Op amp input current noise
Sergio Franco - June 14, 2013
Lately, EDN has published a number of well-written blogs addressing op amp noise and its computer
simulation from a practical and useful standpoint:
Op Amp Noise—the non-inverting amplifier
Op Amp Noise—but what about the feedback?
1/f Noise—the flickering candle
Simulating op amp noise
Visualize op amp noise
Op amp noise revisited - the nuts and bolts
I have called the attention to the fact that when performing noise calculations, we better
treat the inverting-input and noninverting-input current noise densities (call them inn and inp)
separately in order to avoid possible errors. Though this is a definite requirement in the case of
amplifiers with asymmetric inputs such as current-feedback amps and certain audio amplifier
topologies, it may or may not be so in the case of voltage-feedback amplifiers, depending on the
relative sizes of the equivalent external resistances seen by the inverting and noninverting inputs
(call them Rn and Rp, respectively). Let us investigate by means of some SPICE simulations.
The basis is offered by the test circuit of Fig. 1, which uses a Laplace block to simulate a 1MHz op amp configured for a noise gain of NG = (1 + R2/R1) = (1 + 103/10) = 101 V/V. Since the
closed-loop bandwidth is (1 MHz)/101 = 9.9 kHz, the white-noise equivalent bandwidth is NEB =
(π/2)x9.9x103 = 15.55 kHz (for simplicity only white noise is considered here).
The input noise currents are generated separately by playing the Johnson noise currents of Rwp and
Rwn into the current-controlled current sources Fp and Fn, which then establish the densities inp and inn
at the input pins of the op amp. The value of 16.57 kW for Rwp and Rwn has been chosen so as to
ensure inp = inn = 1
. To make current noise prevail, the input pins are terminated on the
deliberately large (1.0 MΩ) resistors Rp and Rn. These resistors themselves contribute Johnson noise
in the amount of
each. However, when combined
in rms fashion with the contribution by either inn or inp, which is (1 MΩ) x(1 pA) = 1,000
,
Johnson noise pales by comparison (this is precisely why Rp and Rn have been chosen so large). By
this artifice, we can ignore also the op amp’s input noise voltage en, which is typically much less than
128.7
.
Figure 1 – Rms output noise with two uncorrelated input noise currents inn and inp (Eno =
18.0 mV rms).
Figure 2 – Rms output noise with a single noise current inacross the inputs (Eno(wrong) = 25.2
mV rms)
Ignoring the Johnson noise of Rp and Rn, we calculate the total rms output noise as
This agrees with PSpice’s value of 18.0 mV, provided we include also Johnson noise. Rerunning
PSpice with either Rp = 0 or Rn = 0, we get Eno = 12.7 mV, which is
(or 3 dB) lower than 18.0 mV.
A common error is to model current noise with a single current in across the inputs as in Fig.
2. In this case Eq. (1) changes to
which is incorrect because it is
(or 3 dB) higher than the correct value of 18 mV. However, for
circuits having either Rp << Rn, or Rp >> Rn, or Rp and Rn very small, the model of Fig. 2 gives
correct results, even though it is conceptually inaccurate. For an example, see p. 11 of the LT1028
data sheets , where Rp = Rn = 100 Ω (very small).
The question has been posed as to whether the input currents i np and i nn are totally
uncorrelated or, if they exhibit a certain degree of correlation, how to express and how to measure
this degree. The extreme case of perfect correlation is depicted in Fig. 3, which uses a single
Johnson source Rw to generate a pair of noise currents denoted by the common symbol inc. In this
case the common-mode components Rpinc and Rninc cancel each other out, leaving only the Johnson
noise of Rp and Rn, so
Figure 3 – Rms output rms noise with perfectly correlated input noise currents inc (Eno = 2.3
mV rms).
Figure 4 – CMOS and bipolar input-stage examples.
Evidently, the Johnson noise of Rp and Rn represents a form of tare for the noise measurements
under consideration. Note, incidentally, that the error posed by the model of Fig. 2 is maximized in
the case of purely correlated noise currents because instead of canceling each other out, they
reinforce each other!
The degree of correlation depends on circuit topology and technology. Consider the two
representative cases of Fig. 4. In the CMOS case, the input noise is established by the reversebiased protective diodes. Even though the dc currents of the diodes associated with the same input
tend to cancel each other out, their noise densities combine in rms fashion. Moreover, they can be
assumed to be totally uncorrelated, as the diodes are physically different. By contrast, in the bipolar
case we have (a) the noise inee generated by the emitter biasing circuitry, which splits equally
between the two matched BJTs to appear as a common-mode input component inc, and (b) the noises
indp and indn generated independently by the BJTs themselves, which will be assumed to be
uncorrelated (should there be any correlation, it will be included in inc). The overall input noise
currents are thus
Figure 5 shows a PSpice circuit designed to simulate Eq. (4). In this example the Johnson
resistances have been chosen for inc = 0.6
and indp = indn = 0.8
, so 0.62 + 0.82 = 1
, as before. With Rp = Rn = 1.0 MΩ, the correlated noise contributions cancel each other out
to give Eno1 = 14.4 mV rms (instead of the value Eno = 18.0 mV of Fig. 1). This represents the
uncorrelated noise contribution, augmented by the tare. However, if we rerun PSpice with Rp = 0,
we no longer have correlated noise cancellation, so the result of Eno2 = 12.7 mV rms represents both
the correlated and uncorrelated noise contributions, again augmented by the tare. We can exploit
the information provided by the two measurements to infer the correlated and uncorrelated noise
components as follows:
●
First, measure the noise Eno1 with both Rn and Rp present. For Rp = Rn we have
Figure 5 – PSpice circuit simulating Eq. (4) with inc = 0.6
.
and indp = indn = 0.8
Figure 6 – (a) Rp = Rn = 1 MΩ, Eno1 = 14.4 mV rms. (b) Rp = 0, Rn = 1 MΩ, Eno2 = 12.7 mV
rms.
Solving for ind gives
●
Next, measure the noise Eno2 with Rp = 0. This noise is
Eliminating
and solving for inc gives
Example: A certain op amp circuit has NG = 50 V/V and NEB = 200 kHz. If the circuit gives Eno1 =
10 mV rms with Rp = Rn = 1.0 MΩ, and Eno2 = 11 mV rms with Rp = 0 and Rn = 1.0 MΩ, find ind, inc,
and in.
Solution:
The above considerations apply to white noise only. As we know, op amps exhibit also 1/f
noise, but we can reduce its effect by performing the rms measurements through a suitable highpass filter. Alternatively, since white noise power increases in proportion to the NEB whereas 1/f
noise power increases with NEB only logarithmically, we can reconfigure the circuit for a sufficiently
higher NEB to render the effect of 1/f noise truly negligible compared to that of white noise.