Lecture 9: Intercept Point, Gain Compression and ... - Ali M. Niknejad

EECS 142Lecture 9: Intercept Point, Gain Compression andBlockingProf. Ali M. NiknejadUniversity of California, BerkeleyCopyright c○ 2005 by Ali M. NiknejadA. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 1/29

Gain CompressiondV odV idV odV iV oV oV iV iThe large signal input/output relation can display gaincompression or expansion. Physically, most amplifierexperience gain compression for large signals.The small-signal gain is related to the slope at a givenpoint. For the graph on the left, the gain decreases forincreasing amplitude.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 2/29

1 dB Compression PointP o,−1dB∣ V o ∣∣∣∣ V iV iP i,−1dBGain compression occurs because eventually theoutput signal (voltage, current, power) limits, due to thesupply voltage or bias current.If we plot the gain (log scale) as a function of the inputpower, we identify the point where the gain has droppedby 1 dB. This is the 1 dB compression point. It’s a veryimportant number to keep in mind.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 3/29

Apparent GainRecall that around a small deviation, the large signalcurve is described by a polynomials o = a 1 s i + a 2 s 2 i + a 3 s 3 i + · · ·For an input s i = S 1 cos(ω 1 t), the cubic term generatesS 3 1 cos 3 (ω 1 t) = S 3 1 cos(ω 1 t) 1 2 (1 + cos(2ω 1t))( 1= S1 3 2 cos(ω 1t) + 2 )4 cos(ω 1t) cos(2ω 1 t)Recall that 2 cosacos b = cos(a + b) + cos(a − b)( 1= S1 3 2 cos(ω 1t) + 1 )4 (cos(ω 1t) + cos(3ω 1 t))A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 4/29

Apparent Gain (cont)Collecting terms( 3= S1 3 4 cos(ω 1t) + 1 )4 cos(3ω 1t)The apparent gain of the system is thereforG = S o,ω 1S i,ω1= a 1S 1 + 3 4 a 3S 3 1S 1= a 1 + 3 4 a 3S 2 1 = a 1(1 + 3 4)a 3S12 a 1= G(S 1 )If a 3 /a 1 < 0, the gain compresses with increasingamplitude.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 5/29

1-dB Compression PointLet’s find the input level where the gain has dropped by1 dB(20 log 1 + 3 )a 3S12 = −1 dB4 a 1S 1 =√4334a 3a 1S 2 1 = −0.11∣ a 1∣∣∣∣ × √ 0.11 = IIP3 − 9.6 dBa 3The term in the square root is called the third-orderintercept point (see next few slides).A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 6/29

Intercept Point IP 2P out(dBm)0OIP 2IP 2-10-20-30Fund2nd20 dBc10 dBc-40-50 -40 -30 -20 -10(dBm)IIP 2P inThe extrapolated point where IM 2 = 0 dBc is known asthe second order intercept point IP 2 .A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 7/29

Properties of Intercept Point IP 2Since the second order IM distortion products increaselike s 2 i , we expect that at some power level the distortionproducts will overtake the fundamental signal.The extrapolated point where the curves of thefundamental signal and second order distortion productsignal meet is the Intercept Point (IP 2 ).At this point, then, by definition IM 2 = 0 dBc.The input power level is known as IIP 2 , and the outputpower when this occurs is the OIP 2 point.Once the IP 2 point is known, the IM 2 at any otherpower level can be calculated. Note that for a dBback-off from the IP 2 point, the IM 2 improves dB for dBA. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 8/29

Intercept Point IP 3P out(dBm)10OIP 3IP 30-10-20-30dBcFund20dBcThird-50 -40 -30 -20 -10(dBm)IIP 3P inThe extrapolated point where IM 3 = 0 dBc is known asthe third-order intercept point IP 3 .A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 9/29

Properties of Intercept Point IP 3Since the third order IM distortion products increase likes 3 i , we expect that at some power level the distortionproducts will overtake the fundamental signal.The extrapolated point where the curves of thefundamental signal and third order distortion productsignal meet is the Intercept Point (IP 3 ).At this point, then, by definition IM 3 = 0 dBc.The input power level is known as IIP 3 , and the outputpower when this occurs is the OIP 3 point.Once the IP 3 point is known, the IM 3 at any otherpower level can be calculated. Note that for a 10 dBback-off from the IP 3 point, the IM 3 improves 20 dB.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 10/29

Intercept Point ExampleFrom the previous graph we see that our amplifier hasan IIP 3 = −10 dBm.What’s the IM 3 for an input power of P in = −20 dBm?Since the IM 3 improves by 20 dB for every 10 dBback-off, it’s clear that IM 3 = 20 dBcWhat’s the IM 3 for an input power of P in = −110 dBm?Since the IM 3 improves by 20 dB for every 10 dBback-off, it’s clear that IM 3 = 200 dBcA. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 11/29

