1
3rd -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. IIP3
OF A
D IFFERENTIAL A MPLIFIER
We want to characterize the non-linearity of a differential amplifier. Its 3rd -order nonlinearity is of most significance.
Assume that the amplifier output characteristic is given by
vout (t)
=
3
α1 vin + α3 vin
(1)
where higher order non-linearities have been neglected.
The 3rd -order intercept point is defined as the point on Pout vs Pin plane where the lines
corresponding to fundamental and the third order inter-modulation product intersect. It can
also be defined on Vout vs Vin plane.
A. Inter-modulation Product: IM
Consider a two-tone test where you have applied
vin
=
A1 cos(ω1 t) + A2 cos(ω2 t)
(2)
to the amplifier. For the given amplifier transfer characteristics, this will generate tones at
the fundamental frequencies (ω1 , ω2 ), their 3rd harmonics (3ω1 , 3ω3 ) and also two tones
at the inter-modulation frequencies — 2ω1 − ω2 and 2ω2 − ω1 .
3
3
3
3
3
2
3
2
vout =
α1 A1 + α3 A1 + α3 A1 A2 cos(ω1 t) + α1 A2 + α3 A2 + α3 A1 A2 cos(ω2 t)
4
2
4
2
α3 3
α3 3
+ A1 cos(3ω1 t) + A2 cos(3ω2 t)
4
4
3
3
2
+ α3 A1 A2 cos((2ω1 − ω2 )t) + α3 A1 A22 cos ((2ω2 − ω1 ) t)
4
4
3
3
2
+ α3 A1 A2 cos ((2ω1 + ω2 ) t) + α3 A1 A22 cos ((2ω2 + ω1 ) t)
(3)
4
4
The following relationships have been used.
1
cos3 (ωt) =
(3 cos(ωt) + cos(3ωt))
(4)
4
1
sin3 (ωt) =
(3 sin(ωt) − sin(3ωt))
(5)
4
1
cos2 (ωt) =
(1 + cos(2ωt))
(6)
2
1
(1 − cos(2ωt))
(7)
sin2 (ωt) =
2
Assuming,
1) α3 α1
2) A1 and A2 are not large enough to compress the gain at ω1 and ω2
3) Tones at 2ω1 +ω2 , 2ω2 +ω1 , 3ω1 and 3ω2 are far away from the band of the amplifier.
then (3) can be approximated as
vout
=
α1 A1 cos(ω1 t) + α1 A2 cos(ω2 t)
3
3
+ α3 A21 A2 cos ((2ω1 − ω2 ) t) + α3 A1 A22 cos ((2ω2 − ω1 ) t)
4
4
(8)
Inter-modulation distortion, IM3 is defined as the ratio of amplitude of output fundamental
tone to the amplitude of the 3rd order product in the output. If A1 = A2 = A, then
α1 A
4 α1
IM3 = 3
=
3 α3 A2
α A3
4 3
(9)
B. Third Order Intercept Points
IM3 is dependent on the amplitude (A) of the input tones. Input 3rd -order intercept point
(IIP3 ) is defined to characterize the linearity, independent of the input amplitude.
r
4α1
IIP3 = A|IM3 =1 =
3α3
2 2
= IM3 A
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Anuranjan
Jha - Columbia University – 2006
2
(10)
(11)
Output Intercept Point OIP3 is given by
r
4α1
(12)
3α3
For SpectreRF pss simulation, having A1 = A2 = A means doing large signal simulation
OIP3
=
α1 IIP3 = α1
for both the tones. This becomes time-consuming. See SpectreRF user-manual. Instead,
assume that A1 A2 , then set the tone at ω1 as large signal and that at ω2 as small signal.
Then do pss and pac simulations.
C. IP3 for modified two-tone test
Assuming A1 A2 , (8) can be further simplified to
vout
=
3
α1 A1 cos(ω1 t) + α1 A2 cos(ω2 t) + α3 A21 A2 cos ((2ω1 − ω2 ) t)
4
(13)
The amplitude corresponding to each of the tones in the output can be simulated or
measured using a spectrum analyzer. Let
VL1
=
α1 A1
(14)
VS1
=
(15)
VS3
=
α1 A2
3
α3 A21 A2
4
(16)
Using (14) and (15) in (16), we get
VS3
=
=
OIP3
=
2
2
3 VL1
VS1
3 α3 VL1
α3 2
=
VS1
4 α1 α1
4 α1 α12
2
1
VL1
VS1
2
V
V
=
S1
L1
2
2
α1 IIP3
OIP32
r
VS1
VL1
VS3
(17)
The assumption, A1 A2 , 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 products.
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 A1 , A2 anyway and would extrapolate.
