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I'm building a circuit that extract the DC component of an input signal, ,divide the DC component by 5, and finally add 2.5V it.

Here is the schematic of the circuit (the last op-amp is to invert the inverted signal):

As you can see, I've added a voltage-follower buffer between the low-pass filter and the voltage-divider sub-circuit, and another similar buffer between the voltage-divider sub-circuit and the rest of the circuit. Without those buffers, the result would be wrong.

I want to build the circuit on a breadboard using LM741 chip as op-amp. So, I was wondering if there another way to implement those buffers that uses less physical space on the breadboard?

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    \$\begingroup\$ How accurate do you need it? What's the lowest frequency you're trying to filter out? \$\endgroup\$
    – horta
    Apr 23, 2016 at 17:55
  • \$\begingroup\$ @horta The lowest frequency to be filtered is 0.0000...1 Hz. I want to extract the DC component of the signal. \$\endgroup\$
    – ammar
    Apr 23, 2016 at 18:03
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    \$\begingroup\$ Not very realistic are you? You've just asked for the impossible. \$\endgroup\$
    – horta
    Apr 23, 2016 at 18:04
  • \$\begingroup\$ I meant that I want to filter the lowest frequency I can filter. I chose R and C of the low-pass filter such that the lowest frequency allowed is 0.016 Hz. \$\endgroup\$
    – ammar
    Apr 23, 2016 at 18:06
  • \$\begingroup\$ And the accuracy of said output after all the math is done? \$\endgroup\$
    – horta
    Apr 23, 2016 at 18:07

3 Answers 3

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An LM324 would be a better option. This has 4 op-amps in a single 14-pin package. This quad op-amp, and its sibling the dual LM358, are very popular general purpose op-amps. In large volume you can buy them for a few cents each. They do have their limitations, and will not do everything, but they are a good replacement for the 741.

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  • \$\begingroup\$ I considered this option, but actually I'm afraid that LM358 is not as good as LM741. I don't know but I feel that LM741 is a robust one. \$\endgroup\$
    – ammar
    Apr 23, 2016 at 18:04
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    \$\begingroup\$ @ammarx Why ask for advice if you don't listen? The '741 is an old op-amp with poor specs, just about every op-amp will perform better (and consequently your circuit will perform better). "..I feel.." is a silly reason to use a '741 when knowledgeable people are telling you not to. \$\endgroup\$
    – uint128_t
    Apr 23, 2016 at 18:20
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    \$\begingroup\$ "Doctor, it burns when I.C.!" \$\endgroup\$ Apr 23, 2016 at 18:31
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    \$\begingroup\$ @ammarx I think your doctor is full of something. Ask him what he actually means by that, because it's ridiculous. \$\endgroup\$
    – pipe
    Apr 23, 2016 at 18:31
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    \$\begingroup\$ If you want to save space on your board, and you must use a part that was first made in 1968, then you could look using at an LM348. This has four 741 op-amps in a single package so it will please your doctor. The price is probably extortionate because professional engineers would not use one. About the only good thing about it is when you want better performance you can replace it with an LM324 because they are pin-for-pin compatible. \$\endgroup\$
    – Steve G
    Apr 23, 2016 at 18:55
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With a 1000 kohm resistor at the front end and with potentially 500 nA bias current required by the dinosauresque and crappy 741 you will immediately get an offset error of 0.5V. That kills it for me. Throw the 741 in the trash can and find something better.

Given also that the 741 has an input resistance of typically 2 Mohms you have now another substantial error. I'm basically just running down the numbers in the data sheet to see how the 741 stacks up in your application.

Supply current ~2mA is very wasteful these days - a lot of modern op-amps have massive DC accuracy and consumes virtually nothing in comparison to the 741 and, in all my very long years of using op-amps I can't remember one occasion on which I caused one to fail by either bad handling, soldering or the occasional misuse in ESD handling (not that I would encourage folk to mishandle any chip).

Input offset voltage could give you a further 5 mV error (piddling against the bias current issue of course).

Also you can combine stage 1, 2 and 3 and use only one op-amp but please don't use the 741 if you want any measure of accuracy.

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  • \$\begingroup\$ "you will immediately get an offset error of 0.5V." The simulation results for an input sine wave of (frequency = 0.1kHz, amplitude = 5V, and DC offset = 5V) show that the output of this circuit is 3.487V which is slightly less than the expected output of 3.5V. Would it be different on the breadboard? \$\endgroup\$
    – ammar
    Apr 23, 2016 at 19:19
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    \$\begingroup\$ It all depends on the sim and the model but the data sheet is what I go by for worst case analysis on dc errors. \$\endgroup\$
    – Andy aka
    Apr 23, 2016 at 19:45
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schematic

simulate this circuit – Schematic created using CircuitLab

Something like this reduces the amount of components you need. First stage does the division, RC filter and inverting buffer all together. Second stage is an inverting voltage level shifter of 2.5V.

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  • \$\begingroup\$ Is this schematic correct? How does the first stage do the division while the gain is 250k/250k=1? Moreover, I don't understand the concept of the second stage. I mean what is the purpose of connecting the non-inverting input to that circuit? What is the configuration of this op-amp? \$\endgroup\$
    – ammar
    Apr 23, 2016 at 19:28
  • \$\begingroup\$ @ammarx the division happens due to the resistors. 250k/(250k+1M) due to the psuedo ground at OA1-. The output of OA1 inverts and mirrors the voltage seen on the cap. The second op-amp's equation is Vout=2.5-Vin. Since Vin is inverted from the last stage, you have Vout=2.5+Vin. Once you add in the voltage division happening out front you have Vout = 2.5+Vin/5. \$\endgroup\$
    – horta
    Apr 23, 2016 at 19:37
  • \$\begingroup\$ You've raised the cut-off filter point by a factor of 4 in frequency. Make the 10uF into 40 uF. Also you haven't quite got the gain right at the front end i.e. you haven't accounted for the 250k being in series with the 1Mohm. Easiest fix is change the 1M to 750k and then make the capacitance ~53 uF \$\endgroup\$
    – Andy aka
    Apr 23, 2016 at 19:58

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