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Texas Instruments Incorporated Data Acquisition Calculating 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 Noise figure (NF) and the third-order intercept point (IP3) are used in receiver link budget analysis as a means 0 f = 80 MSPS to quantify the effects of device noise –20 S fIN =39MHz and nonlinearity on the sensitivity of the –40 SNR = 63.96 radio. Analog-to-digital converters (ADCs) SINAD = 63.3 are used in radio receivers to convert the –60 SFDR = 75.83 from the analog domain to the –80 THD = 71.78 digital domain. NF and IP3 typically are (dB) –100 not specified for the device, but equiva- –120 lent parameters are given whereby they 0 5 10 15 20 25 30 35 40 can be calculated. , f (MHz) ADCs specify signal-to-noise ratio (SNR) and two-tone, third-order inter- modulation (IMD3) under certain input signal and clocking condi- tions. With this information, NF and IP3 can be calculated. NF in ADCs In general, a low NF and high IP3 are desired. The actual There are a couple of ways to go about calculating the values required to meet the design goals depend on the input noise of an ADC, but using the SNR architecture of the system. specification is easy. Review of noise figure To measure SNR, a low-noise signal is input to the ADC, and the output is examined by taking a fast Fourier trans- Noise figure (NF) is the equivalent of noise factor form (FFT) or spectral plot. Figure 1 shows such a plot (F): NF (dB) = 10log(F). from Reference 1. The ratio of the signal to the noise inte- Noise factor of a device is the power ratio of the SNR at grated over half the sampling frequency (f /2) is the SNR. the input (SNR ) divided by the SNR at the output (SNR ): S I O Since the noise of the ADC is—to first-order approxima- SNR tion—independent of signal level, the higher the input F = I . (1) SNR level the better the SNR, up to a point. As the signal O approaches full scale (FS), spurious behavior begins to degrade the SNR. An input signal level 1 dB below full- The output signal (SO) is equal to the input signal (SI) × scale input (–1 dBFS) seems to give good results and is times the gain: SO = SI G. The output noise is equal to commonly used. the noise delivered to the input (NI) from the source plus To find the input , we divide the the input noise of the device (NA) times the gain: × signal level by the SNR divided by half the sampling fre- NO = (NI + NA) G. Substituting into Equation 1 and simplifying, we get quency (since SNR is calculated by dividing the signal by   the noise integrated over fS/2): SI   N (dBm/Hz) = –1 dBFS (dBm) + SNR (dBc) – f /2 (dBHz). SNRI  NI  NA A S F == =+1 . (2) SNR  GS×  N An ADC is a voltage-driven device, so we must choose O  I  I  GNN×+() an input resistance to find the signal power with the for- IA 2 Ω mula P = V /R. Assuming that FS = 2Vp-p and R = 50 , Assuming that the input is terminated in the same the full-scale input is +10 dBm. As an example of how to calculate, consider the following impedance as the source, NI = kT = –174 dBm/Hz, where k is Boltzman’s constant and T = 300 ). Once we find for the ADS5410, a 12-bit ADC. Given that fS = 80 MSPS, Ω the input noise spectral density of the device, it is a simple RIN = 50 , FS = 2Vp-p, and SNR = 63.96, then matter to plug it into Equation 2 and calculate F. NA (dBm/Hz) = +9 dBm – 63.96 dBc – 76.02 dBHz = –130.98 dBm/Hz.

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Analog Applications Journal 4Q 2003 www.ti.com/sc/analogapps Analog and Mixed-Signal Products Data Acquisition Texas Instruments Incorporated

To use Equation 2, we need to use the linear equivalents that the output (y) may have from a constant multiple (m)

of NI and NA: of the input (x) plus any constant offset (b). Expanding the nonlinear transfer functions of basic tran-  −+130. 98 174 FE=+110 = 20045, sistor circuits into a power series is a typical way to quantify   10 distortion products (see Reference 2). For example, typically have an exponential transfer function or NF (dB) = 43.02 dB. (i.e., collector current vs. base emitter voltage), y = ex, Looking at the result, we see that adding the 1 in where x is the input and y is the output. Expanding ex Equation 2 makes very little difference since the noise into a power series around x = 0 results in figure is so high. Therefore, using xxx234 x 5 xn exx =+++++1 ++L . NF (dB) = NA (dBm/Hz) – NI (dBm/Hz) 2 6 24 120 n! introduces little error. Figure 2 shows the function y = ex along with estimates It is common practice to use a transformer or a fully that use progressively more terms of the power series. differential op amp to drive high-performance ADCs differ- The farther x is from 0, the more terms are required to entially. This gives us the opportunity to use higher ADC estimate the value of ex properly. If x < 0.25, the linear input resistance. If NF is calculated based on 50 Ω, it is term 1 + x provides a close estimate of the actual function, reduced by log (impedance ratio). 10 and the circuit is linear. As x becomes larger, progressively For example, if we use a 1:4 impedance ratio (1:2 more terms (quadratic, cubic, and higher-order distortion turns ratio) transformer, the input resistance is 200 Ω to terms) are required to estimate ex properly. match to a 50-Ω drive . The NF is reduced by If the input to this circuit is a sinusoid—i.e., x = Asin(ωt) 10 × log (200/50) = 6 dB. Or, if we use a 1:16 impedance 10 —then the output ratio transformer with 800-Ω input resistance, NF is ω 2 2 ω 3 3 ω ... 12 dB lower. y = K0 + K1Asin( t) + K2A sin ( t) + K3A sin ( t) + ,