Calculated IIP2/IIP3We can also calculate the IIP points directly from ourpower series expansion. By definition, the IIP2 pointoccurs whenIM 2 = 1 = a 2a 1S iSolving for the input signal levelIIP 2 = S i = a 1a 2In a like manner, we can calculate IIP 3IM 3 = 1 = 3 4a 3a 1S 2 i IIP 3 = S i =√43∣ a 1∣∣∣∣a 3A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 12/29

channelInterferenceBlocker or JammerSignalLNAConsider the input spectrum of a weak desired signaland a “blocker”S i = S 1 cos ω 1 t} {{ }Blocker+s 2 cos ω 2 t} {{ }DesiredWe shall show that in the presence of a stronginterferer, the gain of the system for the desired signalis reduced. This is true even if the interference signal isat a substantially difference frequency. We call thisinterference signal a “jammer”.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 13/29

Blocker (II)Obviously, the linear terms do not create any kind ofdesensitization. The second order terms, likewise,generate second harmonic and intermodulation, but notany fundamental signals.In particular, the cubic term a 3 Si 3desensitization termgenerates the jammerS 3 i = S 3 1 cos 3 ω 1 t + s 3 2 cos 3 ω 2 t + 3S 2 1s 2 cos 2 ω 1 t cos ω 2 t+3s 2 1S 2 cos 2 ω 2 t cos ω 1 tThe first two terms generate cubic and third harmonic.The last two terms generate fundamental signals at ω 1and ω 2 . The last term is much smaller, though, sinces 2 ≪ S 1 .A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 14/29

Blocker (III)The blocker term is therefore given bya 3 3S 2 1s 212 cos ω 2tThis term adds or subtracts from the desired signal.Since a 3 < 0 for most systems (compressivenon-linearity), the effect of the blocker is to reduce thegainApp Gain = a 1s 2 + a 332S 2 1 s 2s 2= a 1 + a 332 S2 1 = a 1(1 + 3 2)a 3S12 a 1A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 15/29

Out of Band 3 dB DesensitizationLet’s find the blocker power necessary to desensitizethe amplifier by 3 dB. Solving the above equation(20 log 1 + 3 )a 3S12 = −3 dB2 a 1We find that the blocker power is given byP OB = P −1 dB+ 1.2 dBIt’s now clear that we should avoid operating ouramplifier with any signals in the vicinity of P −1 dB , sincegain reduction occurs if the signals are larger. At thissignal level there is also considerable intermodulationdistortion.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 16/29

Series InversionOften it’s easier to find a power series relation for theinput in terms of the output. In other wordsS i = a 1 S o + a 2 S 2 o + a 3 S 3 o + · · ·But we desire the inverse relationS o = b 1 S i + b 2 S 2 i + b 3 S 3 i + · · ·To find the inverse relation, we can substitute the aboveequation into the original equation and equatecoefficient of like powers.S i = a 1 (b 1 S i + b 2 S 2 i + b 3 S 3 i + · · · ) + a 2 ( ) 2 + a 3 ( ) 3 + · · ·A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 17/29

Inversion (cont)Equating linear terms, we find, as expected, thata 1 b 1 = 1, or b 1 = 1/a 1 .Equating the square terms, we have0 = a 1 b 2 + a 2 b 2 1b 2 = − a 2b 2 1a 1= − a 2a 3 1Finally, equating the cubic terms we have0 = a 1 b 3 + a 2 2b 1 b 2 + a 3 b 3 1b 3 = 2a2 2a 5 1− a 3a 4 1It’s interesting to note that if one power series does nothave cubic, a 3 ≡ 0, the inverse series has cubic due tothe first term above.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 18/29

CascadeIIP2IIP3G A VG A PIIP2 AIIP3 AIIP2 BIIP3 BAnother common situation is that we cascade twonon-linear systems, as shown above. we havey = f(x) = a 1 x + a 2 x 2 + a 3 x 3 + · · ·z = g(y) = b 1 y + b 2 y 2 + b 3 y 3 + · · ·We’d like to find the overall relationz = c 1 x + c 2 x 2 + c 3 x 3 + · · ·A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 19/29

Cascade Power SeriesTo find c 1 , c 2 , · · · , we simply substitute one power seriesinto the other and collect like powers.The linear terms, as expected, are given byc 1 = b 1 a 1 = a 1 b 1The square terms are given byc 2 = b 1 a 2 + b 2 a 2 1The first term is simply the second order distortionproduced by the first amplifier and amplified by thesecond amplifier linear term. The second term is thegeneration of second order by the second amplifier.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 20/29

