Electronic – Is this the correct way of measuring Zi in LTSpice

The 230V is RMS voltage, which for a sine wave, is \$\frac{V_{PEAK}}{\sqrt{2}}\$. SPICE takes the voltage as peak volts, so whatever RMS sine wave voltage you want, just multiply it by ~1.414 and use that.

You may as well get used to converting back and forth between RMS, peak, and peak-to-peak voltage- you'll need to do so early and often.

230V RMS, 325V peak and 650V peak-to-peak all describe the same sine wave.

Frequency is given in Hz rather than radians/second, but you'll also often have to convert between frequency and angular frequency (by multiplying by \$2\cdot\pi\$ for angular frequency since there are \$2\pi\$ radians in a circle ).

Here is an introductory tutorial on LTSPICE from a Canadian University (Laurier).

Given you have "high value" resistors, you are not yet serious about the noise floor.

A 1Kohm resistor produces 4.00 nanoVolts RMS noise /rootHertz, that is in 1Hz bandwidth. At 290 degree Kelvin.

In 10Hz BW, 4*sqrt(10). In 100Hz BW, 4*sqrt(100). In 1,000Hz BW, 4nV * sqrt(1,000).

Lets examine that 8nanoVolt/rtNz noise density. The equivalent resistor inside that opamp, to produce that 8nV, is 4,000 ohms. Vnoise is sqrt(4 * K * T * Bw * R).

If you keep the resistors down to 1Kohm or so, requiring several milliAmps from the opamp, you can easily design a circuit with opamp-noise-density-limited total integrated random noise.

With low Rvalue resistors, you can ignore the current-density. 1kohm * 0.2pA = 0.2nanoVolts, very small compared to 8nanoVolts.

Thus in a 10,000 Hz bandwidth (including the pi/2 factor for 1-pole rolloff) you will have total input referred noise of 4nV * sqrt(10,000) = 400nV = 0.4 microvolt rms.

Since your gain is about ONE, this will also be your output noise. Ignoring power supply trash, ground noise, magnetic field intrusion (20,000 Hz is not shielded or attenuated by standard copper foils), and electric field charge injection.

If you use 20,000Hz bandwidth, 1-pole rolloff, you'll have 20,000 * pi/2 or or 31,000Hz equivalent bandwidth, with the noise voltage being integrated out to infinity as your 1-pole does the rolloff.

The total integrated noise voltage is sqrt(31,000) * 4nanoVolts.

Thus 170*4 == 680nV == 0.68 microvolt RMS.

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After reading the LM4040 datasheet:

The noise bandwidth of the LM4040 is about 40KHz. Thus the total integrated noise will be sqrt(40,000/ 10,000) * 180uVrms, or 360uVrms.

That is divided by R6 and R7. Their own contribution is about 10Kohm equivalent, or 4nV * sqrt(10,000 / 1,000) = 12 nanoVoltsrms/rtHz, with high bandwidth. Assume 1MHz, thus 12nV * sqrt(1,000,000) = 12 uVrms.

The opamp buffer is 8nVrms.

The voltage divider R1 and R2 is about 40,000 equivalent (those 2 in parallel); assume the same 1MHz bandwidth, so use the total integrated noise of R6/R7 and scale up; thus 12uVrms * sqrt(40,000 / 10,000) = 24 uVrms.

The feedback network (gain slightly more than 1) has the same noise contribution, or 24uVrms.

So you have a number of contributions. The largest is the reference diode. Lets filter that, with a 160Hz RC lowpass; we need 1milliSecond time constant TAU. The equivalent resistance on pin#3 of left-most opamp is about 10,000 ohms; install a capacitor in parallel, to ground, with R7, to get a 0.001 second (1e-3 second); a 0.1uF (or 1e-7 farad cap) does this.

Filtering the Reference should be exciting.

Now restrict the output bandwidth to 20KHz, or about 10 microseconds (about 8uS, actually, but lets do easy math).

With R4 of 50Kohm, a 1pF cap in parallel causes tau of 50,000 picoSeconds, or 3MHz. Install 100pF and expect about 30,000Hz bandwidth. [wrong: not for non-inverting circuit use, because the Grounded resistor --- R3 --- prevents gain attenuation below Gain=1. So this 100pF is not a wise approach.]

That should be exciting. [ wrong. The gain will only drop from 1.3 to 1.0 and then not attenuate any more. Thus is not a useful high-frequency lowpass.]

Now use your instrumentation amplifier to examine the (zero output amplitude) of the Function Generator. That should be exciting.

You may need to install 100 ohm resistors in each of the 4 VDD paths. And up the bypass caps to 10uF. This ensures higher frequency noise in the power supply regulator servo loop is filtered DOWN in amplitude.

Let me know what works.

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The righthand opamp U3 is an awkward circuit to convert to low-pass-filter. A capacitor across the Rfeedback merely ensures the high-frequency gain = 1.000, which lets all the opamp noise and the Vnoise of R3 160K to appear on the output.

Assuming no noise entering the Vin-, nor the VDD pins, the opamp Rnoise of 4 Kohm can be added to the 160Kohm, thus predicting 164K ohm. Given 13*13 = 169, we'll scale up 4nV by 13, to 52 nanoVoltrms/rtHz, over 1MHz bandwidth.

The total integrated noise should be 52 microVolts rms. [ error Initially said 52 milliVolts]