Presented by
Snehasish Roychowdhury
Department of Electrical Engineering, IIT Bombay
Design of a Gm-C filter of very low trans-conductance & highly resistant to out-of-
band Electromagnetic Interferences
Application: Gm-C filter of few Hz cut-off
Bio-Medical applications:
o Pulse rate of a human body is 72 pulses/min on average.
o Hence, to isolate and detect our pulses from environmental noises in electronic systems, low pass filters of cut-off frequency in Hz is needed.
o This is how we can reduce the out of band noises and interferences by low pass filtering.
13-Jun-16 Snehasish Roychowdhury 2
Cut-off frequency of Gm-C filter ๐ฃ๐๐ โ ๐ฃ๐๐ข๐ก โ ๐บ๐ = ๐ผ๐๐ข๐ก
Again,
๐ฃ๐๐ข๐ก = ๐ผ๐๐ข๐ก โ (1
๐ ๐ถ๐๐๐๐)
Hence, ๐ฃ๐๐ โ ๐ฃ๐๐ข๐ก โ ๐บ๐ = ๐ ๐ถ๐๐๐๐๐ฃ๐๐ข๐ก
๐ฃ๐๐ โ ๐บ๐ = ๐ ๐ถ๐๐๐๐ + ๐บ๐ โ ๐ฃ๐๐ข๐ก
๐ฃ๐๐ข๐ก
๐ฃ๐๐=
1
1+๐ ๐ถ๐๐๐๐
๐บ๐
oHence, DC gain= 0dB
o Cut-off frequency of the filter is : ๐บ๐
2ฯ๐ถ๐๐๐๐ ๐ป๐ง
13-Jun-16 Snehasish Roychowdhury 3
Fig1: Gm-C filter using voltage follower configuration
Block diagram of Gm-C filter
Cut-off freq=๐บ๐
2๐๐ถ๐๐๐๐
For 70Hz, Cload=1pF ,
Gm=0.47nA/V.
๐บ๐ =๐๐๐ข๐ก
๐ฃ๐๐=
๐๐๐ข๐ก
๐ฃ๐๐โ ๐ด๐ก๐ก๐. ๐๐๐๐ก๐๐
1
๐
Voltage attenuator at input stage
reduces Gm
2nd stage Trans-conductor: Input: vim (voltage), output: Iout (current)
All transistors in Gm-C filter are in sub-threshold region.
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Fig2. Block diagram of Gm-C filter
Iout
Cross-coupled trans-conductor design:
๐ผ๐1 = ๐ผ1 + ๐ผ3 & ๐ผ๐2 = ๐ผ2 + ๐ผ4
๐ผ1
๐ผ2=
exp๐ฃ๐ ๐ 1โ๐ฃ๐1
๐๐๐
exp๐ฃ๐ ๐ 1โ๐ฃ๐2
๐๐๐
= exp (โ๐ฃ๐๐
๐๐๐)
๐ผ3
๐ผ4=
exp๐ฃ๐ ๐ 2โ๐ฃ๐2
๐๐๐
exp๐ฃ๐ ๐ 2โ๐ฃ๐1
๐๐๐
= exp (๐ฃ๐๐
๐๐๐)
Now,
๐ผ1 โ ๐ผ2๐ผ1 + ๐ผ2
=exp
โ๐ฃ๐๐๐๐๐
โ 1
expโ๐ฃ๐๐๐๐๐
+ 1=
expโ๐ฃ๐๐2๐๐๐
โ exp๐ฃ๐๐
2๐๐๐
expโ๐ฃ๐๐12๐๐๐
+ exp๐ฃ๐๐
2๐๐๐
= โtanh (๐ฃ๐๐
2๐๐๐)
๐ผ1 โ ๐ผ2 = โ๐ผ๐ ๐ 1 tanh๐ฃ๐๐
2๐๐๐
๐ผ3 โ ๐ผ4 = ๐ผ๐ ๐ 2 tanh๐ฃ๐๐
2๐๐๐
๐ผ๐๐ = ๐ผ๐1 โ ๐ผ๐2 = ๐ผ๐ ๐ 2 โ ๐ผ๐ ๐ 1 tanh๐ฃ๐๐
2๐๐๐ โฆ . . . (1)
13-Jun-16 Snehasish Roychowdhury 5
Fig3. Cross-coupled Trans-conductor
vim=vm1-vm2
vm1 vm2
Inputs of the trans-conductor: vm1 & vm2
Output current : Iod=Io1-Io2
Voltage attenuator design:
DC Analysis:
Assume all gmโs & Ibiasโs are
equal. Hence,
๐ผ1 = ๐ผ3 = ๐ผ5 & ๐ผ2 = ๐ผ4 = ๐ผ6
Now, for DC analysis, V1=V2
By symmetry,
Va=Vb=Vc &
V1=Vm1=Vm2=V2 โฆโฆโฆโฆโฆ. (2)
13-Jun-16 Snehasish Roychowdhury 6
Fig ref: Sawigun, C.; Pal, D.; Demosthenous, A., "A wide-input linear range sub-threshold transconductor for sub-Hz filtering," in Circuits and Systems (ISCAS), Proceedings of 2010 IEEE International Symposium on , vol., no., pp.1567-1570,2010
Fig4. Normal voltage attenuator
Vm1 Vm2
Voltage attenuator design (Ctd.)
Small signal Analysis:
Assumed all gmโs are equal.