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Anuranjan
Jha - Columbia University – 2006
3
II. E FFECT OF N EGATIVE F EEDBACK ON D ISTORTION [1]
Assume a single-ended amplifier (Fig. 1) with the following characteristics.
y(t)
=
β1 u(t) + β2 u2 (t) + β3 u3 (t)
(18)
1 dn y n! dun u=0
(19)
The coefficients βn are given by
βn
u(t)
Fig. 1.
=
β1u(t) + β2u2(t) + β3u3(t)
y(t)
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 IIP3 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)
+
u(t)
Σ
β1u(t) + β2u2(t) + β3u3(t)
y(t)
v(t)
1
1+T
u(t)
β1u(t) + β2u2(t) + β3u3(t)
y(t)
α
(a)
Fig. 2.
(b)
(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)
=
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Anuranjan
Jha - Columbia University – 2006
v(t) − αy(t)
4
(20)
The output in terms of the input signal v(t) can hence be given by
y(t)
=
κ1 v(t) + κ2 v 2 (t) + κ3 v 3 (t)
=
1 dn y n! dv n v=0
where,
κn
(21)
We want to derive the relationship between κn and βn as a function of the loop factor
T = β1 α (where α is the feedback factor as shown in fig. 2(a)).
Calculation of κ1
κ1
dy dy du =
dv v=0 du dv v=0
=
By defn.
Note that v = 0 ⇔ u = 0, so
κ1
=
=
κ1
=
dy d
× (v − αy)
du u=0 dv
v=0
β1 (1 − ακ1 )
β1
where, T = αβ1
(1 + T )
(22)
We also make note of these relations which we found above:
du
dv
du dv v=0
dy du u=0
dy
dv
=
1−α
=
1
1+T
(24)
=
β1
(25)
(23)
Calculation of κ2
κ2
=
2κ2
=
=
=
1 d2 y By defn.
2! dv 2 v=0
d dy dv dv v=0
du d dy dv du dv v=0
1
d du dy 1 + T du dv du v=0
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Jha - Columbia University – 2006
5
Using eqn. 24
Expanding derivative of a product, we can write
dy d du
du d2 y 2(1 + T )κ2 =
+
du du dv
dv du2 v=0
dy d du
2β2
=
+
Using eqn. 19, 24
du du dv
1 + T u=0
d du 2β2
= β1
+
(26)
du dv u=0 1 + T
du d
To find du dv u=0
d du d
dy =
1−α
Using eqn. 23
du dv u=0
du
dv u=0
d dy = −α
du dv v,u=0
d
d
2
3
= −α
β1 u + β2 u + β3 u + . . . Using eqn. 18
du dv
v,u=0
d
du
du
= −α
β1
+ 2β2 u + . . . du
dv
dv
v,u=0
d du
d
du ≈ −αβ1
− 2αβ2
u
du dv
du
dv v,u=0
d du
d
du (1 + αβ1 )
= −2αβ2
u
du dv
du
dv v,u=0
du
d2 u = −2αβ2
+u 2 dv
dv
v,u=0
−2αβ2
=
Using eqn. 24
1+T
d du −2αβ2
=
(27)
du dv u=0
(1 + T )2
Coming back to eqn. 26 we get,
2(1 + T )κ2
=
(1 + T )κ2
=
=
κ2
=
−2αβ2
2β2
+
2
(1 + T )
1+T
−αβ1 β2
β2
+
2
(1 + T )
1+T
−T β2
β2
β2
T
+
=
1−
(1 + T )2 1 + T
1+T
1+T
β2
(1 + T )3
β1
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Jha - Columbia University – 2006
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(28)
Calculation of κ3
κ3
=
6κ3
=
=
=
=
=
=
d3 y By defn.
dv 3 v=0
d dy
dv dv
v=0
d
d
2
3
(β1 u + β2 u + β3 u )
dv dv
v=0
d
du
(β1 + 2β2 u + 3β3 u2 )
dv
dv
u,v=0
2
d
d
u
du
du du
2
(β1 + 2β2 u + 3β3 u ) 2 + 2β2
+ 6β3 u
dv
dv
dv
dv dv
(
2 )
2
d
d
u
du
(β1 + 2β2 u + 3β3 u2 ) 2 + (2β2 + 6β3 u)
dv
dv
dv
d3 u
du
du d2 u
(β1 + 2β2 u + 3β3 u2 ) 3 + 2β2
+ 6β3 u
dv
dv
dv dv 2
3
du d2 u
du
+ 6β3
+ (2β2 + 6β3 u)2
dv
dv dv 2
1
3!
