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Please refer to the following app note: https://www.ti.com/lit/pdf/sbaa374

It is a resistive circuit in front of a noninverting op amp to scale and level shift an (bipolar) input voltage.

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I get the principle and how to calculate the resistor values but stumbled about the footnote 3:

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In my opinion, one main "charm" of that circuit is that I can use a single-supply op amp like the app note suggests. Now that brings me to two questions:

  1. Why would a high-voltage (even bipolar?) supply increase the input impedance? Does it increase the internal input impedance of the op amp and if so, why?

  2. In the example of the app note, they use rather low-valued resistor values which in any case dominate the input impedance of the circuit? (The op amp input is MUCH higher anyway)?

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  • \$\begingroup\$ Note 3 is talking about the input impedance of the circuit, and you're asking about the impedance of the op amp, which is a different beast. \$\endgroup\$ Commented Nov 18 at 18:19

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No, an op amp's input impedance does not vary significantly with the supply voltage. The reason the app note suggests a high-voltage amplifier buffer with +/-15 V rails is that the design assumes the following about the input:

This circuit document describes how to translate a high-voltage signal (for example, ±10V) to a low voltage ADC input (for example, 0V to 5V).

In order for the buffer (which would be inserted between \$V_\text{in}\$ and \$R_a\$) to work properly, it must have supplies of approximately +/-15 V so that \$V_\text{in}\$ is within the buffer op amp's input voltage range.

The low-valued resistors do dominate in the input impedance calculation compared to the shown OPA320's input impedance, which is why the app note suggests the addition of a high-voltage buffer before \$R_a\$. The input impedance of the overall circuit is just the input impedance of the buffer, in that case.

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  • \$\begingroup\$ Ok I overlooked that they suggest to add another buffer in front of the resistors =( \$\endgroup\$ Commented Nov 18 at 19:02
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From the app note you linked:

"This circuit document describes how to translate a high-voltage signal (for example, ±10V) to a low voltage ADC input (for example, 0V to 5V)."

The reason for the high voltage buffer is because before this scaling circuit (with lower impedance) the signal levels are higher, and possibly bipolar. Therefore, you need enough headroom in your buffer to pass the signals to this circuit without clipping.

The high-voltage comment doesn't relate to the input impedance. The input impedance of the buffer shouldn't vary significantly within its specified supply voltage range.

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The title of the app note is:

Noninverting Circuit for High-to-Low Voltage Level Translation To Drive ADC

In this context, the actual values of V_IN are specified to be ±10V. In the circuit you posted, the impedance of the input is exclusively determined by the resistor network (Ra, Rb, Rc). This is not a high impedance input.

The suggestion in note 3 implies the voltage follower amplifier circuit (below), with Vout driving the level shifting network directly.

For a high-impedance input, use a high-voltage amplifier buffer

By Inductiveload - Own workThis W3C-unspecified vector image was created with Inkscape ., Public Domain, https://commons.wikimedia.org/w/index.php?curid=5794259

The high voltage nature of the op-amp (±15V) is needed because both the input and output voltages of this first buffer amplifier will be ±10V. This amplifier topology is high impedance, or as high impedance as the selected op-amp.

  1. The additional amplifier is in a different topology, which is high-impedance. The supply voltage is a side-effect of the stated input voltage requirements.

  2. Correct, and this is by design so that the resistors are the primary factor to identifying circuit performance

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Why would a high-voltage (even bipolar?) supply increase the input impedance? Does it increase the internal input impedance of the op amp and if so, why?

You have misinterpreted what is being said. All they are saying is add a buffer in front of the circuit shown in your post and, of course, that buffer needs to run from a bipolar supply of 15 volts.

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