Download Calculating noise figure and third-order

Data Acquisition
Texas Instruments Incorporated
Calculating noise figure and third-order
intercept in ADCs
By James Karki (Email: [email protected])
Member, Group Technical Staff, High-Performance Linear
Introduction
Figure 1. ADS5410 FFT or spectral plot from Reference 1
Amplitude (dB)
Noise figure (NF) and the third-order
intercept point (IP3) are used in radio
0
receiver link budget analysis as a means
to quantify the effects of device noise
–20
and nonlinearity on the sensitivity of the
– 40
radio. Analog-to-digital converters (ADCs)
–60
are used in radio receivers to convert the
signal from the analog domain to the
–80
digital domain. NF and IP3 typically are
–100
not specified for the device, but equiva–120
lent parameters are given whereby they
0
can be calculated.
ADCs specify signal-to-noise ratio
(SNR) and two-tone, third-order intermodulation distortion (IMD3) under
certain input signal and clocking conditions. With this information, NF and IP3 can be calculated.
In general, a low NF and high IP3 are desired. The actual
values required to meet the design goals depend on the
architecture of the system.
Review of noise figure
Noise figure (NF) is the decibel equivalent of noise factor
(F): NF (dB) = 10log(F).
Noise factor of a device is the power ratio of the SNR at
the input (SNRI) divided by the SNR at the output (SNRO):
F=
SNRI
SNRO
(1)
.
The output signal (SO) is equal to the input signal (SI)
times the gain: SO = SI × G. The output noise is equal to
the noise delivered to the input (NI) from the source plus
the input noise of the device (NA) times the gain:
NO = (NI + NA) × G. Substituting into Equation 1 and
simplifying, we get
SI




N
I
 = 1 + NA .
F=
=
G
×
S


SNRO
NI
I
 G × (N + N ) 


I
A 
SNRI
(2)
Assuming that the input is terminated in the same
impedance as the source, NI = kT = –174 dBm/Hz, where k
is Boltzman’s constant and T = 300 Kelvin). Once we find
the input noise spectral density of the device, it is a simple
matter to plug it into Equation 2 and calculate F.
fS = 80 MSPS
fIN = 39 MHz
SNR = 63.96
SINAD = 63.3
SFDR = 75.83
THD = 71.78
5
10
15
20
25
30
35
40
Frequency, f (MHz)
NF in ADCs
There are a couple of ways to go about calculating the
input noise spectral density of an ADC, but using the SNR
specification is easy.
To measure SNR, a low-noise signal is input to the ADC,
and the output is examined by taking a fast Fourier transform (FFT) or spectral plot. Figure 1 shows such a plot
from Reference 1. The ratio of the signal to the noise integrated over half the sampling frequency (fS/2) is the SNR.
Since the noise of the ADC is—to first-order approximation—independent of signal level, the higher the input
level the better the SNR, up to a point. As the signal
approaches full scale (FS), spurious behavior begins to
degrade the SNR. An input signal level 1 dB below fullscale input (–1 dBFS) seems to give good results and is
commonly used.
To find the input noise spectral density, we divide the
signal level by the SNR divided by half the sampling frequency (since SNR is calculated by dividing the signal by
the noise integrated over fS/2):
NA (dBm/Hz) = –1 dBFS (dBm) + SNR (dBc) – fS/2 (dBHz).
An ADC is a voltage-driven device, so we must choose
an input resistance to find the signal power with the formula P = V2/R. Assuming that FS = 2Vp-p and R = 50 Ω,
the full-scale input is +10 dBm.
As an example of how to calculate, consider the following
for the ADS5410, a 12-bit ADC. Given that fS = 80 MSPS,
RIN = 50 Ω, FS = 2Vp-p, and SNR = 63.96, then
NA (dBm/Hz) = +9 dBm – 63.96 dBc – 76.02 dBHz
= –130.98 dBm/Hz.
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To use Equation 2, we need to use the linear equivalents
of NI and NA:
NF (dB) = NA (dBm/Hz) – NI (dBm/Hz)
that the output (y) may have from a constant multiple (m)
of the input (x) plus any constant offset (b).
Expanding the nonlinear transfer functions of basic transistor circuits into a power series is a typical way to quantify
distortion products (see Reference 2). For example,
transistors typically have an exponential transfer function
(i.e., collector current vs. base emitter voltage), y = ex,
where x is the input and y is the output. Expanding ex
into a power series around x = 0 results in
x 2 x 3 x 4 x5
xn
ex = 1 + x +
+
+
+
+L +
.
2
6 24 120
n!
introduces little error.
It is common practice to use a transformer or a fully
differential op amp to drive high-performance ADCs differentially. This gives us the opportunity to use higher ADC
input resistance. If NF is calculated based on 50 Ω, it is
reduced by log10(impedance ratio).
For example, if we use a 1:4 impedance ratio (1:2
turns ratio) transformer, the input resistance is 200 Ω to
match to a 50-Ω drive amplifier. The NF is reduced by
10 × log10(200/50) = 6 dB. Or, if we use a 1:16 impedance
ratio transformer with 800-Ω input resistance, NF is
12 dB lower.
Figure 2 shows the function y = ex along with estimates
that use progressively more terms of the power series.
The farther x is from 0, the more terms are required to
estimate the value of ex properly. If x < 0.25, the linear
term 1 + x provides a close estimate of the actual function,
and the circuit is linear. As x becomes larger, progressively
more terms (quadratic, cubic, and higher-order distortion
terms) are required to estimate ex properly.
If the input to this circuit is a sinusoid—i.e., x = Asin(ωt)
—then the output
y = K + K Asin(ωt) + K A2sin2(ωt) + K A3sin3(ωt) + ...,
 −130.98 + 174 
F = 1 + 10E
 = 20045,

