Jump to content

Quiet Resistor


Enthalpy

Recommended Posts

Hello dear friends!

I believe it is known, but not universally: a quiet amplifier can synthesize a quiet resistor that produces less noise than a physical resistor at the same temperature.

Amplifiers can be very quiet. Under S/N=3dB they add less noise than a resistive source at 300K produces. Good RF preamplifiers of usual technology at room temperature can offer S/N=0.8dB, adding 20% to the noise power of a resistive source. Cooled or exotic RF preamplifiers can be much better. Audio amplifiers can add even less noise, especially FET amplifiers on a high-impedance source.

Here the amplifier with gain -A shall have a high input impedance to simplify the computations. This is consistent with an AF bipolar transistor biassed for best noise or with a FET. A feedback resistor of value (A+1)R results by Miller effect in a stage input resistance of R, the synthesized quiet resistor.

QuietResistor.png.be388e414ce843d188ed3fa64ef42ea9.png

The noise resulting from the feedback resistor is sqrt(A+1) times smaller than what a physical resistor R produces, both in voltage and current.

The setup increases the noise injected by the amplifier in the input. It adds quadratically en/R to the amplifier's in and in*R to en, but their contributions to the setup's En and In are fully correlated. If the synthesized resistance R is the amplifier's optimum en/in, then the setup nearly quadruples the amplifier's noise power. With an amplifier better than S/N=1.0dB, the setup can be quiter than a physical resistor.

Noise computations are slippery, so double-checkers are even more welcome than usually!
Marc Schaefer, aka Enthalpy

Link to comment
Share on other sites

If I wanted to implement that I would need an amplifier with a well defined gain "A".

In practical terms, I'd do that with an op amp and two resistors.

In order not to load the rest of the system, those resistors would need to be reasonably large.

 

What's the effect of those large resistors (with correspondingly large noise voltages) on the overall performance of the circuit?

Link to comment
Share on other sites

Hi JC, thanks for your interest!

With op amps, I'd use two of them (hence beware the stability). The first non-inverting, so its feedback doesn't load the input at all, the second inverting.

Using at the first stage an amplifier stable on capacitive loads, you can have a mainly capacitive feedback divider that adds no noise. Such amplifiers are common within IC to make switched capacitor filters.

After significant gain at the first stage, the noise of the second stage matters little.

Though, I'd rather consider a lone bipolar transistor in common emitter as the amplifier. With ideality factor =1, the voltage gain is just the voltage drop in the load resistor divided by 26mW, so some bias help can stabilize the gain. A small resistor at the emitter would define the gain better. Other stages can reduce the output impedance.

Edited by Enthalpy
Link to comment
Share on other sites

2 hours ago, Enthalpy said:

Good RF preamplifiers of usual technology at room temperature

 

1 hour ago, John Cuthber said:

What's the effect of those large resistors (with correspondingly large noise voltages)

 

Both good points that bring out the fact that noise in resistors doesn't only depend upon temperature.

Vn = √(4kTR∆f)

So low resistance, narrow frequency bands also contribute to low noise.

RF amps are usually very low impedance.

Link to comment
Share on other sites

1 hour ago, Enthalpy said:

Though, I'd rather consider a lone bipolar transistor in common emitter as the amplifier.

I wanted to recommend a current feedback op amp or a transconductance feedback op amp, which enable low resistance value in the feedback loop and have a internal polarization that make the transconductance independent of the temperature.

Alas, the IC I've seen are noisy. The designers had other priorities.

One can still design a bias current proportional to the absolute temperature (PTAT) for a discrete bipolar transistor. Then, the voltage gain is not badly defined, and the noise minimum.

23 minutes ago, studiot said:

Vn = √(4kTR∆f)

So low resistance, narrow frequency bands also contribute to low noise.

RF amps are usually very low impedance.

A low resistance value increases the noise current
In2 = 4kTB/R
so there is an optimum in the impedances giving a minimum noise.

We have little choice in the RF impedances. Too high, and stray capacitances limit the speed. Too low, stray inductances do. Around 50-100ohm, the impedance of usual printed lines, both stray capacitances and inductance act the same, and with proper design they add a constant propagation delay instead of slowing down the frequency response.

