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enter image description here

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This is a picture of my Differentiator and Integrator combinded, in a working delta sigma modulator.

My question here is: Many other integrator and differntiators have the input signal on input 2. Opposite of what I have. And input 3 down to ground.

Will it have anything to say for the result ? Is my circut wrong ?

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    \$\begingroup\$ Is there anything missing from circuit? \$\endgroup\$ – Chu Nov 18 '15 at 12:45
  • \$\begingroup\$ At the moment, despite the R and C feedback, all you have is a unity gain amplifier. \$\endgroup\$ – Andy aka Nov 18 '15 at 12:54
  • \$\begingroup\$ I've updated the picture. \$\endgroup\$ – Eivindo Nov 18 '15 at 13:06
  • \$\begingroup\$ Why does Q1 buffer your oscillator output and generally, what does the circuit do functionally and where is the output? \$\endgroup\$ – Andy aka Nov 18 '15 at 14:23
  • \$\begingroup\$ This circut is a working delta sigma modulator. AC sinus in, 5V P-P and digial pulse out. Tested ok, simulated ok and produced. Sorry for removing the output, its Q invers, the 6 output on the D-Latch. But my integrator/Differentiator is quite different from others i've seen, so I wonder what this has to say for the output signal. \$\endgroup\$ – Eivindo Nov 18 '15 at 14:32
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At a guess, the "other" circuits you've seen use U1A as an integrator in the "normal" configuration, with the + input tied to ground.

Your circuit works like this. Assume the output of U1A is positive. Then the output of U1B is positive, and the input to U3A is 0. So the Q* output goes high after the next clock, and the U2A output will go high. This will drive the integrator towards zero. The integrator will then tend to wander around zero, and the circuit will work more or less as a sigma-delta convertor.

However, because you've driven the + input of the integrator, rather than using a standard resistor to -, you've produced a converter which will be distinctly non-linear. This won't be a problem for small input swings and low precision, but it will be a real problem for large inputs. Consider the correction current in the standard configuration - it is independent of the input voltage. Assuming the op amps give +/- 4 volts out when driven into saturation, the correction current into the integrator will be +/- 4V divided by 100K, or 40 uA, regardless of input voltage. In your circuit, a 0 volt input will produce a correction current of 40 uA, but a 2 volt input will produce a correction current of (4-2)/100K, or 20 uA. Like I say, this will probably do OK if the input swings are small, especially if the downstream circuitry only attempts to zero the input, but it will provide inaccurate readings if you're trying to measure the input.

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