10
or NF (dB) = 43.02 dB.
Looking at the result, we see that adding the 1 in
Equation 2 makes very little difference since the noise
figure is so high. Therefore, using
Review of third-order intercept point (IP3)
Due to nonlinearity in the transfer function of all electronics, distortion is generated. With reference to the formula
of a straight line, y = b + mx, nonlinearity is any deviation
0
1
2
3
where K0, K1, etc. are constant scaling factors. Using the
trigonometric identities
1 − cos(2ωt)
sin2(ωt) =
and
2
Figure 2. Function y = ex and its power series estimates
x2 + ––
x5
x4 + –––
x3 + ––
y5(x) = 1 + x + ––
2
6 24 120
8
x
y(x) = e
7
x3 + ––
x2 + ––
x4
y4(x) = 1 + x + ––
6 24
2
2
3
x
x + ––
y3(x) = 1 + x + ––
6
2
6
x2
y2(x) = 1 + x + ––
2
5
y
4
y1(x) = 1 + x
3
2
1
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
x
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sin3(ωt) =
shows that the quadratic terms give rise to HD2 and
second-order intermodulation distortion (IMD2).
Expanding the fourth term, we get
3 sin(ωt) − sin(3ωt)
4
shows that the quadratic and cubic terms give rise to
second- and third-order harmonic distortion (HD2 and
HD3, respectively). Similarly, higher-order terms give rise
to higher-order harmonic distortion.
If the input is comprised of two tones—i.e.,
x = A1sin(ω1t) + A2sin(ω2t)—then the output
y = K0
+ K1[A1sin(ω1t) + A2sin(ω2t)]
+ K2[A1sin(ω1t) + A2sin(ω2t)]2
+ K3[A1sin(ω1t) + A2sin(ω2t)]3
+ ...,
Using the trigonometric identities
3 sin(ωt) − sin(3ωt)
and
sin3(ωt) =
4
sin 2(ω1t)sin(ω2t) =
where K0, K1, etc. are constant scaling factors.
Expanding the third term, we get
= K2[A1sin(ω1t) + A2sin(ω2t)]2
= K2[A12sin2(ω1t) + 2A1A2sin(ω1t)sin(ω2t) + A22sin2(ω2t)].
Using the trigonometric identities
1 − cos(2ωt)
and
sin2(ωt) =
2
sin(ω1t)sin(ω2t) =
= K3[A1sin(ω1t) + A2sin(ω2t)]3
= K3[A13sin3(ω1t) + 3A1A2sin2(ω1t)sin(ω2t)
+ 3A1A2sin(ω1t)sin2(ω2t) + A23sin3(ω2t)].
shows that the cubic terms give rise to HD3 and third-order
intermodulation distortion (IMD3). Similarly, higher-order
terms give rise to higher-order harmonic and intermodulation distortion.
Table 1 shows the frequencies of the distortion products
that will be generated due to second- and third-order nonlinearity, given a two-tone input at frequencies f1 and f2.
SECOND-ORDER FREQUENCIES
2f1
2f2
f1 – f2
f1 + f2
2
Figure 3. Input and output two-tone and intermodulation distortion
–1-dB
Compression
Point, P1
IP3 (dBm)
OIP2 (dBm)
IP2 (dBm)
OIP3 (dBm)
1 dB
POUT (dBm)
Fundamental
1
1
3
2
IMD 2
4
Table 1. Distortion product frequencies due to
second- and third-order nonlinearity
cos(ω1t − ω2t) − cos(ω1t + ω2t)
1
2sin(ω2t) − sin(2ω1t + ω2t) − sin(2ω2t + ω1t)
1
IMD3
IIP2 (dBm)
IIP3 (dBm)
THIRD-ORDER FREQUENCIES
3f1
3f2
2f1 – f2
2f2 – f1
2f1 + f2
2f2 + f1
So now the question arises: Why
is all this important? The answer
is that radio specifications for GSM,
CDMA2000, WCDMA, and the like
all call for sensitivity requirements
to be met with two interfering
signals spaced in the frequency
domain such that their third-order
intermodulation product will fall
on top of the signal of interest.
The third-order intermodulation
point is used to quantify how
much distortion is generated.
Referring this to the antenna
input provides an easy method to
determine whether or not the
spec can be met.
If the input and output power
of two tones applied to a device
and their intermodulation products are graphed on a log-log
scale as shown in Figure 3, the
fundamental tones have a slope of
1, the second-order product has a
slope of 2, and the third-order
products have a slope of 3. The
PIN (dBm)
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device will go into compression before the lines intersect.
The point where the output power is reduced by 1 dB
from what is expected is called the 1-dB compression
point (P1). By extending the lines, the second- and thirdorder intercept points (IP2 and IP3, respectively) can be
found. If they are referred to the input, they are called
input intercept points (IIP2, IIP3); and if they are referred
