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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?

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    \$\begingroup\$ What you're most likely referring to is called Nyquist noise, or Johnson-Nyquist noise. Wikipedia has a section on how it affects reactive impedances here. \$\endgroup\$ – Hearth Jul 25 '18 at 0:04
  • \$\begingroup\$ you have to consider the input amplifier impedance. When you make a 1M:100ohm divider, the noise resistance is only 100 ohm. \$\endgroup\$ – Henry Crun Jul 25 '18 at 0:09
  • \$\begingroup\$ only if the reactance resonates or passes more noise from source more than signal \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 25 '18 at 0:13
  • \$\begingroup\$ So if I understand it correctly,resistors just add thermal noise while ideal inductors and capacitors wont? That means pure reactance doesnt cause noise. \$\endgroup\$ – wav scientist Jul 25 '18 at 1:07
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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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  • \$\begingroup\$ "purely reactive element theoretically doesn't generate noise by itself" That mean in my example,the signal that passed through resistor would be noisier than the ones that gone through capacitor or inductor,correct? \$\endgroup\$ – wav scientist Jul 25 '18 at 20:21
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    \$\begingroup\$ In your example, there is theoretically no noise source if you only add capacitors and/or inductors. \$\endgroup\$ – Sven B Jul 25 '18 at 20:44
  • \$\begingroup\$ So resistor = noise,inductor/capacitor no noise.... thank you \$\endgroup\$ – wav scientist Jul 25 '18 at 22:45
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Not directly.

Reactive currents don't result in real power dissipated in a load or or transferred to the load. There will be no noise energy transferred. No energy = no noise, from the reactance

But like power factor issues, reactance may be shunting signal current through real resistances, causing real power loss, and thus a real reduction in signal-noise ratio.

Another view is that reactances represent a mismatch, where some signal energy is not transferred into the amplifier - again worse SNR.

If you put a giant reactance in series with the source, clearly very little signal power will get through to the load - bad SNR.

Now the second issue is that your amplifier has input noise currents. When these are forced through a reactance, they will become real noise voltages. (I think - open for debate).


If you simply divide the voltage, by what ever means, then you will worsen SNR as power is reduces by V^2. If you want to try and preserve SNR, then you must do an impedance transformation, so that current is increased as voltage is dropped, and thus power is constant. When you put a 1Mohm resistor or reactance divided with 100ohm reistor, then the thermal noise resistance is 100 ohms regardless

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  • \$\begingroup\$ What is noise energy? Noise and signal connected,if something decrease noise energy,it must decrease signal energy same about becose both really are just electrical energy living together in same voltage/current waveform."Another view is that reactances represent a mismatch, where some signal energy is not transferred into the amplifier - again worse SNR" But the transfer is decreased by same amount no matter if it is resistance or reactance,1MΩ at 100 Hz is 1MΩ at 100 Hz no matter if resistance or reactance is decreasing the signal. \$\endgroup\$ – wav scientist Jul 25 '18 at 0:12
  • \$\begingroup\$ No, every (real) resistance makes a fixed amount of noise energy at a given temperature and bandwidth. It is -174dBm/hz at room temperature. This is the bottom line of noise performance. You can only improve it by lowering temperature, or reducing signal bandwidth \$\endgroup\$ – Henry Crun Jul 25 '18 at 0:14
  • \$\begingroup\$ I dont mean that,I mean if we have AC voltage divider,it will divide the TRANSFERED energy onto two loads equaly if the impedances are same between resistor variant and reactance variant.Ohm is ohm from energy transfer perspective. \$\endgroup\$ – wav scientist Jul 25 '18 at 0:17
  • \$\begingroup\$ The SNR effect will be basically the same: after the divider, the noise is coming from the input amplifier and input resistances, regardless of the type of divider. The degradation of SNR is simply caused by the reduction of signal power that makes it into the input amplifier - using R or C doesn't change that. Using a transformer, or a resonant matching network can preserve SNR. Note that as you change the apparent source impedance, you need a different amplifier circuit to get best noise. Each type of amplifier has an "equivalent noise resistance" where it can give best SNR. \$\endgroup\$ – Henry Crun Jul 25 '18 at 0:35
  • \$\begingroup\$ BTW, noise is unintuitive to get your head around at first, and much of what is written on the 'webs is badly wrong. Get a copy of Art of Electronics and read the low noise sections. \$\endgroup\$ – Henry Crun Jul 25 '18 at 0:39
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Your application is unclear, but I'll assume you're talking about transmitting a signal through a long enough transmission line for it it to be subjected to real noise then running that signal into an amplifier or buffer.

Basically resistance just reduces your signal power and nothing can be done to compensate for it, thus you have to amplify your signal, and along with it the noise.

Reactance is telling you how much the the circuit is resistant to state change and is what determines how much of a signal is reflected back toward your transmitter as it transitions carriers, be that a transmission line, or the output/input of a circuit. Since the reactance can be tuned in the input and output circuit it is theoretically possible to have a perfect (reflectionless) match where the loss is purely resistive. In practice the reflection will always be non-zero.

To get more to your point, if your circuit is tuned you'll have fairly less loss from your reactance than from your resistance, however you'll still have to boost the signal from both sources of power loss (i.e. resistance & reflection loss... also commonly called insertion loss).

Hope that helps

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  • \$\begingroup\$ No,I dont mean transmission lines. \$\endgroup\$ – wav scientist Jul 25 '18 at 0:52
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Yes. Absolutely. As you said, the resistance at specific frequencies will be largely just like a resistor. There are many sources of noise that will be more or less dominant depending on the circuit but if you have an high gain amplifier with an RC filter in front of it and it presents a higher source impedance at higher / lower frequencies, that noise will be amplified.

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