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This op amp circuit is used to offset stepped-down AC voltage by 2.5V(approx), so that the RMS of the voltage can be calculated. I can not change the configuration of the circuit, so that is ruled out.

This looks like a differential configuration, but i have not been able to understand certain decisions:

  1. Why the 2.5V DC offset is fed into using a 10Kohm resistor and why there is a 1Kohm following it.

  2. Why the + and - input of the op am p has this 1Kohm across it.

  3. Why go to the trouble of adding the resistor network consisting of 4.7K and 100K when the AC is stepped down to 5V RMS.

  4. Whether the AC source must be connected to ground or not. (Connecting and disconnecting gives different results).

Snapshot of the circuit during a simulation.

UPDATE: I noticed something fishy today. When i ramp up the frequency to 50Hz, the mains frequency in India, i see that the Voltage at both the points are equal in magnitude and in DIRECTION! Isn't that 0V across the VSINE then? This happens only when the op amp is connected with the DC offset. When i remove the op amp and DC offset, the voltages are equal in magnitude and opposite in polarity. Is this because of the negative feedback; removing the negative feedback also gives similar results but lower magnitude voltages. See picture.enter image description here

I confirmed that the resistor network before the op amp was only to bring the voltage down. See picture. enter image description here

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Using an offset places high accuracy and stability requirement on its source. The classical 4 resistor amp (you are using) places extreme matching accuracy requirement for the resistors. Linear Technology offers LT5400 series matched resistors assembly (SOI8). A resistor across OP AMP input pins is weird. Attenuator could be made by splitting the front resistors, but can achieve the gain by the ratio of resistors. Now you have 10k/100k i.e. gain of 0.1. You can go as far as say 1k/1 Meg i.e. gain of 0.001. Pay attention to layout: keep input nodes very small. Place the pair of resistors associated with them right in front of the pin they are connected to (on either side) so the longer track is not associated with inputs, but with GND, Output, Vref i.e. low ohmic sources. You can also guard the input solder pads by thin whiskers around them, connected to GND.

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I think I can help partially, but I'm not sure.

1) The reason for providing 2.5VDC offset has to do with the way the OP is feeded. In this case it's not simetrical but goes from [0;5] instead of [-5;+5]. From this circuit I understand that the objective is to have a sine signal but always possitive. 3) The objective of 4.7K and 100K resistor is to reduce the initial sine Wave which I suppose is 5V.

What I suppose from this is the initial process is to reduce the signal and then shift to only possitive voltage.

Hope this helps you. As for the 2nd quiestion it's a differential resistor, but can't see it's use without a simulation.

PD: It's a fun circuit.

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  • \$\begingroup\$ R20 is just loading the input, it's not reducing the voltage. R1, R21 and R2 are the potential divider which reduce the input to the op amp on which the gain resistor R3 operates. \$\endgroup\$ – Ian Bland Aug 3 '17 at 12:29
  • \$\begingroup\$ You are right about R20. I forgot about Ground and yes, R2 with R29 work as a divider. In this case the 2nd devider is R1, R21,R2. What I don't understand is why this distribution if it could be done much easier. \$\endgroup\$ – Dimitri Aug 3 '17 at 12:48
  • \$\begingroup\$ I might be being stupid here, but where is the output of the circuit meant to be? \$\endgroup\$ – Ian Bland Aug 3 '17 at 13:04
  • \$\begingroup\$ The output is across C1. I agree with the loading caused by R20, and one important thing i noticed is the amount of DC shift is assumed to be 2.5V, so i must remove the ground on the VSINE generator, and then you would see the voltage at the + input is exactly 2.4999V (notice how it is 2.21V here because of the ground i believe). Something i am seeing for the first time is that the feedback network is only a single resistor and not a divider(?). And any thoughts on point 4? \$\endgroup\$ – Prithvi Raj Prakash Aug 4 '17 at 13:45
  • \$\begingroup\$ As I see it it's a simulation so I "-" of the SINE is the same as GND. If you don't add a GND to "-" it's the same as having a circuit with different returns. If you delete the GND from the circuit your generator doesn't have any return so that's the reason for having different results. \$\endgroup\$ – Dimitri Aug 7 '17 at 9:57

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