to the output, they are called output intercept points
(OIP2, OIP3).
Since we are interested in intermodulation distortion
relative to the carriers, why should we concern ourselves
with some fictitious point that the amplifier will never reach?
The answer is that there is a mathematical relationship
between the two. Given the intercept point, we can calculate the intermodulation product for any input/output power.
Given that the slopes are known, equations for slopes L1
and L3 are written as shown in Figure 4.
Subtracting two arbitrary points on each line and
rearranging gives us
⇒ y2 = y3 + x2 – x3 for L1, and
y2 – y3 = x2 – x3
y1 – y3 = 3x2 – 3x3 ⇒ y1 = y3 + 3x2 – 3x3 for L3.
Subtracting again results in
y2 = y3 + x2 – x3
–(y1 = y3 + 3x2 – 3x3)
y2 – y1 = 2(x3 – x2).
↑
↑
IMD3 (dBc)
↑
IIP3 (dBm) PIN
From this it is seen that
IIP3 (dBm) = PIN (dBm) −
IMD3 (dBc)
2
.
(3)
Once we find IMD3 and know the input power, it is a simple
matter to plug them into Equation 3 and calculate IIP3.
Figure 4. Straight-line relationship between IMD3 and the fundamental
L3: y = 3x + b3
L1: y = x + b1
y3
OIP3 (dBm)
IP3 (dBm)
y2
POUT = y
IMD3 (dBc)
1
3
1
y1
IIP3 (dBm)
1
x2
x3
PIN = x
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Figure 5. Two-tone FFT or spectral plot from Reference 1
Amplitude (dB)
0
fS = 80 MSPS
fIN1 = 15.17 MHz
fIN2 = 15.89 MHz
IMD = 77 dB
–20
– 40
–60
–80
–100
–120
0
5
10
15
20
25
30
35
40
Frequency, f (MHz)
IP3 in ADCs
In ADC testing, two tones are applied to the ADC, and the
output is examined by taking an FFT or spectral plot to
find the two-tone IMD3. Figure 5 shows such a plot from
Reference 1. The ratio of each of the tones to the IMD3
product(s) is the two-tone IMD3 in dBc. IMD3 depends on
signal level. Too high a signal level results in excessive
distortion, and too low a signal level makes distortion hard
to detect in the presence of noise and other spurious
components. An input signal level 7 dB below full-scale
input (–7 dBFS) seems to give good results and is
commonly used.
Since an ADC is a voltage-driven device, we must
choose an input resistance to find the signal power with
the formula P = V2/R. Assuming that FS = 2Vp-p and
R = 50 Ω, the full-scale input is +10 dBm. With the input
power and the IMD3, Equation 3 is used to find IIP3.
As an example of how to calculate, consider the following for the ADS5410, a 12-bit ADC. Given that FS = 2Vp-p,
RIN = 50 Ω, and IMD3 = 77 dBc, then
IIP3 = 3 dBm −
−77 dBc
= 41.5 dBm .
2
As mentioned earlier, it is common practice to use a
transformer or a fully differential op amp to drive highperformance ADCs differentially. If 50 Ω is originally used
as shown in the example, then the IIP3 is reduced by
10 × log10(impedance ratio).
For example, if we use a 1:4 impedance ratio (1:2
turns ratio) transformer, the input resistance is 200 Ω to
match to a 50-Ω drive amplifier. The IIP3 is reduced by
10 × log10(200/50) = 6 dB. Or, if we use a 1:16 impedance
ratio transformer with 800-Ω input resistance, IIP3 is
12 dB lower.
Conclusion
We have examined typical ADC noise and distortion specifications to see how they relate to NF and IP3. It is seen
that the required information to calculate NF and IP3 is
contained in a typical ADC data sheet.
A key point to remember is that an ADC is a voltagedriven device, whereas NF and IP3 are associated with
power. Thus, in order for the calculations to proceed, an
impedance is imposed on the ADC input to find the corresponding power levels.
References
For more information related to this article, you can download an Acrobat Reader file at www-s.ti.com/sc/techlit/
litnumber and replace “litnumber” with the TI Lit. # for
the materials listed below.
Document Title
TI Lit. #
TM
1. “12-Bit, 80 MSPS CommsADC Analogto-Digital Converter,” Data Sheet, p. 7 . . . . . . . .slas346
2. Piet Wambacq and Willy Sansen, Distortion
Analysis of Analog Integrated Circuits
(Kluwer Academic Publishers, 1998).
Related Web sites
analog.ti.com
www.ti.com/sc/device/ADS5410
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