(๐ฃ๐โ๐ฃ1) + (๐ฃ๐ โ ๐ฃ2) = 0
& ๐๐
4๐ฃ๐ โ ๐ฃ๐ = ๐๐ ๐ฃ๐ โ ๐ฃ2
(๐ฃ๐โ๐ฃ๐) =2
3(๐ฃ1 โ ๐ฃ2) โฆ.. (3)
Again,๐๐
4๐ฃ๐ โ ๐ฃ๐ = ๐๐ ๐ฃ๐ โ ๐ฃ๐1 = ๐๐(๐ฃ๐2 โ ๐ฃ๐)
By solving: ๐ฃ๐1 โ ๐ฃ๐2 =๐ฃ๐โ๐ฃ๐
2 โฆโฆ. (4)
From (3) & (4), ๐ฃ๐1 โ ๐ฃ๐2 = ๐ฃ๐๐ =๐ฃ1โ๐ฃ2
3=
๐ฃ๐๐
k โฆโฆโฆโฆ. (5)
Attenuation factor for one stage attenuator = (1๐ ).
13-Jun-16 Snehasish Roychowdhury 7
Fig5. ac equivalent of voltage attenuator
vm1 vm2
Overall trans-conductor From (1) & (5),
๐ผ๐๐ = ๐ผ๐ ๐ 2 โ ๐ผ๐ ๐ 1 tanh๐ฃ๐๐
2๐๐๐๐ &
๐บ๐ =(๐ผ๐ ๐ 2โ๐ผ๐ ๐ 1)
2๐๐๐๐sech2 ๐ฃ๐๐
2๐๐๐๐ โฆ(6)
Voltage Attenuator takes role in x-axis compression too.
Cross-coupled trans-conductor reduces Gm.
From Simulation: ๐ผ๐ ๐ 1 = 4๐๐ด, ๐ผ๐ ๐ 2 = 5.23๐๐ด, ๐ = 3
From eqn(6):Gmโ0.47nA/V. From simulation,
Gm has almost constant value:0.4637nS upto 140mVpp.
8 Fig6. Gm variation with input amplitude
(140mVpp, 0.463nA/V)
y=sech2(x)
1st order Gm-C filter response
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Fig7. Closed loop Gm-C filter
Fig8. ac response of the Gm-C filter
Cut-off freq of unity gain
Closed loop confign: 71.18Hz
Roll off: -20dB/decade
Cut-off frequency, ๐๐ =๐๐
2ฯ๐ถ๐ฟ๐๐๐
For Gm=0.463 nA/V , Cload=1pF , Theoretically cut-off frequency, fo=70.53Hz.
HF linearity issue (coherent sampling)
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Fig10 : FFT Analysis (Log Magnitude vs frequency) with coherent sampling NWINDOW=29, NRECOED=512
Freq=10Hz
Amp=20mVpp
DC offset=78uV
Freq=1MHz
Amp=20mVpp
DC offset=0.24mV
High frequency Interferences
Input stage offset is critical, as it is amplified at the output stage. So, we want to reduce offset at input stage, Voltage Attenuator.
Unity gain configura-
tion at high frequency
๐๐๐ข๐ก = ๐๐ = ๐๐ท๐ถ
Hence, CM interference,
๐ฃ๐๐ =๐ฃ๐+๐ฃ๐
2= ๐๐ท๐ถ +
๐ฃ๐๐๐
2
๐ฃ๐๐ = ๐ฃ๐ โ ๐ฃ๐ = ๐ฃ๐๐๐
It behaves like open loop
With both CM & DM
Interferences.DM inter-
ference canโt be avoided.
CM interference effect at output can be made zero by proper biasing technique.
13-Jun-16 Snehasish Roychowdhury 11
Fig11. High frequency equivalent circuit
CM interference
DM interference
CM & DM Interferences
๐ป๐๐ =๐ฃ๐1,๐๐
๐ฃ๐๐ ๐ฃ๐๐=0
๐ป๐๐ =๐ฃ๐1,๐๐
๐ฃ๐๐/2 ๐ฃ๐๐=0
Hence, ๐ฃ๐1 = ๐ป๐๐๐ฃ๐๐ + ๐ป๐๐ฃ๐
& ๐ฃ๐2 = ๐ป๐๐๐ฃ๐๐ โ ๐ป๐๐ฃ๐
13-Jun-16 Snehasish Roychowdhury 12
CM half circuit
DM half circuit
Fig12: Voltage attenuator
vo1 vo2 va vb
Offset issues in front end Attenuator
Let, ๐ = ๐จ๐๐๐(ฯ๐)
For 2nd order harmonics at output:
๐๐ = ๐จ๐๐๐๐๐ ฯ๐ =๐จ๐
๐(๐ โ ๐๐๐(๐ฯ๐))
Hence, 2nd order harmonics give DC offset that canโt be removed by low-pass filtering.
Offset 1 :
due to second order term : a2(Hcmvcm+Hdvd )2 ,
Offset 2 :
due to second order term : a2(Hcmvcm-Hdvd )2 ,
Hence, Offset 1 & Offset 2 are significantly different values from each other.
13
Offset issues in front end Attenuator
This offset will increase further in second stage, where differential input is amplified by a high gain factor.
Let, net output vout=G*{(Offset1-Offset2)+2Hdvd} , G: gain of second stage= high
Draws huge offset at output.
Hence, elimination of offset
at the 1st stage is important.
Offset 1 :
due to : a2(Hcmvcm+Hdvd )2 ,
Offset 2 :
due to : a2(Hcmvcm-Hdvd )2 ,
Solution : as Hdโ 0, we can make Hcm=0 for Offset1 = Offset 2.
By this, vout gets rid of any DC offset.
