What is nV/vHz Noise?

When
you look at data sheets for operational amplifiers you'll notice specifications
for 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 denominator
seems 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 effect
of temperature on the electrons in a resistive device, which pretty much
includes everything. Second, the spec identifies the spectral density of
noise as a root-mean-square (rms) value over a given bandwidth.

Say
you 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 take
the square root of the 19 kHz bandwidth (138 vHz). Multiply 3.5 nV/vHz by 138
vHz and you get 482 nV. If you operate the op-amp with a gain of 25, the noise
gets multiplied by 25 also, so you have 12.1 µV. Use a 1V output (0 dBV) from
the 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-amp
data sheets might include graphs of nV/vHz versus supply voltage, ambient temperature,
source resistance and other variables. You can use this information to refine
noise calculations based on your requirements.

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

Engineers
often rewrite Nyquist's equation for thermal electromotive force:

E2dv =
4kRTdv  as  E 2
= 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 E 2
equation with k, R and T in MKS units, and drop the unit-less 4, you obtain
Equation 1.

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

E2
= (V 2•s) = V 2 / (1/s),

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

E
= V/vHz

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

What is nV/√Hz Noise?

Comments (0)

Please log in or register to post comments.
By submitting this form, you accept the Mollom privacy policy.
  • Oldest First
  • Newest First
Loading Comments...