3 rd -order Intercept Point

Transcription

1 1 3 rd -order Intercept Point Anuranjan Jha, Columbia Integrated Systems Lab, Department of Electrical Engineering, Columbia University, New York, NY Last Revised: September 12, 2006 These derivations are for my documentation. I. IIP 3 OF A DIFFERENTIAL AMPLIFIER We want to characterize the non-linearity of a differential amplifier. Its 3 rd -order nonlinearity is of most significance. Assume that the amplifier output characteristic is given by v out (t) = α 1 v in + α 3 vin 3 (1) where higher order non-linearities have been neglected. The 3 rd -order intercept point is defined as the point on P out vs P in plane where the lines corresponding to fundamental and the third order inter-molation proct intersect. It can also be defined on V out vs V in plane. A. Inter-molation Proct: IM Consider a two-tone test where you have applied v in = A 1 cos(ω 1 t) + A 2 cos(ω 2 t) (2) to the amplifier. For the given amplifier transfer characteristics, this will generate tones at the fundamental frequencies (ω 1, ω 2 ), their 3 rd harmonics (3ω 1, 3ω 3 ) and also two tones

2 at the inter-molation frequencies 2ω 1 ω 2 and 2ω 2 ω 1. ( v out = α 1 A α 3A ) 2 α 3A 1 A 2 2 cos(ω 1 t) + (α 1 A α 3A α 3A 21A ) 2 cos(ω 2 t) + α 3 4 A3 1cos(3ω 1 t) + α 3 4 A3 2cos(3ω 2 t) α 3A 2 1A 2 cos((2ω 1 ω 2 )t) α 3A 1 A 2 2 cos ((2ω 2 ω 1 ) t) α 3A 2 1A 2 cos ((2ω 1 + ω 2 ) t) α 3A 1 A 2 2 cos ((2ω 2 + ω 1 ) t) (3) The following relationships have been used. Assuming, cos 3 (ωt) = 1 (3 cos(ωt) + cos(3ωt)) (4) 4 sin 3 (ωt) = 1 (3 sin(ωt) sin(3ωt)) (5) 4 cos 2 (ωt) = 1 (1 + cos(2ωt)) (6) 2 sin 2 (ωt) = 1 (1 cos(2ωt)) (7) 2 1) α 3 α 1 2) A 1 and A 2 are not large enough to compress the gain at ω 1 and ω 2 3) Tones at 2ω 1 +ω 2, 2ω 2 +ω 1, 3ω 1 and 3ω 2 are far away from the band of the amplifier. then (3) can be approximated as v out = α 1 A 1 cos(ω 1 t) + α 1 A 2 cos(ω 2 t) α 3A 2 1A 2 cos ((2ω 1 ω 2 ) t) α 3A 1 A 2 2 cos ((2ω 2 ω 1 ) t) (8) Inter-molation distortion, IM 3 is defined as the ratio of amplitude of output fundamental tone to the amplitude of the 3 rd order proct in the output. If A 1 = A 2 = A, then α 1 A IM 3 = 3 α 4 3A = 4 α 1 (9) 3 3 α 3 A 2 B. Third Order Intercept Points IM 3 is dependent on the amplitude (A) of the input tones. Input 3 rd -order intercept point (IIP 3 ) is defined to characterize the linearity, independent of the input amplitude. 4α1 IIP 3 = A IM3 =1 = (10) 3α 3 = IM3 2 A 2 (11) c Anuranjan Jha - Columbia University

3 Output Intercept Point OIP 3 is given by OIP 3 = α 1 IIP 3 = α 1 4α1 3α 3 (12) For SpectreRF pss simulation, having A 1 = A 2 = A means doing large signal simulation for both the tones. This becomes time-consuming. See SpectreRF user-manual. Instead, assume that A 1 A 2, then set the tone at ω 1 as large signal and that at ω 2 as small signal. Then do pss and pac simulations. C. IP 3 for modified two-tone test Assuming A 1 A 2, (8) can be further simplified to v out = α 1 A 1 cos(ω 1 t) + α 1 A 2 cos(ω 2 t) α 3A 2 1A 2 cos ((2ω 1 ω 2 ) t) (13) The amplitude corresponding to each of the tones in the output can be simulated or measured using a spectrum analyzer. Let Using (14) and (15) in (16), we get V L1 = α 1 A 1 (14) V S1 = α 1 A 2 (15) V S3 = 3 4 α 3A 2 1A 2 (16) V S3 = 3 4 α VL1 2 3 = 1 α 2 1 α 2 1IIP 2 3 V S1 α 3 = 3 VL1 2 α 1 4 α 1 V α1 2 S1 V 2 L1V S1 = V 2 L1 V S1 OIP 2 3 OIP 3 = V L1 VS1 V S3 (17) The assumption, A 1 A 2, is used in the following ways: 1) Second signal does not cause the nonlinearities. So the approximations made earlier are even more valid! 2) We need not bother about other harmonics/im procts. 3) We can use PSS and PAC for the simulation. So, in my opinion, this would be a better way even in measurements. But there are other factors also which cause non-linearities/cancellation of non-linearities. So we would (not) use very large A 1, A 2 anyway and would extrapolate. c Anuranjan Jha - Columbia University