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Reduce CM interference:
Outputs of attenuator: ๐ฃ๐1 = ๐ป๐๐๐ฃ๐ = ๐ฃ๐2 = ๐ฃ๐ (say) {as vdm=0 here}
KCL at node s1 (7), vo1 (8), s2 (9):
๐ ๐ถ๐๐ 1 + ๐๐ ๐ฃ๐ โ ๐ฃ๐ 1 + ๐๐ + ๐ ๐ถ๐๐ 2 ๐ฃ๐ โ ๐ฃ๐ 1 = ๐ ๐ถ๐1๐ฃ๐ 1 (7)
๐๐ + ๐ ๐ถ๐๐ 2 ๐ฃ๐ โ ๐ฃ๐ 1 + ๐๐ + ๐ ๐ถ๐๐ 3 ๐ฃ๐ โ ๐ฃ๐ 2 + ๐ ๐ถ๐2๐ฃ๐ = 0 (8)
๐๐ + ๐ ๐ถ๐๐ 3 ๐ฃ๐ = ๐ฃ๐ 2(๐๐ + ๐ (๐ถ๐๐ 3 + ๐ถ๐3)) (9)
From (9), find vs2 & Put in (8):
๐ฃ๐ 1 =
๐ฃ๐1 2๐๐ + ๐ ๐ถ๐๐ 2 + ๐ถ๐๐ 3 + ๐ถ๐2 โ๐๐ + ๐ ๐ถ๐๐ 3
2
๐๐ + ๐ (๐ถ๐๐ 3 + ๐ถ๐3)
(๐๐+๐ ๐ถ๐๐ 2)
Put vs1 in (7): ๐ป๐๐ =๐ฃ๐1
๐ฃ๐๐=
(๐ ๐ถ๐๐ 1)
(๐๐+๐ ๐ถ๐๐ 2)โ 2๐๐1+๐ ๐ถ๐1+๐ถ๐๐ 1+๐ถ๐๐ 2
๐๐+๐ ๐ถ๐๐ 22๐๐+๐ ๐ถ๐๐ 2+๐ถ๐๐ 3+๐ถ๐2 โ
๐๐+๐ ๐ถ๐๐ 32
๐๐+๐ ๐ถ๐๐ 3+๐ถ๐3
Observation: Hcm decreases if CT1 & CT3 decreases. We want to minimize CT1 & CT3.
CT1=Cdb+Csb1+Csb2 , CT3=Csb3+ Cdb/2
13-Jun-16 Snehasish Roychowdhury 15
vcm vcm
vo1 vo2 s1 s2
Reduction of CM interference(Ctd.)
CT1=Cdb+Csb1+Csb2 , CT3=Csb3+ Cdb/2
We want Csb=0, but in twin-tub CMOS process, shorting
S to B causes high well capacitance (CGND) at source.
Hence, one auxiliary pair source node(Vs1) is connected to
to bulk of main pair. This is Source-buffered structure.
Csb1+Csb2 sees almost same potential across it. Though
these two potentials are not exactly equal, but voltage
across Csb1+Csb2 is very small, causing it to be
virtually shorted.
In advantage, S & B of main pair are decoupled.
13-Jun-16 Snehasish Roychowdhury 16
Fig14: vertical p-MOS, Fig15: Source-buffering
Fig14 Ref: Jingjing Yu; Amer, A.; Sanchez-Sinencio, E., "Electromagnetic Interference Resisting Operational Amplifier," in Circuits and Systems I :Regular Papers, IEEE Transactions on , vol.61, no.7, pp.1917-1927, July 2014
Common mode Transfer Function (Hcm)
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Fig17: Proposed Attenuator structure
Fig18: ac equivalent model of Half
Circuit for Common mode inputs
๐๐๐๐๐ = ๐๐๐๐๐๐ + ๐๐๐๐๐๐๐
๐๐๐๐๐ = ๐๐๐๐๐๐ + ๐๐๐๐๐๐๐
๐๐๐๐๐ = ๐๐๐๐๐๐ + ๐๐๐๐๐๐๐
๐ถ1 = ๐ถ๐๐ 1 + ๐ถ๐๐ฅ๐ก
Calculation for CMTF (Hcm)
Applying KCL at A,B,C,D,E,X nodes, we get six equations.
By solving these equations, we get:
๐ฏ๐๐(โ) =๐๐
๐๐=
๐ถ๐ษฃ๐โ๐ถ๐ษฃ๐
๐ถ๐๐ท๐โ๐ถ๐๐ท๐
Here, ฮฑ1, ฮฑ2 ,ฮฒ1, ฮฒ2 ,ษฃ1, ษฃ2 are constant values at very high
Frequency, coming from parasitic capacitances & C1.
By making , Hcm(โ)=0, we find a quadratic equation of C1.
By solving the equation , (C1=Cgs1+Cext ) we find:
๐ช๐๐๐ = ๐๐๐๐ญ
Adding this amount of external capacitance will make Hcm= 0 at very high frequency.
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Fig18: ac equivalent model of Half
Circuit for Common mode inputs
Comparison I (100mVpp)
13-Jun-16 Snehasish Roychowdhury 19
In frequency range [10Hz-100MHz], Maximum EMI induced offset (Magnitude): (i) Voltage attenuator with
cross-coupled transconductor based Gm-C filter:: 1.996% (ii) Proposed Gm-C filter:: 0.459%
Upto 100KHz, offsets are
almost same as parasitic effect
comes at high frequencies.
Upto 100KHz, EMI induced
Offset is very low(~uV). This
proves good linearity of the fil-
ter at low frequency.
Proposed Gm-C filter shows
small offset even at high freq
due to its immunity to CM inter
ference.