Cascade CubicFinally, the cubic terms are given byc 3 = b 1 a 3 + b 2 2a 1 a 2 + b 3 a 3 1The first and last term have a very clear origin. Themiddle terms, though, are more interesting. They arisedue to second harmonic interaction. The second orderdistortion of the first amplifier can interact with the linearterm through the second order non-linearity to producecubic distortion.Even if both amplifiers have negligible cubic,a 3 = b 3 ≡ 0, we see the overall amplifier can generatecubic through this mechanism.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 21/29

Cascade ExampleIn the above amplifier, we can decompose thenon-linearity as a cascade of two non-linearities, the G mnon-linearityi d = G m1 v in + G m2 v 2 in + G m3 v 3 in + · · ·And the output impedance non-linearityv o = R 1 i d + R 2 i 2 d + R 3i 3 d + · · ·The output impedance can be a non-linear resistor load(such as a current mirror) or simply the load of thedevice itself, which has a non-linear component.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 22/29

IIP2 CascadeCommonly we’d like to know the performance of acascade in terms of the overall IIP2. To do this, notethat IIP2 = c 1 /c 2This leads toc 2c 1= b 1a 2 + b 2 a 2 1b 1 a 1= a 2a 1+ b 2b 1a 11IIP2 = 1IIP2 A + a 1IIP2 BThis is a very intuitive result, since it simply says thatwe can input refer the IIP2 of the second amplifier tothe input by the voltage gain of the first amplifier.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 23/29

IIP2 Cascade ExampleExample 1: Suppose the input amplifiers of a cascadehas IIP2 A = +0 dBm and a voltage gain of 20 dB. Thesecond amplifier has IIP2 B = +10 dBm.The input referred IIP2 B i = 10 dBm − 20 dB = −10 dBmThis is a much smaller signal than the IIP2 A , so clearlythe second amplifier dominates the distortion. Theoverall distortion is given by IIP2 ≈ −12 dB.Example 2: Now suppose IIP2 B = +20 dBm. SinceIIP2 B i = 20 dBm − 20 dB = 0 dBm, we cannot assumethat either amplifier dominates.Using the formula, we see the actual IIP2 of thecascade is a factor of 2 down, IIP2 = −3 dBm.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 24/29

IIP3 CascadeUsing the same approach, let’s start withc 3c 1= b 1a 3 + b 2 a 1 a 2 2 + b 3 a 3 1b a a 1=(a3a 1+ b 3b 1a 2 1 + b 2b 12a 2)The last term, the second harmonic interaction term,will be neglected for simplicity. Then we have1IIP3 2 = 1IIP3 2 A+ a2 1IIP3 2 BWhich shows that the IIP3 of the second amplifier isinput referred by the voltage gain squared, or the powergain.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 25/29

LNA/Mixer ExampleA common situation is an LNA and mixer cascade. Themixer can be characterized as a non-linear block with agiven IIP2 and IIP3.In the above example, the LNA has anIIP3 A = −10 dBm and a power gain of 20 dB. The mixerhas an IIP3 B = −20 dBm.If we input refer the mixer, we haveIIP3 B i = −20 dBm − 20 dB = −40 dBm.The mixer will dominate the overall IIP3 of the system.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 26/29

Example: Disto in Long-Ch. MOS AmpI D = I Q + i oI D =2 1µC oxv iV Qi o +I Q = 1 2 µC oxWL (V GS − V T ) 2WL (V Q+v i −V T ) 2Ignoring the output impedance we have= 1 2 µC ox= I Q}{{}dcWL{(VQ − V T ) 2 + v 2 i + 2v i (V Q − V T ) }+µC oxWL v i(V Q − V T )} {{ }linearWL v2 i} {{ }quadratic+ 1 2 µC oxA. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 27/29

Ideal Square Law DeviceAn ideal square law device only generates 2nd orderdistortioni o = g m v i + 1 2 µC WoxL v2 ia 2 = 1 2 µC oxa 1 = g mWL = 1 2a 3 ≡ 0The harmonic distortion is given byHD 2 = 1 2a 2a 1v i = 1 4g mV Q − V Tg m 1v i = 1 V Q − V T g m 4HD 3 = 0v iV Q − V TA. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 28/29

MobilityReal MOSFET Device1412Triode CLM DIBL SCBE600Rout kΩ1086440020020 1 2 3 4V ds (V)0Effective FieldThe real MOSFET device generates higher orderdistortionThe output impedance is non-linear. The mobility µ isnot a constant but a function of the vertical andhorizontal electric fieldWe may also bias the device at moderate or weakinversion, where the device behavior is moreexponentialThere is also internal feedbackA. M. Niknejad University of California, Berkeley EECS 142 Lecture 9 p. 29/29