d
dv
d
dv
d
dv
Putting u = 0 above, we get
6κ3
=
=
6κ3
=
To find, d2 u/dv 2 |u,v=0 :
d2 u dv 2 u,v=0
3
d3 u
du d2 u
du
du d2 u
β1 3 + 2β2
+
6β
+
4β
3
2
dv
dv dv 2
dv
dv dv 2
(
)
3
d3 u
du
du d2 u β1 3 + 6β3
+ 6β2
dv
dv
dv dv 2 u=0
β1
d3 u
6β3
6β2 d2 u
+
+
dv 3 (1 + T )3 1 + T dv 2
=
=
=
Using eqn. 24
d du
d2 y = −α 2 Using eqn. 23
dv dv
dv u,v=0
−2ακ2
Using eqn. 21
−2αβ2
Using eqn. 28
(1 + T )3
To find d3 u/dv 3 |u,v=0 use eqns. 30 and 21:
d3 u d3 y = −α 3 = −6ακ3
dv 3 u,v=0
dv u,v=0
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Jha - Columbia University – 2006
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(29)
(30)
(31)
(32)
So, eqn. (29) becomes,
6κ3
=
=
(1 + T )κ3
=
κ3
=
κ3
=
6β2 −2αβ2
6β3
+
3
(1 + T )
1 + T (1 + T )3
−12αβ22
6β3
−6T κ3 +
+
(1 + T )3 (1 + T )4
−2αβ22
β3
+
(1 + T )3 (1 + T )4
−2αβ22
β3
+
(1 + T )4 (1 + T )5
(1 + T )β3 − 2αβ22
(1 + T )5
−6αβ1 κ3 +
So, the IIP3 for the feedback system is,
r
4κ1
IIP3f bk =
3κ
s 3
4β1 3((1 + T )β3 − 2αβ22 )
=
/
1+T
(1 + T )5
s
4β1 (1 + T )4
=
3 ((1 + T )β3 − 2αβ22 )
s
4β1 (1 + T )3
≈
for (β3 (1 + T ) 2αβ22 )
3β3
IIP3f bk
=
3
IIP3 × (1 + T ) 2
c
Anuranjan
Jha - Columbia University – 2006
8
(33)
III. H OW
TO SIMULATE
IIP3
USING
S PECTRE RF?
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 VGS 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 3rd -order amplitude in units of dBV. The pss results are shown in fig. 5. I tabulated
the results for different VGS cases and plotted them using MATLAB. The intercept point
is shown in fig. 8(a). The IIP3 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 =
[1 2 5 10 20 30 40 50 75 100 150]*1e-3;
in =
[-54.02 -48 -40.04 -34.02 -28 -24.48 -21.98 ...
-20.04 -16.52 -14.02 -10.5];
out10M = [-36.54 -30.52 -22.57 -16.6 -10.79 -7.663 ...
-5.99 -5.09 -4.064 -3.63 -3.262];
out8M
= [-127.8 -109.74 -85.9 -67.98 -50.28 -38.52 ...
-28.71 -23.59 -18.3 -16.09 -14.35];
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Jha - Columbia University – 2006
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(a)
(b)
Fig. 3.
Test setup for IIP3 measurement of (a) Common-Source Amplifier (b) Source-degenerated amplifier.
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Jha - Columbia University – 2006
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(a)
Fig. 4.
(b)
(a) Properties of the port used in the test setup of fig. 3(a) (b) PSS analysis form.
im3 = out10M - out8M;
inp =
[-54.02 -48 -40.04 -34.02 -28 -24.48 -21.98 ...
-20.04 -16.52 -14.02 -10.5 -7.5];
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;
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Jha - Columbia University – 2006
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Fig. 5.
SpectreRF results for smallest applied VGS .
Fig. 6.
Properties of the vsin source used in the test setup.
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Jha - Columbia University – 2006
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SpectreRF results for largest applied VGS for fig. 3(b)
Fig. 7.
ω1
0
2ω1 − ω2
−25
−25
−50
−50
Vout, dBV
Vout, dBV
0
−75
−100
−100
−125
−150
−150
−50
−40
−30
Vgs, dBV
−20
−10
−175
−60
0
(a)
Fig. 8.
2ω1 − ω2
−75
−125
−175
−60
ω1
−50
−40
−30
Vgs, dBV
−20
(b)
IIP3 comparison for (a) Common-Source amplifier and (b) Source degenerated amplifier.
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Anuranjan
Jha - Columbia University – 2006
13
−10
0
set(gca,’ylim’,[-175 15],’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.
R EFERENCES
[1] W. Sansen, “Distortion in Elementary transistor circuits,” IEEE Trans. on Circuits and Systems - II: Analog and
Digital Signal Processing, vol. 46, pp. 315–324, Mar. 1999.
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Jha - Columbia University – 2006
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