Acc to calculation, offset will
reach zero at infinite frequency
Fig19: EMI induced DC offset variation
with frequency for 100mVpp input signal
Comparison II ( FFT: 50mVpp, 1MHz)
13-Jun-16 Snehasish Roychowdhury 20
Fig 20:Voltage attenuator with
cross-coupled trans-conductor
based Gm-C filter
DC offset=1.908mV
Fig 21: Proposed EMI
Resisting Gm-C Filter
DC offset=0.46mV
Coherent sampling: fin=1MHz, NWINDOW=29, NRECORD=512 : irreducible. DC Offset is reduced in EMI resisting filter.
Comparison III (1MHz)
13-Jun-16 Snehasish Roychowdhury 21
With input amplitude,EMI induced DC offset increases. Rate of increment is less in proposed Filter.
For non-lineatity of the filter
at high frequency, DC offset
is high. Let, vin=Asin(ฯt). ie.
๐ฃ๐๐2 =
๐ด2
21 โ cos ฯt .This DC
Offset is high when amplitude
is high as seen in figure.
But, by making Hcm(โ)=0,
we are reducing non-linearity
of the filter at high frequency.
This is why at ๐ โซ ๐บ๐ต๐
(f=1MHz here), this proposed
filter is much immune to
Electromagnetic Interference.
(CM interference precisely)
Fig22: EMI Induced DC offset variation
with input EMI amplitude at 1MHz
Conclusion
Advantage:
oFor out-of band EMI frequencies, when parasitic capacitances come in picture, EMI induced DC offset becomes a challenging issue. Hence, This approach for EMI resisting highly linear sub-kilohertz Gm-C filter is noble in this field.
Disadvantage:
o Added power dissipation
o Transistor area increment
13-Jun-16 Snehasish Roychowdhury 22
THANK YOU
13-Jun-16 Snehasish Roychowdhury 23
Source buffering We want to make Hcm=0 & accordingly weโll choose C1
KCL at node C(10),E(11),D(12):
๐ ๐ถ๐๐ + ๐๐1 ๐ฃ1 + ๐ฃ๐ธ โ 2๐ฃ๐ด โ 2๐ฃ๐๐ 1 = ๐ ๐ถ๐1๐ฃ๐ด + ๐ (๐ถ๐๐ 1 + ๐ถ๐1)๐ฃ๐๐ 1 .(10)
2๐ฃ๐ธ โ ๐ฃ๐ท โ ๐ฃ๐ถ ๐ ๐ถ๐๐ + ๐๐1 + ๐ ๐ถ๐2๐ฃ๐ธ = 0 .(11)
๐ฃ๐ธ โ ๐ฃ๐ต โ ๐ฃ๐๐ 2 ๐ ๐ถ๐๐ + ๐๐1 = ๐ ๐ถ๐๐ 2+๐ถ๐3
2๐ฃ๐๐ 2 + ๐
๐ถ๐3
2๐ฃ๐ต .(12)
Solving (11) & (12) to omit vE:
๐๐ + ๐ ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
22๐๐ + ๐ 2๐ถ๐๐ 2 + ๐ถ๐2 โ ๐๐ + ๐ ๐ถ๐๐
2โ ๐ฃ๐๐ 2 + ๐ฃ๐ต = (๐ฃ๐ด + ๐ฃ๐๐ 1) ๐๐ + ๐ ๐ถ๐๐
2 (13)
Put vE value from(12) in (10):
๐ ๐ถ๐๐ + ๐๐1 ๐ฃ1 + ๐ฃ๐๐ 2 ๐๐ + ๐ ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2+ ๐ฃ๐ต ๐๐ + ๐ ๐ถ๐๐ +
๐ถ๐3
2=
๐ฃ๐ด 2๐๐ + ๐ 2๐ถ๐๐ + ๐ถ๐1 + ๐ฃ๐๐ 1 2๐๐ + ๐ 2๐ถ๐๐ + ๐ถ๐1 + ๐ถ๐ ๐1 (14)
13-Jun-16 Snehasish Roychowdhury 24
Fig12: ac equivalent of the CM Half circuit
Now, KCL at A(15), B(16), X(17):
๐ ๐ถ1 + ๐๐ ๐ฃ1 + ๐ฃ๐ฅ + 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐ฃ๐๐ 1 = 2๐๐ + ๐ 2๐ถ1 + ๐ถ๐1 ๐ฃ๐ด โฆ.โฆ. (15)
๐ ๐ถ1 + ๐๐ ๐ฃ๐ฅ + ๐๐๐1 +๐ ๐ถ๐ ๐2
2๐ฃ๐ ๐2 = ๐๐ + ๐ ๐ถ1 +
๐ถ๐3
2๐ฃ๐ต โฆโฆ. (16)