Departing from the best impedances is avoided wherever possible. It was necessary for high-power RF amplifiers using bipolars because the transit time limits the epitaxy's thickness hence the breakdown voltage and the supply voltage. MOS do that better for being fast and allowing higher voltages, so the load impedance can be less low. But you still see very broad striplines in power amplifiers, and difficult emitter or source grounding.

Link to comment
Share on other sites

On 9/22/2019 at 10:04 AM, Enthalpy said:

[...] Noise computations are slippery [...]

Indeed. I computed En and In from the open-circuit voltage and the short-circuit current. The setup can be modelled by a noise voltage or current source, not both, and a noiseless resistor R. The resistance and the noise voltage source give the short-circuit noise current and reciprocally.

QuietResistorErratum.png.f34e6e8f264a0db8bcb00c4902dddc22.png

That's why I had found fully correlated noise and voltage sources. They were the same noise.

So the setup doesn't double the amplifier's current and voltage noises, it just keeps them. Nice !

Link to comment
Share on other sites

This slightly unusual circuit combines low noise with constant gain from discrete bipolars.

QuietResistorDiscrete.png.a741cd5d21a31667631cbc846bb01c2d.png

The first stage amplifies voltages by 100. It's cascoded because I believed a big G22 in a datasheet. The follower brings only 0.8 voltage gain.

Stability is uncertain at some source impedances. A capacitor at the cascode's output, or the general input, or very few pF across the feedback resistor would help.

The junctions at the follower increase at heat the voltage drop across the cascode's load. With 7.2V supply, the current is proportional to the absolute temperature, and the transconductance and voltage gain constant. It's more a gadget here, but fun.

Voltage regulators use to be extremely noisy, like 100nV/sqrt(Hz). The capacitor and split resistor at the cascode's load bring some 45dB @10kHz, power supply rejection by the closed loop does the rest.

The second and third transistors add very little noise, which results essentially from the lone first transistor, advantage over an IC. At 10kHz and reasonably higher, it corresponds to Ic=30µA, Ib=10nA and 170ohm base spread resistance. The cascode's load resistor and the feedback add a bit. For 17.1kohm synthesized resistance, I find 4.5nV/sqrt(Hz) in open circuit, while a physical resistor would have 16.5nV/sqrt(Hz). At room temperature, the circuit is as quiet as a resistor at 22K.

The emitter of the follower can provide an amplified output, to measure the circuit's impedance and noise, alone or connected to a similar one or a physical resistor.

Marc Schaefer, aka Enthalpy

Link to comment
Share on other sites

On 9/22/2019 at 10:18 AM, Enthalpy said:

With op amps, I'd use two of them (hence beware the stability). The first non-inverting, so its feedback doesn't load the input at all, the second inverting.

If the 1st op amp doesn't load the circuit, doesn't that imply a high input resistance with a corresponding high noise?

Link to comment
Share on other sites

On 9/22/2019 at 10:39 AM, John Cuthber said:

If I wanted to implement that I would need an amplifier with a well defined gain "A".

In practical terms, I'd do that with an op amp and two resistors.

In order not to load the rest of the system, those resistors would need to be reasonably large.

What's the effect of those large resistors (with correspondingly large noise voltages) on the overall performance of the circuit?

With two op amp stages, the non-inverting first stage can use low resistances, since the output provides abundant current. This reduces the voltage noise, and at this location the current noise is unimportant.

Differential instrumentation amplifiers with low voltage noise, typically meant to amplify thermocouple signal, use resistor values under 100ohm for that reason.

Link to comment
Share on other sites

The source impedance of a thermocouple is, literally, that of a bit of wire, so 100 ohms doesn't load it significantly.

But it doesn't meet this criterion.

On 9/22/2019 at 9:04 AM, Enthalpy said:

Here the amplifier with gain -A shall have a high input impedance to simplify the computations.


 

Link to comment
Share on other sites

On 9/29/2019 at 6:22 PM, John Cuthber said:

If the 1st op amp doesn't load the circuit, doesn't that imply a high input resistance with a corresponding high noise?

