What is nV/vHz Noise?

Whenyou look at data sheets for operational amplifiers you'll notice specificationsfor input noise voltage and input noise current expressed in units of nV/vHz(nanovolts per root Hertz) or as pA/vHz. At first glance, the vHz denominatorseems odd. After all, how can vHz have any physical meaning?

First,the nV/vHz spec refers to thermal noise, or Johnson noise, caused by the effectof temperature on the electrons in a resistive device, which pretty muchincludes everything. Second, the spec identifies the spectral density ofnoise as a root-mean-square (rms) value over a given bandwidth.

Sayyou have a Texas Instruments OP27A op-amp with an input noise voltage (V_n)of 3.5 nV/vHz. You'll operate the op-amp from 1,000 Hz to 20 kHz, so you takethe square root of the 19 kHz bandwidth (138 vHz). Multiply 3.5 nV/vHz by 138vHz and you get 482 nV. If you operate the op-amp with a gain of 25, the noisegets multiplied by 25 also, so you have 12.1 µV. Use a 1V output (0 dBV) fromthe op-amp and you calculate the signal-to-noise ratio: SNR = 20 * log (1V /12.1*10^-6V) = 98 dB. (See For More Information, below)

Op-ampdata sheets might include graphs of nV/vHz versus supply voltage, ambient temperature,source resistance and other variables. You can use this information to refinenoise calculations based on your requirements.

Buthow does the vHz unit get into a noise value in the first place? In 1928, JohnB. Johnson at Bell Telephone Labs explained the creation of noise in conductorsand Harry Nyquist, also at Bell Labs, followed with a paper (worth reading)that provided a theoretical basis for this noise and related it to temperatureand resistance alone.

Engineersoften rewrite Nyquist's equation for thermal electromotive force:

E²dv =4kRTdv  as  E²= 4kRT or as E = v4kRT

Where:k = Boltzman's constant in joules/Kelvin, R = resistance in ohms, T =temperature in Kelvin, and E represents the noise voltage. If you write the E²equation with k, R and T in MKS units, and drop the unit-less 4, you obtainEquation 1.

InEquation 2, the Kelvin units cancel out, the Amperes (A) separate, andmultiplying by s/s yields Equation 3. The (kg•m²)/(A•s³)unit equals volts (V), so:

E²= (V²•s) = V² / (1/s),

E= V / v1/s) and 1/s = Hz, so

E= V/vHz

Thesesteps show the source of the V/vHz unit and that it has a empirical andtheoretical origin. Keep in mind other sources can contribute noise, too.

What is nV/√Hz Noise?_A