2๐๐ + ๐ 2๐ถ1 + ๐ถ๐2 ๐ฃ๐ฅ + ๐๐๐ ๐ฃ๐๐ 1 + ๐ฃ๐๐ 2 = (๐ฃ๐ด + ๐ฃ๐ต)(๐๐ + ๐ ๐ถ1) โฆโฆ. (17)
For minimizing,we assume:
๐ ๐ฃ๐๐ 2 + ๐ฃ๐ต = ๐ ๐ฃ๐ด + ๐ฃ๐๐ 1 โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..โฆ.โฆโฆ (13.a) , ๐ = ๐๐ + ๐ ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
22๐๐ + ๐ 2๐ถ๐๐ 2 + ๐ถ๐2 โ ๐๐ + ๐ ๐ถ๐๐
2 ,
๐ = ๐๐ + ๐ ๐ถ๐๐ 2
๐๐ฃ1 + ๐๐ฃ๐๐ 2 + ๐ ๐ฃ๐ต = ๐๐ฃ๐ด + ๐๐ฃ๐๐ 1 โฆโฆโฆโฆโฆโฆโฆโฆ.โฆโฆโฆ. (14.a) ๐ = ๐ ๐ถ๐๐ + ๐๐1 , ๐ = ๐๐ + ๐ ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2, ๐ = ๐๐ + ๐ ๐ถ๐๐ +
๐ถ๐3
2
๐ด ๐ฃ1 + ๐ฃ๐ + ๐ต๐ฃ๐๐ 1 = ๐ถ๐ฃ๐ดโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.โฆ... (15.a) ๐ = 2๐๐ + ๐ 2๐ถ๐๐ + ๐ถ๐1 , ๐ = 2๐๐ + ๐ 2๐ถ๐๐ + ๐ถ๐1 + ๐ถ๐ ๐1 , ๐ด = ๐ ๐ถ1 + ๐๐ , ๐ต = 2๐๐๐ + ๐ ๐ถ๐๐ 1
๐ท๐ฃ๐ + ๐ธ๐ฃ๐๐ 2 = ๐น๐ฃ๐ต โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.... (16.a) ๐ถ = 2๐๐ + ๐ 2๐ถ1 + ๐ถ๐1 , ๐ท = ๐ ๐ถ1 + ๐๐ , ๐ธ = ๐๐๐1 +๐ ๐ถ๐ ๐2
2,
๐น = ๐๐ + ๐ ๐ถ1 +๐ถ๐3
2
๐บ๐ฃ๐ + ๐ป ๐ฃ๐๐ 1 + ๐ฃ๐๐ 2 = ๐ผ(๐ฃ๐ด + ๐ฃ๐ต) โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.. (17.a) ๐บ = 2๐๐ + ๐ 2๐ถ1 + ๐ถ๐2 , ๐ป = ๐๐๐, ๐ผ = (๐๐ + ๐ ๐ถ1)
Solving (13.a) & (14.a):
๐ฃ๐๐ 2 = ๐2๐ฃ1 + ๐2๐ฃ๐ต + ๐2๐ฃ๐ด where, ๐2 =๐๐
(๐๐โ๐๐), ๐2 =
โ(๐โ๐)
(๐๐โ๐๐), ๐2 =
โ(๐๐โ๐๐ )
(๐๐โ๐๐) โฆโฆโฆ(18)
๐ฃ๐๐ 1 = ๐1๐ฃ1 + ๐1๐ฃ๐ต + ๐1๐ฃ๐ด where , ๐1 =๐๐
(๐๐โ๐๐), ๐1 =
๐(๐ โ๐)
(๐๐โ๐๐), ๐1 = โ 1 +
๐ ๐โ๐
๐๐โ๐๐ โฆ.โฆ.. (19)
13-Jun-16 Snehasish Roychowdhury 25
At high frequency (sโ), we find the following:
๐2 =๐๐
(๐๐โ๐๐)=
๐ถ๐๐ 3
2๐ถ๐๐ +๐ถ๐1+๐ถ๐ ๐1 ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
22๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐
2 โ๐ถ๐๐ 2 (๐ถ๐๐ +
๐ถ๐ ๐2+๐ถ๐32
)
๐2 =โ๐(๐โ๐)
(๐๐โ๐๐)=
๐ถ๐ ๐1โ๐ถ๐๐ 2
2๐ถ๐๐ +๐ถ๐1+๐ถ๐ ๐1 ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
22๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐
2 โ๐ถ๐๐ 2 (๐ถ๐๐ +
๐ถ๐ ๐2+๐ถ๐32
)
๐2 =โ ๐๐โ๐๐
(๐๐โ๐๐)=
๐ถ๐๐ 2 (๐ถ๐๐ +
๐ถ๐32
)โ 2๐ถ๐๐ +๐ถ๐1+๐ถ๐ ๐1 ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
22๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐
2
2๐ถ๐๐ +๐ถ๐1+๐ถ๐ ๐1 ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
22๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐
2 โ๐ถ๐๐ 2 (๐ถ๐๐ +
๐ถ๐ ๐2+๐ถ๐32
)
๐1 =๐๐
๐๐โ๐๐=
๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2 2๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐ 2 ๐ถ๐๐
2๐ถ๐๐ +๐ถ๐1+๐ถ๐ ๐1 ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2 2๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐ 2 โ๐ถ๐๐
2 (๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2 )
๐1 =๐(๐ โ๐)
๐๐โ๐๐=
โ ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2 2๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐ 2 โ
