Question 8.14: Estimate the output noise PSD and the RMS value for the inte......

Estimate the output noise PSD and the RMS value for the integrator circuit seen in Fig. 8.29 (which includes the only noise source, that is, the thermal noise from the resistor). Assume that the op-amp is ideal (noiseless, infinite gain, and infinite bandwidth).

8.29
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

The op-amp will keep the inverting input at the same potential as the noninverting input, that is, ground. This means that all of the thermal noise from the resistor will flow through the feedback capacitor (both sides of R are at ground). The integrator’s output noise PSD is then

V_{o n o i s e}^{2}(f)=\left|\frac{1}{j\omega C}\right|^{2}\cdot\frac{4k T}{R}=\frac{4k T}{R}\cdot\frac{1}{\left(2\pi C\right)^{2}}\cdot\frac{1}{f^{2}}\mathrm{~units~of~}V^{2}/Hz

Noise with a 1/f² PSD shape is often called red noise, Fig. 8.30. Looking at this spectrum, we see that the noise PSD becomes infinite as we approach DC.

Remember, from our discussion at the beginning of the chapter concerning how a spectrum analyzer operates, that a point representing the PSD at a particular frequency is the measured power, V^2_{RMS}, divided by the resolution bandwidth of the spectrum analyzer, f_{res} (that is, V^2_{RMS}/f_{res}). To make the low-frequency measurements, f_{res} must decrease, e.g., go from 1 Hz to 0.01 Hz to 0.00001 Hz, etc. A DC signal (or a sinewave for that matter, see Fig. 8.5) results in an infinite PSD point if we take f_{res}\rightarrow 0.

Let’s use Eq. (8.10) to determine the RMS output noise voltage for this circuit

V_{RMS}=\sqrt{\int\limits_{fL}^{fH}{V^2_{noise}(f)\cdot df} }      Volts          (8.10)

V_{o n o i s e,R M S}^{2}=\int\limits_{fL}^{fH}{V^2_{noise}(f)\cdot df}=\frac{4kT}{R}\cdot \frac{1}{(2\pi C)^2}\cdot \left\lgroup\frac{1}{f_L}-\frac{1}{f_H} \right\rgroup           (8.50)

If we take f_H\rightarrow \infty , this equation becomes

V_{o n o i s e,R M S}={\sqrt{\frac{4k T}{R}}}\cdot{\frac{1}{2\pi C}}\cdot{\frac{1}{\sqrt{f_{L}}}}          (8.51)

If we call the length of time we integrate (or measure an input signal since integrators are often used for sensing) T_{meas}, then we can get an approximation for lower integration frequency as

T_{m e a s}\approx{\frac{1}{f_{L}}}          (8.52)

or, rewriting Eq. (8.51),

V_{o n o i s e,R M S}\approx\sqrt{\frac{4k T}{R}}\,\cdot\frac{1}{2\pi C}\cdot\sqrt{T_{m e a s}}          (8.53)

This result is practically very important. If we average (the same as integrating, which is just summing a variable) a signal with a 1/f² noise spectrum (integrated thermal noise), the RMS output noise voltage actually gets bigger the longer we average (unlike averaging thermal noise, see Eq. [8.48]). A signal containing 1/f³ noise (e.g., integrating flicker noise) shows a linear increase in its RMS noise voltage with measurement time. The result is an SNR that doesn’t get better, and may get worse, by increasing the measurement time. These results pose a limiting factor when imaging in astronomy. The night sky contains this type of noise (flicker, red, etc.). We can make images brighter by exposing our imaging system for longer periods of time but the images don’t get clearer.

V_{R M S}^{2}=\frac{4k T R B}{K}~\mathrm{with~units~of~}V^{2}          (8.48)

8.30
8.5

Related Answered Questions

Question: 8.4

Verified Answer:

The term kT has units of Joules (or Watt-sec) and ...
Question: 8.11

Verified Answer:

Again, as determined in Ex. 8.5, the total output ...
Question: 8.18

Verified Answer:

The schematic used for simulations is seen in Fig....
Question: 8.17

Verified Answer:

This amplifier is a transimpedance amplifier, that...