A high input resistance doesn't always result in high noise.

In a Jfet or Mosfet for instance, we use to take the inverse of the transconductance to compute the voltage noise produced by the transistor itself, at least at frequencies where 1/F noise gets negligible.

Then you have the noise at the input impedance, which sees the signal source in parallel (often a lower resistance than the amplifier), and which can be shorted by the varied capacitances, depending on the frequency and the compared values.

With piezoelectric or with capacitive sensors, at frequencies not very low, the capacitance uses to dominate. Then the voltage noise drops strongly. One can use for noise voltage and currents the standard computations. A big resistance provides a small noise current sqrt(4kTB/R) that gets shorted by the parallel capacitance.

Then, the noise is minimized by a bigger resistance that produces a smaller noise current. This is the reason for my proposed bias or feedback using photodiodes there
https://www.scienceforums.net/topic/88763-photocurrent-bias/
when no bias correction is needed, the conductance is essentially zero, removing this noise source.

18 minutes ago, John Cuthber said:

The source impedance of a thermocouple is, literally, that of a bit of wire, so 100 ohms doesn't load it significantly.

But it doesn't meet this criterion: [Enthalpy's " Here the amplifier with gain -A shall have a high input impedance to simplify the computations."]

In an instrumentation differential amplifier, the resistances defining the gain are not connected to the sensor, and their noise is not shorted by the sensor's impedance. Their noise add to the source's and amplifier's ones. That's why application notes tell to use small ohmic values.

When the amplifier's input is shorted by the signal source, the smaller resistance applies, or better, the combination of both. This often means that the noise of the source remains, and the amplifier adds some. The goal of a low-noise amplifier is to add little noise.

Sometimes the source produces less noise than its equivalent resistance at 300K. This is typically the case for antennas over roughly 300MHz. Having low losses, they convert little feed power to heat as a transmitter, so their temperature produces little noise as a receiver (thermodynamics). Most of their noise is picked from the surrounding and converted efficiently. A dish antenna at 6 or 10GHz pointing to the sky picks little noise from its main lobe, and while it may show resistive 75ohm or 300ohm, its "noise temperature" is much lower than 300K, rather 100K.

Than a quiet preamplifier is much more important. 1dB rather than 2dB noise figure (where 0dB refers to 300K) makes more than 1dB difference in the noise. Then we prefer to indicate a "noise temperature" in K rather than a noise figure.

This background noise depends much on how visible our warm Earth is for the antenna. This explains why UHF TV antennas better point slightly upwards, to remove some Earth from their main lobe while losing little signal near the maximum. More importantly, dishes for satellite TV have their primary source below the dish's axis to point strongly upwards, so the part of the primary's radiation that misses the dish aims at the sky rather than Earth. More refined low-noise antennas exist that minimize the secondary lobes of the primary source - but absorption by some material at 300K isn't an option, only reflection injects no reception noise.

Edited by Enthalpy
Link to comment
Share on other sites

  • 3 weeks later...
On 9/22/2019 at 10:39 AM, John Cuthber said:

[Op amp and reasonably large resistors defining the gain. Effect on the noise?]

 

On 9/22/2019 at 11:18 AM, Enthalpy said:

[I'd use two op amps, the first non-inverting, so its feedback doesn't load the input, the second inverting.]

Now with a diagram :

QuietResistanceOpamps.png.2cdc9b1de42e06605d44c4ebeeea79f2.png

The LT1028 produces beyond 1kHz typically 0.85nV/sqrt(Hz), and as seen previously, its 1pA/sqrt(Hz) convert to 1nV/sqrt(Hz) in the synthesized 990ohm, and the 100kohm add 0.4nV/sqrt(Hz) only. 9.9ohm in the feedback add 0.4nV/sqrt(Hz). This sums to 1.43nV/sqrt(Hz), as much as a 990ohm resistor at 39K.

Obtaining all the gain from the first stage preserves the loop's phase margin and makes the second stage's noise negligible.

Marc Schaefer, aka Enthalpy

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.