๐ถ๐32
2๐ถ๐๐ +๐ถ๐1+๐ถ๐ ๐1 ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2 2๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐ 2 โ๐ถ๐๐
2 (๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2 )
๐1 =๐(๐โ๐)
๐๐โ๐๐=
๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2 2๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐ 2 โ๐ถ๐ ๐1
2๐ถ๐๐ +๐ถ๐1+๐ถ๐ ๐1 ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2 2๐ถ๐๐ +๐ถ๐2 โ๐ถ๐๐ 2 โ๐ถ๐๐
2 ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2
โ 1
All terms are independent of C1
13-Jun-16 Snehasish Roychowdhury 26
Put equation(19) in (15.a): ๐ฃ1 ๐ด + ๐ต๐1 + ๐ต๐1๐ฃ๐ต + ๐ด๐ฃ๐ = ๐ถ โ ๐ต๐1 ๐ฃ๐ด
Put equation(18) in (16.a): ๐ท๐ฃ๐ + ๐ธ๐2๐ฃ1 + ๐ธ๐2๐ฃ๐ด = ๐น โ ๐ธ๐2 ๐ฃ๐ต
Put value of vA from first equation in second equation: ๐ผ1๐ฃ๐ต = ๐ฝ1๐ฃ๐ + ษฃ1๐ฃ1 (20)
where, ๐ผ1 = ๐น โ ๐ธ๐2 ๐ถ โ ๐ต๐1 โ ๐ธ๐2๐ต๐1 , ๐ฝ1 = ๐ท ๐ถ โ ๐ต๐1 + ๐ธ๐2๐ด , ษฃ1 = ๐ธ๐2 ๐ถ โ ๐ต๐1 + ๐ธ๐2 ๐ด + ๐ต๐1
From (18) & (19), put values vbs1 & vbs2 in (17.a): ๐ผ2๐ฃ๐ต = ๐ฝ2๐ฃ๐ + ษฃ2๐ฃ1 (21)
where, ๐ผ2 = ๐ผ +๐ผ๐ต๐1
๐ถโ๐ต๐1โ ๐ป ๐1 + ๐2 โ
๐ป ๐1+๐2 ๐ต๐1
๐ถโ๐ต๐1 , ๐ฝ2 = ๐บ +
๐ด๐ป ๐1+๐2
๐ถโ๐ต๐1โ
๐ด๐ผ
๐ถโ๐ต๐1 , ษฃ2 = ๐ป ๐1 + ๐2 +
๐ป ๐1+๐2 ๐ด+๐ต๐1
๐ถโ๐ต๐1โ
๐ด+๐ต๐1 ๐ผ
๐ถโ๐ต๐1
From (20) & (21): ๐ผ2
๐ผ1๐ฝ1๐ฃ๐ + ษฃ1๐ฃ1 = ๐ฝ2๐ฃ๐ + ษฃ2๐ฃ1
๐ฏ๐๐ =๐ถ๐ษฃ๐ โ ๐ถ๐ษฃ๐
๐ถ๐๐ท๐ โ ๐ถ๐๐ท๐
For Hcm=0, we find: ๐ถ๐ษฃ๐ โ ๐ถ๐ษฃ๐=0
๐ผ(๐ถ โ ๐ต๐1) + ๐ผ๐ต๐1 โ ๐ป ๐1 + ๐2 (๐ถ โ ๐ต๐1) โ ๐ป ๐1 + ๐2 ๐ต๐1 โ ๐ธ๐2 ๐ถ โ ๐ต๐1 + ๐ธ๐2 ๐ด + ๐ต๐1 = ๐น โ ๐ธ๐2 ๐ถ โ ๐ต๐1 โ ๐ธ๐2๐ต๐1 โ ๐ป ๐1 + ๐2 (๐ถ โ ๐ต๐1) + ๐ป ๐1 + ๐2 ๐ด + ๐ต๐1 โ ๐ด + ๐ต๐1 ๐ผ
1st term: ๐ผ(๐ถ โ ๐ต๐1) + ๐ผ๐ต๐1 โ ๐ป ๐1 + ๐2 (๐ถ โ ๐ต๐1) โ ๐ป ๐1 + ๐2 ๐ต๐1 divide numerator and denominator by [order of
max(num, den) = 2] , weโll find the term is simplified to: (๐ถ1) 2๐ถ1 + ๐ถ๐1 โ ๐ถ๐๐ 1๐1 + ๐ถ1๐ถ๐๐ 1๐1
2nd term: ๐ธ๐2 ๐ถ โ ๐ต๐1 + ๐ธ๐2 ๐ด + ๐ต๐1 divide numerator and denominator by s2 gives:
๐ถ๐ ๐2
2๐2 2๐ถ1 + ๐ถ๐1 โ ๐ถ๐๐ 1๐1 +
๐ถ๐ ๐2
2๐2 C1 + Cbs1๐1
13-Jun-16 Snehasish Roychowdhury 27
3rd term: ๐น โ ๐ธ๐2 ๐ถ โ ๐ต๐1 โ ๐ธ๐2๐ต๐1 is simplified by dividing num & den by s2 and put sโ
: ๐ถ1 +๐ถ๐3
2โ
๐ถ๐ ๐2
2โ ๐2 2๐ถ1 + ๐ถ๐1 โ ๐ถ๐๐ 1๐1 โ
๐ถ๐ ๐2
2๐ถ๐๐ 1 ๐2๐1
4th term: ๐ป ๐1 + ๐2 (๐ถ โ ๐ต๐1) + ๐ป ๐1 + ๐2 ๐ด + ๐ต๐1 โ ๐ด + ๐ต๐1 ๐ผ is simplifed by dividing num & den by s2 and put sโ : โ C1 + ๐ถ๐๐ 1๐1 (๐ถ1)
๐ช๐๐ ๐๐ฟ๐ + ๐๐๐ โ ๐ +
๐ช๐ ๐๐ช๐ป๐๐ฟ๐ โ ๐๐ช๐๐๐๐๐๐ฟ๐ + ๐๐๐๐ฌ๐๐ฟ๐๐๐ + ๐ช๐ป๐ โ ๐ช๐๐๐๐๐ + ๐ช๐๐๐๐๐ ๐๐ฟ๐ + ๐๐ โ ๐ช๐ป๐ โ ๐ช๐๐๐๐๐ โ ๐ช๐๐๐๐๐๐๐ + ๐๐ช๐๐๐๐ฟ๐ +๐ช๐ป๐ โ ๐ช๐๐๐๐๐ + ๐ช๐๐๐๐๐ ๐ช๐ป๐๐ฟ๐ โ ๐ช๐๐๐๐๐๐ฟ๐ + ๐๐๐ฌ๐๐ฟ๐๐๐ โ ๐ช๐๐๐๐ฟ๐(๐ช๐ป๐ โ ๐ช๐๐๐๐๐ โ ๐ช๐๐๐๐๐๐๐) = 0
This follows the form: ๐๐ถ12 + ๐๐ถ1 + ๐ = 0
