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In the first stage the signal is converted to single-ended, and it is amplified using a digital potentiometer in the second stage to be able to change the gain with a microcontroller.

In the last stage I have converted the signal back to differential again.

Thd ADC range is 0-3.3 V and the max. sampling frequency is 5.6 MHz, or 3.3 MHz will be used.

The reason for the differential input is because it has 6 to 8 dB better SNR performance. The signal frequency range of interest is 1 kHz to 1.5 MHz or so.

If you can answer each question specifically it would be great.

ADA4807 Datasheet

Schematic

My questions are:

  1. Is this schematic correct in the case of cascade connections and especially the part where the signal converted back to differential again (the last stage)? The gain is adjusted in the second stage, so the first and last stages are just for proper conversion. Please correct any wrong resistance value I selected, if any.

  2. What should be the values of R12 and R14? One reference design uses 1 kΩ, another one uses 10 kΩ. Is it important?

  3. Does R10 (10 kΩ) have to be the same as R13 (1 kΩ)? Another way of asking is, is the value of R10 important in this conversion?

  4. Do I need R9 and R19 before the DC bias voltage connection?

  5. SAR ADC recommends an RC filter at the input for filtering and capacitance recharging, so I have placed 20 Ω and 2.7 nF. Is this correct?

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  • \$\begingroup\$ Do the IFF+ and IFF– come from a capacitively coupled source? \$\endgroup\$ Commented Jan 10 at 12:55
  • \$\begingroup\$ @RohatKılıç IFF+ and IFF- are coming from LT6232 opamp output which is for sallen key active high pass filtering. \$\endgroup\$
    – Cenk
    Commented Jan 10 at 13:02
  • \$\begingroup\$ This doesn't answer my question. Is it capacitively coupled i.e. AC-coupled or direct coupled? Do the IFFs have any DC offset? Do the IFFs connect directly to R7 & R8? \$\endgroup\$ Commented Jan 10 at 13:20
  • \$\begingroup\$ @RohatKılıç What i meant is that the output of the LT6232 opamp directly connected R7 and R8. There no capacitor in series etc. Offset is 1.65V. The signal is mixed sinusoidal with multiple frequencies on a 1.65v dc offset. \$\endgroup\$
    – Cenk
    Commented Jan 10 at 13:30
  • \$\begingroup\$ The differential input resistance of ADA4807 is as low as 35 kohm. This has an important impact on the gain calculations. \$\endgroup\$
    – Jens
    Commented Jan 10 at 13:42

1 Answer 1

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I am answering your question based upon what I believe you are apparently trying to accomplish, rather than what you ask directly.

I believe what you are trying to accomplish is to amplify a differential signal, with a variable gain based upon a single variable resistance. The following circuit will do that job:

schematic

simulate this circuit – Schematic created using CircuitLab

R1 and R2 should be precision resistors and have the same value. The differential mode gain will be:

$$A_{V(diff)} = 1 + 2\frac{R1}{R{gain}}$$

You may want to put a fixed resistor in series with the variable resistor to prevent the total resistance going to near 0, and the gain being unreasonably high.

The common mode input voltage is "passed through" this circuit. Thus, if the input signals are centered around 1V, instead of 0V, the output signals will also be centered around 1V.

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  • \$\begingroup\$ The perennial problem: "based upon what I believe you are apparently trying to accomplish, rather than what you ask directly". +1 just for that. I'd give you another +1 for the rest of the answer, if SE let me. \$\endgroup\$ Commented Jan 10 at 14:03
  • \$\begingroup\$ I have updated the circuit and added image of it to the question as later update. Could you check it please. Is it correct. \$\endgroup\$
    – Cenk
    Commented Jan 10 at 16:14
  • \$\begingroup\$ @Cenk I am going to revert your question, because, on this site, you shouldn't change the substance of questions after they have been answered. However, I am wondering in your "new" circuit, why you have a separate "differential gain" stage. It actually amplifies both the differential signal, and the common mode signal, which is OK if the common mode signal has no noise, (i.e. unwanted signal), but otherwise can be problematic. \$\endgroup\$ Commented Jan 10 at 17:17

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