Unformatted text preview: creased by adding the M6
commonsource stage.
Output Stages 1027 Analog ICs; JiehTsorng Wu Parallel CommonSource Conﬁguration
EP1 VOS VDD
M1 VDD
V2 V3 EP2 M11
Io
Vo Vi
M2
VOS RL
EN2 M12
VSS VSS EN1
VDD IB2 VDD M5 M3 M4 V1 Output Stages M21 M22
V3 VB M23 EP2 Amplifier VSS 1028 VB M24 M25 IB1
EP1 Amplifier V2
V2 V2 V2
V1 IB2 M6 M26
VSS Analog ICs; JiehTsorng Wu Parallel CommonSource Conﬁguration
• Want turn oﬀ M11 and M12 when Vo ≈ Vi = 0, so that AE P 2 and AE N 2 have high gain,
and AE P 1 and AE N 1 have low gain.
• VOS of EP1 is introduced by making (W/L)3
ID3 = IB1 · AE P 1 = 0.8(W/L)4 . When Vo ≈ Vi = 0, (W/L)3 ID1 = (ID3 − IB2 ) · (W/L)3 + (W/L)4 gm3
=
gm5 ID3
kn (W/L)3
·
·
kp (W/L)5 ID3 − IB2 (W/L)1
(W/L)3 AE P 2 ≈ gm22ro22 • When Vi  is small, and M11 and M12 are not turned on, the output is
Vo = Output Stages Vi
1 + 1/(A1gm1RL)
1029 Analog ICs; JiehTsorng Wu Parallel CommonSource Conﬁguration
• When Vi is large, M11 can be turned on, and the output becomes
Vo ≈ Vi − VOS if AE P 2 → ∞ • When V2+ = V2− at EP2, deﬁne V3 = Vov 25 + Vt25 + VSS = VK + VSS . Then
V3 = [Vo − (Vi − VOS )]AE P 1AE P 2 + VK + VSS
Deﬁne Vi (mi n) as the minimum input to turn on M11. Let V3 = VDD − Vtp11, we have
Vi (mi n) = VOS (1 + AE P 1gm1RL) − (VDD − VSS − VK − Vtp11 )(1 + AE P 1gm1RL)
AE P 1AE P 2 Vi (mi n) = VOS (1 + AE P 1gm1RL) if AE P 2 → ∞ M11 and M12 remain oﬀ for only a small range of input voltages.
Output Stages 1030 Analog ICs; JiehTsorng Wu Noise Analysis and Modeling JiehTsorng Wu ES A December 5, 2002 1896 National ChiaoTung University
Department of Electronics Engineering Noise in Time Domain
n(t) PDF t 0 n
0 1
Mean = n =
T T n(t )d t = 0
0 1
Noise Power = n2 =
T Root Mean Square = nrms = n2 T n 2 (t )d t
0 1/2 • T is a suitable averaging time interval. Typically, a longer T gives a more accurate
measurement.
Noise 112 Analog ICs; JiehTsorng Wu Probability Density Function
• The probability that the noise lies between values n and n + d n at any time is given
by P (n)d n. P (n) is the probability density function (PDF).
• The PDF of a random noise is usually Gaussian, i.e.,
1 P (n) = √
2πσ 2
− n2
e 2σ We have
+∞
−∞ PDF(n)d n = 1 and
Variance = Noise +∞
−∞ n2 · PDF(n)d n = n2 = σ 2 113 Analog ICs; JiehTsorng Wu Noise in Frequency Domain
BPF Onesided power spectral density n
f
f n2(f )
SD(f ) = lim
∆f →0 ∆f Power
Meter Spectral Density Onesided root spectral density 2 V
Hz RD(f ) = (SD)1/2
log f The total noise power is
Root Spectral Density
∞ V
√
Hz SDn(f )d f = n2 0 log f Noise 114 Analog ICs; JiehTsorng Wu Filtered Noise
ni no H(s) SDno (f ) = SDni (f ) × H (j 2πf )2 If SDni (f ) = N is a constant (white noise), then
n2
o = ∞
0 SDni (f ) · H (j 2πf ) d f = N ·
2 ∞ H (j 2πf )2d f = N · Bn 0 • Bn is called the noise bandwidth of the ﬁlter.
• For a singlepole ﬁlter H (s) =
Bn = ∞ 1
,
1+s/ωo H (j 2πf )2d f = 0 Noise ∞
0 115 1
1+ f
fo 2 df = π
·f
2o Analog ICs; JiehTsorng Wu Noise Summation
n i1 H 1 (s) n i2 H 2 (s) n i3 H 3 (s) n i1
n o1 n o2 n i2 If two noises, ni and nj , are uncorrelated then, i.e., ni · nj = 0. Then
n21 = (ni 1 + ni 2)2 = n21 + n22 + 2 · ni 1ni 2 = n21 + n22
o
i
i
i
i
SDno2 (f ) = H1(j 2πf )2SDni 1 + H2(j 2πf )2SDni 2 + H3(j 2πf )2 SDni 3 Noise 116 Analog ICs; JiehTsorng Wu Piecewise Integration of Noise
200
(nV)2 Hz 20
2 2
2 ∝
2
0
10 1
10 2
10 N1 f 3
10
N2 1
f 4
10
N3 5
10 6
10 7
10 N4 The noise power in each frequency region is
102 PN1 = 100 2 200
102
2
d f = 200 ln(f )100 = 1.84 × 105 (nV)2
f
103 PN2 =
Noise 20 d f = 20
2 102 2 103
f 102 117 = 3.6 × 105 (nV)2
Analog ICs; JiehTsorng Wu Piecewise Integration of Noise
104 PN3 = PN4 103 = = 20
103
∞
104 200 2 f df = 1+
2 20
103 2 200 2 f
105 2 2 = ∞
0 13
f
3
200 1+ 104 = 1.33 × 108 (nV)2
103
10 4 2 f
105 2 df − 2002d f
0 π
105 − 2002 · 104 = 5.88 × 109 (nV)2
2 Total rms of the noise is
nrms = PN1 + PN2 + PN3 + PN4 1/2 = 77.5 µV rms • 1/f noise tangent principle: Lower a 1/f line until it touches the spectral density
curve; the total noise can be approximated by the noise in the vicinity of the 1/f line.
Noise 118 Analog ICs; JiehTsorng Wu Thermal Noise
R
R R i2 v2 v2
= 4kT R
∆f 1
i2
= 4kT
∆f
R f =0∼∞ T = Absolute Temperature in Kelvins k = 1.38 × 10−23 watt/KHz (Boltzmann’s Constant) ∆f = Bandwidth per Hertz • Thermal noise is a white noise, i.e., its power spectral density v 2/∆f is independent
of frequency, and its amplitude distribution is Gaussian.
◦ • For a 1 kΩ resistor at 300 K,
Noise v 2/∆f √
2
≈ (4 nV/ Hz) .
119 Analog ICs; JiehTsorng Wu Thermal Noise with Loading
2
vo Pn
R R
RL v2 C v2 • The RL load receives the maximum power if RL = R . Thus the available noise power
for RL is
1
· v 2 · Bn = kT Bn
Bn = Noise Bandwidth
Pn =
4R
• For the RC lowpass network
1
1
π
=
Bn = ·
2 2πRC 4RC
◦ 2
vo kT
1
=
= 4kT R ·
4RC
C 2 2
If C = 1 pF a...
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 Winter '09
 Choma
 Integrated Circuit, Transistor, The Land, Bipolar junction transistor, VDS, Analog ICs, JiehTsorng Wu

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