From DC simulation: Cgs=7.05fF , Csb1=17.14fF , Csb2=17.02fF, CT1=10.46fF , CT2=10.47fF , CT3=10.47fF , gm=42.4nA/V
Hence, putting the values in all equations, ๐ = โ4.03, ๐ = โ2.53 โ 10โ15 & ๐ = 4.12 โ 10โ26
๐ช๐ = โ๐. ๐๐ โ ๐๐โ๐๐ โ โ๐. ๐๐ โ ๐๐โ๐๐ ๐ + ๐ โ ๐. ๐๐ โ ๐. ๐๐ โ ๐๐โ๐๐
๐ โ ๐. ๐๐= ๐๐๐. ๐๐ ๐๐ญ
๐ถ1 = ๐ถ๐๐ + ๐ถ๐๐ฅ๐ก , Here, Cgs=7.05fF , ๐ถ๐๐ฅ๐ก = 93.74๐๐น
We consider, ๐ช๐๐๐ = ๐๐๐๐ญ
13-Jun-16 Snehasish Roychowdhury 28
RETURN
Infinite frequency for 1st term:
1st term:
๐ผ(๐ถ โ ๐ต๐1) + ๐ผ๐ต๐1 โ ๐ป ๐1 + ๐2 (๐ถ โ ๐ต๐1) โ ๐ป ๐1 + ๐2 ๐ต๐1
= ๐๐ + ๐ ๐ถ1 2๐๐ + ๐ 2๐ถ1 + ๐ถ๐1 โ 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐1 + 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐1
โ ๐๐๐ ๐1 + ๐2 2๐๐ + ๐ 2๐ถ1 + ๐ถ๐1 โ ,๐๐๐ ๐1 + ๐2 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐1-
The critical frequencies above which (at least 10 times) we are considering the approximation of sโ is valid:
๐ ๐๐
๐ถ1=
42.2๐
100.79๐ = 67.48 ๐พ๐ป๐ง
๐๐2๐๐
2๐ถ1 + ๐ถ๐1= 64.27 ๐พ๐ป๐ง
๐๐๐2๐๐๐
๐ถ๐๐ 1= 74.9 ๐พ๐ป๐ง
13-Jun-16 Snehasish Roychowdhury 29
Infinite frequency for 2nd term:
2nd term:
๐ธ๐2 ๐ถ โ ๐ต๐1 + ๐ธ๐2 ๐ด + ๐ต๐1
= ๐๐๐1 +๐ ๐ถ๐๐ 2
2,๐2 2๐๐ + ๐ 2๐ถ1 + ๐ถ๐1 โ 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐1 + ๐2* ๐๐ + ๐ ๐ถ1 + 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐1+-
The critical frequencies above which (at least 10 times) we are considering the approximation of sโ is valid:
๐2๐๐๐2 โ 2๐๐๐๐1
2๐ถ1 + ๐ถ๐1 ๐2 โ ๐ถ๐๐ 1๐1= 71.33 ๐พ๐ป๐ง
๐๐(๐๐+2๐๐๐๐1)
๐ถ1 + ๐ถ๐๐ 1๐1= 67.16 ๐พ๐ป๐ง
13-Jun-16 Snehasish Roychowdhury 30
Infinite frequency for 3rd term:
3rd term:
๐น โ ๐ธ๐2 ๐ถ โ ๐ต๐1 โ ๐ธ๐2๐ต๐1
= ๐๐ + ๐๐๐๐2 + ๐ ๐ถ1 +๐ถ๐3
2โ
๐ถ๐๐ 2
2๐2 2๐๐ โ 2๐๐๐ + ๐ 2๐ถ1 + ๐ถ๐1 โ ๐ถ๐๐ 1๐1 โ ๐2๐1(๐๐๐
+๐ ๐ถ๐๐ 2
2)(2๐๐๐ + ๐ ๐ถ๐๐ 1)
The critical frequencies above which (at least 10 times) we are considering the approximation of sโ is valid:
๐๐๐ + ๐๐๐๐2
๐ถ1 +๐ถ๐32 โ
๐ถ๐๐ 22 ๐2
= 64.18 ๐พ๐ป๐ง
๐๐2(๐๐ โ ๐๐๐)
2๐ถ1 + ๐ถ๐1 โ ๐ถ๐๐ 1๐1= 54.6 ๐พ๐ป๐ง
13-Jun-16 Snehasish Roychowdhury 31
Infinite frequency for 4th term:
4th term:
๐ป ๐1 + ๐2 (๐ถ โ ๐ต๐1) + ๐ป ๐1 + ๐2 ๐ด + ๐ต๐1 โ ๐ด + ๐ต๐1 ๐ผ
= ๐๐๐ ๐1 + ๐2 2๐๐ + ๐ (2๐ถ1+๐ถ๐1 โ 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐1- + ๐๐๐(๐1 + ๐2),๐๐ + ๐ ๐ถ1 + 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐1-
โ (๐๐ + ๐ ๐ถ1),๐๐ + ๐ ๐ถ1 + 2๐๐๐ + ๐ ๐ถ๐๐ 1 ๐1-
The critical frequencies above which (at least 10 times) we are considering the approximation of sโ is valid:
๐๐๐ + 2๐๐๐๐1
๐ถ1 + ๐ถ๐๐ 1๐1= 67.16 ๐พ๐ป๐ง
๐๐2(๐๐ โ ๐๐๐๐1)
2๐ถ1 + ๐ถ๐1 โ ๐ถ๐๐ 1๐1= 64.37 ๐พ๐ป๐ง
13-Jun-16 Snehasish Roychowdhury 32
๐ = ๐๐ + ๐ ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