4 II. EFFECT OF NEGATIVE FEEDBACK ON DISTORTION [1] Assume a single-ended amplifier (Fig. 1) with the following characteristics. y(t) = β 1 u(t) + β 2 u 2 (t) + β 3 u 3 (t) (18) The coefficients β n are given by β n = 1 n! d n y n (19) u=0 u(t) β 1 u(t) + β 2 u 2 (t) + β 3 u 3 (t) y(t) Fig. 1. Amplifier with weak non-linearity. In the presence of feedback, the system looks like fig. 2(a). This is different from the case where the input is scaled down in which case the IIP 3 of the system will just scale up by the inverse of the attenuation factor. When there is a feedback the input signal seen by v(t) + Σ - u(t) β 1 u(t) + β 2 u 2 (t) + β 3 u 3 (t) y(t) v(t) 1 u(t) β 1 u(t) + β 2 u 2 (t) + β 3 u 3 (t) y(t) α (a) (b) Fig. 2. (a) Amplifier with weak non-linearity now in a negative feedback system (b) Attenuated input applied to the same amplifier. the amplifier is given by u(t) = v(t) αy(t) (20) c Anuranjan Jha - Columbia University

5 The output in terms of the input signal v(t) can hence be given by y(t) = κ 1 v(t) + κ 2 v 2 (t) + κ 3 v 3 (t) where, κ n = 1 d n y n! n (21) v=0 We want to derive the relationship between κ n and β n as a function of the loop factor T = β 1 α (where α is the feedback factor as shown in fig. 2(a)). Calculation of κ 1 κ 1 = dy = dy v=0 v=0 By defn. Note that v = 0 u = 0, so κ 1 = dy d (v αy) u=0 v=0 = β 1 (1 ακ 1 ) β 1 κ 1 = where, T = αβ 1 (22) ( ) We also make note of these relations which we found above: = 1 α dy (23) 1 = (24) v=0 dy = β 1 (25) u=0 Calculation of κ 2 κ 2 = 1 d 2 y 2! 2 By defn. v=0 2κ 2 = d ( ) dy v=0 = ( ) d dy ( v=0 1 d dy = ) v=0 Using eqn. 24 c Anuranjan Jha - Columbia University

6 Expanding derivative of a proct, we can write { ( ) dy d 2( )κ 2 = + { ( ) dy d = ( = β 1 d ) u=0 d 2 y 2 } v=0 + 2β 2 } u=0 + 2β 2 Using eqn. 19, 24 (26) To find d ( ) u=0 ( ) d u=0 (1 + αβ 1 ) d = d ( = α d = α d = α d αβ 1 d = 2αβ 2 d 1 α dy ) ( u=0 dy Using eqn. 23 ) { v,u=0 d ( β1 u + β 2 u 2 + β 3 u )} { v,u=0 β 1 + 2β 2u v,u=0 +...} ( ) ( d 2αβ 2 u ) v,u=0 ( u ) v,u=0 = 2αβ 2 ( + ud2 u 2 ) v,u=0 Using eqn. 18 d = 2αβ 2 Using eqn. 24 2αβ 2 = (27) u=0 ( ) 2 Coming back to eqn. 26 we get, 2( )κ 2 = β 1 2αβ 2 ( ) 2 + 2β 2 ( )κ 2 = αβ 1β 2 ( ) + β 2 2 T β 2 = ( ) + β 2 2 = β 2 κ 2 = ( 1 T ) β 2 ( ) 3 (28) c Anuranjan Jha - Columbia University

7 Calculation of κ 3 κ 3 = 1 d 3 y 3! 3 By defn. v=0 6κ 3 = d { ( )} d dy v=0 = d { ( d d (β 1u + β 2 u 2 + β 3 u ))} v=0 3 = d { ( d (β 1 + 2β 2 u + 3β 3 u 2 ) )} u,v=0 = d { ( (β 1 + 2β 2 u + 3β 3 u 2 ) d2 u + 2β β 3u ) } { = d ( ) } 2 (β 1 + 2β 2 u + 3β 3 u 2 ) d2 u + (2β β 3 u) ( = (β 1 + 2β 2 u + 3β 3 u 2 ) d3 u + 2β β 3u ) d 2 u 2 ( ) 3 + 6β 3 + (2β 2 + 6β 3 u)2 d 2 u 2 Putting u = 0 above, we get d 3 u 6κ 3 = β 1 + 2β 3 2 { d 3 u = β 1 + 6β 3 3 ( ) d 2 3 u + 6β β 2 ) } 3 d 2 u + 6β 2 2 u=0 ( 6κ 3 = β 1 d 3 u 3 + 6β 3 ( ) 3 + 6β 2 d 2 u 2 d 2 u 2 Using eqn. 24 (29) To find, d 2 u/ 2 u,v=0 : d 2 u 2 = d u,v=0 = y αd2 2 Using eqn. 23 (30) u,v=0 = 2ακ 2 Using eqn. 21 = 2αβ 2 ( ) 3 Using eqn. 28 (31) To find d 3 u/ 3 u,v=0 use eqns. 30 and 21: d 3 u 3 = α d3 y u,v=0 3 = 6ακ 3 (32) u,v=0 c Anuranjan Jha - Columbia University

