Voltage vs. Current Noise Theory

[solis365]

The white noise voltage spectral density of a resistor is given by

Vn^2 = 4*k*T*R

where
k = boltzmann constant (1.38 E -23)
T = temp, assume 300 K
R = resistance in ohms

It is also well known that the CURRENT noise spectral density is given by

In^2 = (4*k*T)/R

This makes perfect sense, given ohms law, or general circuit intuition. If you have a high-output-impedance current source, and you are looking at output current, then noise voltage fluctuations at the output wont change the output current. they get divided by the output resistance. the same goes for a low-output-impedance voltage source. noise current variations wont affect the output voltage much, as they are multiplied by a small R.

my question is twofold.

Does this mean that, in a current-mode circuit; that is, a circuit where the signal remains a current (current mode opamps for example), you are free to use large resistors and still keep things low noise? since adding a larger resistance "reduces" current noise.

second, in the following datasheet, for the CA3094 operational transconductance amplifier
https://www.electro-tech-online.com/custompdfs/2010/06/CA309428AB29.pdf

down on pages 4 and 5 they talk about their noise test setup. for testing output CURRENT noise they use a large resistor (1MΩ), and when testing output VOLTAGE noise they use a 0Ω resistance on the source. in terms of voltage noise, 0Ω resistor would contribute no noise. In terms of current noise, a 1MΩ resistor would contribute very little as well.

am I thinking about this right? does someone wanna explain this?
it seems like a paradox that increasing resistance "decreases" noise. it sort of makes sense since, in a current mode circuit, you dont care what the voltage fluctuations are, but I am not sure if I trust my intuition.

much thanks to anyone that can shed some light on this.

 

[crutschow]

You intuition is correct. Voltage noise increases with resistance and current noise decreases.

But working in the current mode you should note that the circuit may also respond to the resistor noise voltage as well as the current noise. For example, an op amp connected as a trans-impedance current amplifier will still have noise at the output due to the voltage noise of the feedback resistor. That voltage will be amplified by the ratio of the the feedback resistance to the current source impedance feeding the amp input. Thus for minimum noise, you want the current source impedance to be much higher than the feedback resistance.

 

[solis365]

thank you. good explanation.

a transimpedance amp will be affected by the voltage noise of its feedback resistor (as well as current noise of the current source feeding it)...

but, for example, if you have a current mirror, and you wish the output to have low current noise (for whatever reason)... you would be safe with a large resistor on the input side, since its current noise will be low. this large resistor will contribute a "negligible" amount of noise to the input, and thus the only noise you are left with at the output is the noise intrinsic to the current mirror itself (and whatever you stick on the output).

i think your explanation just made something click, im gonna go re-read that datasheet and I think i can figure out why they changed resistors now.