22๐๐ + ๐ 2๐ถ๐๐ 2 + ๐ถ๐2 โ ๐๐ + ๐ ๐ถ๐๐
2 ,
๐ = ๐๐ + ๐ ๐ถ๐๐ 2, ๐ = ๐ ๐ถ๐๐ + ๐๐1 ,
๐ = ๐๐ + ๐ ๐ถ๐๐ +๐ถ๐ ๐2+๐ถ๐3
2, ๐ = ๐๐ + ๐ ๐ถ๐๐ +
๐ถ๐3
2
๐ = 2๐๐ + ๐ 2๐ถ๐๐ + ๐ถ๐1 , ๐ = 2๐๐ + ๐ 2๐ถ๐๐ + ๐ถ๐1 + ๐ถ๐ ๐1 ,
๐ด = ๐ ๐ถ1 + ๐๐ , ๐ต = 2๐๐๐ + ๐ ๐ถ๐๐ 1
๐ถ = 2๐๐ + ๐ 2๐ถ1 + ๐ถ๐1 , ๐ท = ๐ ๐ถ1 + ๐๐ ,
๐ธ = ๐๐๐1 +๐ ๐ถ๐ ๐2
2, ๐น = ๐๐ + ๐ ๐ถ1 +
๐ถ๐3
2
๐บ = 2๐๐ + ๐ 2๐ถ1 + ๐ถ๐2 , ๐ป = ๐๐๐, ๐ผ = (๐๐ + ๐ ๐ถ1)
๐2 =๐๐
(๐๐โ๐๐), ๐2 =
โ๐(๐โ๐)
(๐๐โ๐๐), ๐2 =
โ(๐๐โ๐๐ )
(๐๐โ๐๐)
๐1 =๐๐
(๐๐โ๐๐), ๐1 =
๐(๐ โ๐)
(๐๐โ๐๐), ๐1 = โ 1 +
๐ ๐โ๐
๐๐โ๐๐
13-Jun-16 Snehasish Roychowdhury 33
For X1,2 , Y1,2, Z1,2 , the frequencies above which we can consider๐ โ , are given as following:
M,Q: (i) ๐๐
๐ถ๐๐ +๐ถ๐๐ 2+๐ถ๐3
2
= 324.5๐พ๐ป๐ง
(ii) 2๐๐
2๐ถ๐๐ +๐ถ๐2= 549.3๐พ๐ป๐ง
N,P : ๐๐
๐ถ๐๐ = 957.2๐พ๐ป๐ง
R: 2๐๐
2๐ถ๐๐ +๐ถ๐3= 549.3๐พ๐ป๐ง
T: ๐๐
๐ถ๐๐ +๐ถ๐1+๐ถ๐๐ 1
2
< 549๐พ๐ป๐ง
S: 2๐๐
2๐ถ๐๐ +๐ถ๐1= 549.3๐พ๐ป๐ง
Conclusion: All the terms are considered individually to find the frequency above which we can consider our assumption is valid. Ie Hcm is zero. From all the terms above, we find highest frequency =957.2 KHz. Hence, above almost 10 times, of it ie 9-10 MHz, we can Assume frequency to be infinite. All our assumptions are valid above this frequency. This is valid as 10MHz is considered typically in EMI frequency range.
Comparison (100mVpp)
13-Jun-16 Snehasish Roychowdhury 34
In frequency range [10Hz-100MHz], Maximum EMI induced offset (Magnitude): (i) Voltage attenuator with
cross-coupled transconductor based Gm-C filter:: 1.996% (ii) Proposed Gm-C filter:: 0.459%
Here, we see the EMI
Induced DC offset is reduced
significantly for this proposed
Gm-C filter compared to
Uncompesated Gm-C filter
From the frequency near
10MHz, which is considered to
Be the minimum value of
Infinty while calculating CMTF,
Hcm(โ)=0.
Fig19: EMI induced DC offset variation
with frequency for 100mVpp input signal
DC offset reduction at high frequency
13-Jun-16 Snehasish Roychowdhury 35
๐ฃ๐1 = ๐ป๐๐๐ฃ๐๐ + ๐ป๐๐๐ฃ๐๐
2 &
๐ฃ๐2 = ๐ป๐๐๐ฃ๐๐ โ ๐ป๐๐๐ฃ๐๐
2
And, ๐๐1 = ๐ฃ๐1 + ๐๐ท๐ถ & ๐๐2 = ๐ฃ๐2 + ๐๐ท๐ถ
CASE1: If Common mode interference
comes at attenuator output, then from
equations above: |๐ฃ๐1| โ |๐ฃ๐2| Unequal capacitive division occurs
across gate-source of M1 & M2 due to
CT,tail and as a result, |vgs,M1|โ |vgs,M2| ie
finite offset current IOS at output.
CASE2: Instead, if we make Hcm=0
somehow, ie no CM Interference at
attenuator output, ๐ฃ๐1 = |๐ฃ๐2|
In this case, P acts as virtual ground as second stage doesnโt see any common mode input. ie Ios=0.
This is why we need to minimize Hcm of the first stage to reduce DC offset at output.