8 So, eqn. (29) becomes, 6κ 3 = 6αβ 1 κ 3 + 6β 3 ( ) 3 + 6β 2 = 6T κ 3 + 6β 3 ( ) + 12αβ2 2 3 ( ) 4 ( )κ 3 = β 3 ( ) + 2αβ2 2 3 ( ) 4 κ 3 = β 3 ( ) 4 + 2αβ2 2 ( ) 5 κ 3 = ( )β 3 2αβ 2 2 ( ) 5 2αβ 2 ( ) 3 So, the IIP 3 for the feedback system is, 4κ1 IIP 3fbk = 3κ 3 = 4β 1 /3(( )β 3 2αβ2) 2 ( ) 5 = 4β 1 ( ) 4 3 (( )β 3 2αβ2) 2 4β 1 ( ) 3 3β 3 for (β 3 ( ) 2αβ2) 2 IIP 3fbk = IIP 3 ( ) 3 2 (33) c Anuranjan Jha - Columbia University

9 III. HOW TO SIMULATE IIP 3 USING SPECTRERF? A. Low frequency case: Where you do not need 50 ohm source and input matching. Let s begin with the simplest example a common-source amplifier with resistive load with no degeneration and no input matching. We will deal in units of dbv. Fig. 3(a) presents a test setup for measuring the IIP3 of my amplifier. First I set the twotones needed for the test as 10 MHz and 12 MHz (see Fig. 4(a) and 4(b)). The input is specified in terms of peak amplitude. Even though I am using a port in the simulation, it is not required here. In the results you will note that I have taken the V GS voltage as my input reference. With ports, unless there is a perfect matching with its input resistance, the voltage you get from it is dependent on the load. For example, since the port is driving a capacitive load here, the voltage at the gate will be 40 mvpp for Vpk set to 20 mvpp in the port properties. I can now run pss to find the steady state response and then proceed to find the 3 rd -order amplitude in units of dbv. The pss results are shown in fig. 5. I tabulated the results for different V GS cases and plotted them using MATLAB. The intercept point is shown in fig. 8(a). The IIP 3 for this Common-Source amplifier is about 7.5 dbv. Next I simulated an amplifier with same nfet but with a source degeneration resistance of 500 Ω and load of 5 kω. Figs. 3(b), 6, 7 shows the test setup, the properties of the vsin source used in the setup and the SpectreRF results when two-tone signals of amplitude 350 mv are applied, respectively. The IIP 3 result is shown in Fig. 8(b). It has improved to about 2.5 dbv. The data and MATLAB code for Common-Source amplifier is attached below. % Results of two-tone test of Common-Source Amplifier % Circuit: /home/anu/cadence/ee6314_f05/csampl vrf = [ ]*1e-3; in = [ ]; out10m = [ ]; out8m = [ ]; c Anuranjan Jha - Columbia University

10 (a) (b) Fig. 3. Test setup for IIP3 measurement of (a) Common-Source Amplifier (b) Source-degenerated amplifier. c Anuranjan Jha - Columbia University

11 (a) (b) Fig. 4. (a) Properties of the port used in the test setup of fig. 3(a) (b) PSS analysis form. im3 = out10m - out8m; inp = [ ]; int1 = out10m(1) + (inp-inp(1)); int3 = out8m(1) + 3*(inp-inp(1)); figure; plot(in,out10m, k-x, linewidth,2, markersize,12); hold on; plot(in,out8m, k-o, linewidth,2, markersize,12); plot(inp,int1, r-x, linewidth,2, markersize,8); plot(inp,int3, g-o, linewidth,2, markersize,8); hold off; c Anuranjan Jha - Columbia University

12 Fig. 5. SpectreRF results for smallest applied V GS. Fig. 6. Properties of the vsin source used in the test setup. c Anuranjan Jha - Columbia University

13 Fig. 7. SpectreRF results for largest applied V GS for fig. 3(b) 0 ω 1 2ω 1 ω 2 0 ω 1 2ω 1 ω V out, dbv V out, dbv V gs, dbv V gs, dbv (a) (b) Fig. 8. IIP 3 comparison for (a) Common-Source amplifier and (b) Source degenerated amplifier. c Anuranjan Jha - Columbia University

14 set(gca, ylim,[ ], ytick,[-175:25:0], fontsize,16); L = legend( \omega_1, 2\omega_1 - \omega_2 ); set(l, box, off, fontsize,16, location, northwest ); xlabel( V_{gs}, dbv, fontsize,16); ylabel( V_{out}, dbv, fontsize,16); % Results in an IIP3 of -7.5 dbv print -depsc CSamplIIP3.eps B. RF case: Where you do need 50 ohm source and input matching. REFERENCES [1] W. Sansen, Distortion in Elementary transistor circuits, IEEE Trans. on Circuits and Systems - II: Analog and Digital Signal Processing, vol. 46, pp , Mar c Anuranjan Jha - Columbia University