where K0, K1, etc. are constant scaling factors. Using the Review of third-order intercept point (IP3) trigonometric identities Due to nonlinearity in the transfer function of all electron- 12− cos(ωt ) ics, distortion is generated. With reference to the formula sin2()ωt = and of a straight line, y = b + mx, nonlinearity is any deviation 2

Figure 2. Function y = ex and its power series estimates

2 3 4 5 y5(x)=1+x+––+––+––+–––x x x x 8 2 6 24 120

x 2 3 4 y(x) = e y4(x)=1+x+––+––+––x x x 7 2 6 24

2 3 y3(x)=1+x+––+––x x 2 6 6

2 5 y2(x)=1+x+––x 2 y

4

3 y1(x)=1+x

2

1 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2

x

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Analog and Mixed-Signal Products www.ti.com/sc/analogapps 4Q 2003 Analog Applications Journal Texas Instruments Incorporated Data Acquisition

33sin(ωωtt )− sin( ) shows that the quadratic terms give rise to HD2 and sin3 (ωt ) = second-order intermodulation distortion (IMD ). 4 2 Expanding the fourth term, we get shows that the quadratic and cubic terms give rise to ω ω 3 = K3[A1sin( 1t) + A2sin( 2t)] second- and third-order distortion (HD and 3 3 ω 2 ω ω 2 = K3[A1sin ( 1t) + 3A1A2sin ( 1t)sin( 2t) HD3, respectively). Similarly, higher-order terms give rise + 3A A sin(ω t)sin2(ω t) + A3sin3(ω t)]. to higher-order harmonic distortion. 1 2 1 2 2 2 Using the trigonometric identities If the input is comprised of two tones—i.e., ω ω 33sin(ωωtt )− sin( ) x = A1sin( 1t) + A2sin( 2t)—then the output sin3()ωt = and 4 y = K0 ω ω + K1[A1sin( 1t) + A2sin( 2t)] 2sin(ωωωωω t)−+−+ sin(2 t t) sin(2 t t) ω ω 2 sin2 (ωω t)sin( t) = 21221 + K2[A1sin( 1t) + A2sin( 2t)] 12 ω ω 3 4 + K3[A1sin( 1t) + A2sin( 2t)] + ..., shows that the cubic terms give rise to HD3 and third-order intermodulation distortion (IMD ). Similarly, higher-order where K , K , etc. are constant scaling factors. 3 0 1 terms give rise to higher-order harmonic and intermodula- Expanding the third term, we get tion distortion. ω ω 2 = K2[A1sin( 1t) + A2sin( 2t)] Table 1 shows the of the distortion products 2 2 ω ω ω 2 2 ω = K2[A1sin ( 1t) + 2A1A2sin( 1t)sin( 2t) + A2sin ( 2t)]. that will be generated due to second- and third-order non- Using the trigonometric identities linearity, given a two-tone input at frequencies f1 and f2. 12− cos(ωt ) sin2()ωt = and 2 Table 1. Distortion product frequencies due to second- and third-order nonlinearity

cos(tωω−− t) cos (t ωω + t) SECOND-ORDER FREQUENCIES THIRD-ORDER FREQUENCIES sin(ωω t)sin( t) = 12 12 2f 2f 3f 3f 12 2 1 2 1 2 f1 – f2 f1 + f2 2f1 – f2 2f2 – f1 2f1 + f2 2f2 + f1

So now the question arises: Why Figure 3. Input and output two-tone and intermodulation distortion is all this important? The answer is that radio specifications for GSM, –1-dB CDMA2000, WCDMA, and the like Compression all call for sensitivity requirements Point, P1 IP3 (dBm) to be met with two interfering spaced in the frequency

OIP2 (dBm) IP2 (dBm) domain such that their third-order intermodulation product will fall OIP (dBm) 3 on top of the signal of interest. 1dB 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 POUT (dBm) determine whether or not the spec can be met. If the input and output power Fundamental 1 of two tones applied to a device 1 and their intermodulation prod- ucts are graphed on a log-log 1 2 3 scale as shown in Figure 3, the fundamental tones have a slope of 1 IMD2 1, the second-order product has a IMD3 slope of 2, and the third-order products have a slope of 3. The IIP2 (dBm) IIP3 (dBm)

PIN (dBm)

