Electronic – Does reactance increase noise

capacitancenoisereactanceresistance

Impedance is made of resistance and reactance which is capacitance and inductance. I read that increasing resistance increases noise, but what if capacitance or inductance is increased, does the noise increase too?

The way I see it,capacitance and inductance act like resistance too, they too resist, only difference is that their "resistance" is frequency variable unlike normal resistance. Also normal resistance dissipates energy, while capacitance and inductance store energy. I am not sure if this have impact on noise.

Example: I have three identical signal sources outputing 100 Hz sinewave, I want to attenuate it by some amount. One sinewave goes through 1MΩ resistor, second through capacitor that have 1MΩ capacitive reactance at 100 Hz and third sinewave goes through inductor with 1MΩ capacitive reactance at 100 Hz. How high will be noise at each case? Which of the three sinewaves would suffer worst SNR degradation?

Best Answer

I don't think there is a way of telling which way the noise will go just from the statement increasing/decreasing resistance or something alike. It severely depends on the application.

For example, in a voltage divider, the output is a voltage and the noise contributions can be calculated using

$$\overline{dv_R^2} = 4k_BT\cdot R\cdot df$$

However, when building a transimpedance amplifier for example (turning an input current into an output voltage), we need to look at how much current noise the resistor generated, in which case we now have to look at the equation:

$$\overline{di_R^2} = 4k_BT\cdot \frac{1}{R} df$$

Suddenly, the noise goes down with \$R\$, just because the information we're looking at is different.

A similar reasoning can be made for reactances. While a purely reactive element theoretically doesn't generate noise by itself, they can affect the noise both ways indirectly.

For example, looking at the total integrated output noise of a simple RC lowpass filter, you can find that

$$\overline{v_n^2} = \frac{k_BT}{C}$$

Having a large capacitance will decrease the noise on the output voltage! On the flip side, in imagers you are often interested in the amount of charge on the capacitor, in some cases you will want to almost count the number of electrons. In such a case, the total integrated noise is

$$\overline{Q_n^2} = k_BTC$$

In this application you want to have a minimum capacitance.

You might think that this is contradictory, but it isn't. If you are interested in the amount of charges, you want to have the largest voltage change possible per electron stored on the capacitor. So a few electrons more or less on the capacitor will lead to a very large voltage swing if the capacitance is small (ie. the voltage noise is large), but we can very accurately measure the amount of charge. However, if we just want to keep the voltage stable, then we don't want those few electrons to have a large effect, and so the capacitance is chosen very large in that case.

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