

But after the last op-amp the voltage should be in the 0-5 V range, according to the publication, which also says that the last op-amp introduces a 2.5 V DC offset to the signal. I think I should be able to measure about 2.5 V at the red arrow, but I consistently measure -0.6 V. Why? Also, I am using a -12 V source with a 72Kohm resistor,but this shouldn't matter I suppose (-12/72 * 15 = -2.5)?


the circuit

  • 2
    \$\begingroup\$ I agree with your calculation at the end. If you are using \$-12\:\textrm{V}\$ instead and changed the resistor to \$72\:\textrm{k}\Omega\$, then the same bias would seem to be added as in the article's circuit. However, you are using a multimeter, which may be ill-suited. Remove the \$1.5\:\textrm{k}\Omega\$ resistor between the last two opamps and measure the output to the ADC. Should read \$2.5\:\textrm{V}\$ then. Check that it does. (Cool article, by the way. Thanks!) \$\endgroup\$ – jonk Nov 13 '17 at 6:40
  • \$\begingroup\$ Thanks, I'll try that later today. Also, I noticed I'm actually missing the antiparallel diodes at the beginning of the circuit. \$\endgroup\$ – ClayP Nov 13 '17 at 7:35
  • \$\begingroup\$ Don't worry about getting 2.5 at the red arrow until you have verified that you have 0 at the black. And are you sure you are using both plus and minus 12 for your op amp supplies? \$\endgroup\$ – WhatRoughBeast Nov 13 '17 at 14:27
  • \$\begingroup\$ @jonk I isolated the last op-amp, as you suggested, but the voltage is still -0.6 V. I thought the LT1007 could be defective, but when I tried with another one I got the same result. I'm quite lost at this point. Any chance I could use the V+ rail (pin 7) to adjust the offset? \$\endgroup\$ – ClayP Nov 13 '17 at 14:28
  • \$\begingroup\$ @WhatRoughBeast The voltage at the black arrow is effectively 0. \$\endgroup\$ – ClayP Nov 13 '17 at 14:29

The fact that you are using ground as the positive supply is your problem. On the one hand, an op amp cannot possibly produce an output higher than the + power supply nor lower than the -, so hoping for 2.5 from your circuit is a no-go. On the other, in general it can't get all that close to those supplies either. There are exceptions to the latter, called rail-to-rail op amps, but the LT1007 is not one of these. This is why you're only getting -0.6 rather than 0. I'm actually fairly impressed that the LT1007 gets that close to V+. Linear Technology makes good stuff.

The first thing you should do is look at the data sheet. While you aren't an expert in these things, you should notice that the LT1007 is only specified for operation at +/- 15 volts. In fact, you can get away with +/- 12 volts (which is what I recommend). And, depending on the op amp, you can often use even lower voltages - but I wouldn't recommend it in this case. A repurposed PC supply will have +12 available. In a pinch, you could have a go at using +5. Just be aware that if you do, you are operating way below the specified conditions, and if it doesn't work you have no complaint.

  • \$\begingroup\$ Thank you very much for your answer. If I understood correctly, the circuit I should make is this one ? Should the power rails in the other op-amps (e.g. bandpass filter) also be connected to the power supply in a similar way? \$\endgroup\$ – ClayP Nov 13 '17 at 16:08
  • 1
    \$\begingroup\$ @ClayP - Yes to both. If you don't, each op amp will be unable to amplify the positive part of the AC signal coming in, and this will grossly distort the signal. \$\endgroup\$ – WhatRoughBeast Nov 13 '17 at 16:24
  • \$\begingroup\$ ATX power supplies have -12V too (blue cable), but Im not too sure about current capabilities. \$\endgroup\$ – Jan Dorniak Nov 13 '17 at 23:00
  • \$\begingroup\$ @WhatRoughBeast It worked perfectly. Thanks again. \$\endgroup\$ – ClayP Nov 14 '17 at 12:05

Since the 2nd Op Amp is ground referenced on + in the output will be 0 and not work with a single supply.

Any bipolar supply that can produce a 5V swing for the ADC will work and +/-15 is probably what was being used in the article... is overkill.

If you want to use a single supply then U2 +in must be connected to ADV Vref/2 to make its output the same DC voltage and that becomes the 0 Voltage level in the ADC.

The -15V injects an offset with an inverting gain of;

Av=-Rf/Rin thus offset injected by -15V = (-15V)* -15k/91K = +2.47V which is close to 2.5V.

You dont need this if your ADC Vref is 5V then you create 2.5V for U2in+ with an equal R divider and filter cap.

enter image description here

  • \$\begingroup\$ The article itself has that particular LT1007 inputs arranged oppositely. The inverting input is grounded in the article itself. Also, I suspect the OP is using the \$\pm 12\:\textrm{V}\$ rails available in a PC power supply, as well as the \$+5\:\textrm{V}\$ supply rail, since the article refers to the idea of re-purposing a PC power supply for the project. I think this is why the OP changed the resistor on that final opamp. I don't think the OP is stuck with a single supply rail. \$\endgroup\$ – jonk Nov 13 '17 at 7:21
  • \$\begingroup\$ @jonk The author of the article later published errata, mentioning that the LT1007 inputs in the paper should be upside down. Hence mine are corrected following his own corrections. \$\endgroup\$ – ClayP Nov 13 '17 at 7:34
  • \$\begingroup\$ @ClayP Thanks. Where did you pick up the errata? \$\endgroup\$ – jonk Nov 13 '17 at 7:41
  • \$\begingroup\$ @ClayP Ah. Never mind. I found the revised article, I think. ( researchgate.net/publication/… ) \$\endgroup\$ – jonk Nov 13 '17 at 7:44
  • \$\begingroup\$ @ClayP , so the schematic error you saw caused confusion and you understand the inverting gain solution ok now? \$\endgroup\$ – Tony Stewart EE75 Nov 13 '17 at 12:42

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.