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Analog Applications Journal 4Q 2003 www.ti.com/sc/analogapps Analog and Mixed-Signal Products Data Acquisition Texas Instruments Incorporated

device will go into compression before the lines intersect. Subtracting two arbitrary points on each line and The point where the output power is reduced by 1 dB rearranging gives us from what is expected is called the 1-dB compression ⇒ y2 – y3 = x2 – x3 y2 = y3 + x2 – x3 for L1, and point (P ). By extending the lines, the second- and third- ⇒ 1 y1 – y3 = 3x2 – 3x3 y1 = y3 + 3x2 – 3x3 for L3. order intercept points (IP2 and IP3, respectively) can be Subtracting again results in found. If they are referred to the input, they are called y2 = y3 + x2 – x3 input intercept points (IIP2, IIP3); and if they are referred to the output, they are called output intercept points –(y1 = y3 + 3x2 – 3x3) y – y = 2(x – x ). (OIP2, OIP3). 2 1 3 2 Since we are interested in intermodulation distortion ↑ ↑ ↑ relative to the carriers, why should we concern ourselves IMD3 (dBc) IIP3 (dBm) PIN with some fictitious point that the amplifier will never reach? The answer is that there is a mathematical relationship From this it is seen that between the two. Given the intercept point, we can calcu- IMD3 () dBc IIP() dBm=− P () dBm . (3) late the intermodulation product for any input/output power. 3 IN 2 Given that the slopes are known, equations for slopes L1 Once we find IMD3 and know the input power, it is a simple and L3 are written as shown in Figure 4. 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

OIP3 (dBm) y 3 IP3 (dBm)

y 2

P=yOUT

IMD3 (dBc)

1

1 3

y 1 IIP3 (dBm) 1

x x 2 3

P=xIN

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Analog and Mixed-Signal Products www.ti.com/sc/analogapps 4Q 2003 Analog Applications Journal Texas Instruments Incorporated Data Acquisition

Figure 5. Two-tone FFT or spectral plot from Reference 1

0

–20 fS = 80 MSPS f 1 = 15.17 MHz –40 IN fIN2 = 15.89 MHz –60 IMD=77dB –80

Amplitude (dB) –100 –120 0 5 10 15 20 25 30 35 40 Frequency, f (MHz)

× IP in ADCs 10 log10(200/50) = 6 dB. Or, if we use a 1:16 impedance 3 Ω In ADC testing, two tones are applied to the ADC, and the ratio transformer with 800- input resistance, IIP3 is output is examined by taking an FFT or spectral plot to 12 dB lower. find the two-tone IMD . Figure 5 shows such a plot from 3 Conclusion Reference 1. The ratio of each of the tones to the IMD 3 We have examined typical ADC noise and distortion speci- product(s) is the two-tone IMD in dBc. IMD depends on 3 3 fications to see how they relate to NF and IP . It is seen signal level. Too high a signal level results in excessive 3 that the required information to calculate NF and IP is distortion, and too low a signal level makes distortion hard 3 contained in a typical ADC data sheet. to detect in the presence of noise and other spurious A key point to remember is that an ADC is a voltage- components. An input signal level 7 dB below full-scale driven device, whereas NF and IP are associated with input (–7 dBFS) seems to give good results and is 3 power. Thus, in order for the calculations to proceed, an commonly used. impedance is imposed on the ADC input to find the corre- Since an ADC is a voltage-driven device, we must sponding power levels. choose an input resistance to find the signal power with 2 the formula P = V /R. Assuming that FS = 2Vp-p and References Ω R = 50 , the full-scale input is +10 dBm. With the input For more information related to this article, you can down- power and the IMD , Equation 3 is used to find IIP . 3 3 load an Acrobat Reader file at www-s.ti.com/sc/techlit/ As an example of how to calculate, consider the follow- litnumber and replace “litnumber” with the TI Lit. # for ing for the ADS5410, a 12-bit ADC. Given that FS = 2V , p-p the materials listed below. Ω RIN = 50 , and IMD3 = 77 dBc, then Document Title TI Lit. # −77 dBc IIP=−3 dBm = 41.. 5 dBm 1. “12-Bit, 80 MSPS CommsADCTM Analog- 3 2 to-Digital Converter,” Data Sheet, p. 7 ...... slas346 As mentioned earlier, it is common practice to use a 2. Piet Wambacq and Willy Sansen, Distortion transformer or a fully differential op amp to drive high- Analysis of Analog Integrated Circuits performance ADCs differentially. If 50 Ω is originally used (Kluwer Academic Publishers, 1998). as shown in the example, then the IIP3 is reduced by × Related Web sites 10 log10(impedance ratio). For example, if we use a 1:4 impedance ratio (1:2 analog.ti.com turns ratio) transformer, the input resistance is 200 Ω to www.ti.com/sc/device/ADS5410 Ω match to a 50- drive amplifier. The IIP3 is reduced by

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