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 π£ππ β π£ππ’π‘ β πΊπ = πΌππ’π‘
Again,
π£ππ’π‘ = πΌππ’π‘ β (1
π πΆππππ)
Hence, π£ππ β π£ππ’π‘ β πΊπ = π πΆπππππ£ππ’π‘
π£ππ β πΊπ = π πΆππππ + πΊπ β π£ππ’π‘
π£ππ’π‘
π£ππ=
1
1+π πΆππππ
πΊπ
oHence, DC gain= 0dB
o Cut-off frequency of the filter is : πΊπ
2ΟπΆππππ π»π§
13-Jun-16 Snehasish Roychowdhury 3
Fig1: Gm-C filter using voltage follower configuration
Block diagram of Gm-C filter
Cut-off freq=πΊπ
2ππΆππππ
For 70Hz, Cload=1pF ,
Gm=0.47nA/V.
πΊπ =πππ’π‘
π£ππ=
πππ’π‘
π£ππβ π΄π‘π‘π. ππππ‘ππ
1
π
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.
13-Jun-16 Snehasish Roychowdhury 4
Fig2. Block diagram of Gm-C filter
Iout
Cross-coupled trans-conductor design:
πΌπ1 = πΌ1 + πΌ3 & πΌπ2 = πΌ2 + πΌ4
πΌ1
πΌ2=
expπ£π π 1βπ£π1
πππ
expπ£π π 1βπ£π2
πππ
= exp (βπ£ππ
πππ)
πΌ3
πΌ4=
expπ£π π 2βπ£π2
πππ
expπ£π π 2βπ£π1
πππ
= exp (π£ππ
πππ)
Now,
πΌ1 β πΌ2πΌ1 + πΌ2
=exp
βπ£πππππ
β 1
expβπ£πππππ
+ 1=
expβπ£ππ2πππ
β expπ£ππ
2πππ
expβπ£ππ12πππ
+ expπ£ππ
2πππ
= βtanh (π£ππ
2πππ)
πΌ1 β πΌ2 = βπΌπ π 1 tanhπ£ππ
2πππ
πΌ3 β πΌ4 = πΌπ π 2 tanhπ£ππ
2πππ
πΌππ = πΌπ1 β πΌπ2 = πΌπ π 2 β πΌπ π 1 tanhπ£ππ
2πππ β¦ . . . (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 gmβs & Ibiasβs are
equal. Hence,
πΌ1 = πΌ3 = πΌ5 & πΌ2 = πΌ4 = πΌ6
Now, for DC analysis, V1=V2
By symmetry,
Va=Vb=Vc &
V1=Vm1=Vm2=V2 β¦β¦β¦β¦β¦. (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 gmβs are equal.
(π£πβπ£1) + (π£π β π£2) = 0
& ππ
4π£π β π£π = ππ π£π β π£2
(π£πβπ£π) =2
3(π£1 β π£2) β¦.. (3)
Again,ππ
4π£π β π£π = ππ π£π β π£π1 = ππ(π£π2 β π£π)
By solving: π£π1 β π£π2 =π£πβπ£π
2 β¦β¦. (4)
From (3) & (4), π£π1 β π£π2 = π£ππ =π£1βπ£2
3=
π£ππ
k β¦β¦β¦β¦. (5)
Attenuation factor for one stage attenuator = (1π ).
13-Jun-16 Snehasish Roychowdhury 7
Fig5. ac equivalent of voltage attenuator
vm1 vm2
Overall trans-conductor From (1) & (5),
πΌππ = πΌπ π 2 β πΌπ π 1 tanhπ£ππ
2ππππ &
πΊπ =(πΌπ π 2βπΌπ π 1)
2ππππsech2 π£ππ
2ππππ β¦(6)
Voltage Attenuator takes role in x-axis compression too.
Cross-coupled trans-conductor reduces Gm.
From Simulation: πΌπ π 1 = 4ππ΄, πΌπ π 2 = 5.23ππ΄, π = 3
From eqn(6):Gmβ0.47nA/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
13-Jun-16 Snehasish Roychowdhury 9
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, ππ =ππ
2ΟπΆπΏπππ
For Gm=0.463 nA/V , Cload=1pF , Theoretically cut-off frequency, fo=70.53Hz.
HF linearity issue (coherent sampling)
13-Jun-16 Snehasish Roychowdhury 10
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
πππ’π‘ = ππ = ππ·πΆ
Hence, CM interference,
π£ππ =π£π+π£π
2= ππ·πΆ +
π£πππ
2
π£ππ = π£π β π£π = π£πππ
It behaves like open loop
With both CM & DM
Interferences.DM inter-
ference canβt 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
π»ππ =π£π1,ππ
π£ππ π£ππ=0
π»ππ =π£π1,ππ
π£ππ/2 π£ππ=0
Hence, π£π1 = π»πππ£ππ + π»ππ£π
& π£π2 = π»πππ£ππ β π»ππ£π
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, π = π¨πππ(Οπ)
For 2nd order harmonics at output:
ππ = π¨πππππ Οπ =π¨π
π(π β πππ(πΟπ))
Hence, 2nd order harmonics give DC offset that canβt 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 Hdβ 0, we can make Hcm=0 for Offset1 = Offset 2.
By this, vout gets rid of any DC offset.
13-Jun-16 Snehasish Roychowdhury 14
Reduce CM interference:
Outputs of attenuator: π£π1 = π»πππ£π = π£π2 = π£π (say) {as vdm=0 here}
KCL at node s1 (7), vo1 (8), s2 (9):
π πΆππ 1 + ππ π£π β π£π 1 + ππ + π πΆππ 2 π£π β π£π 1 = π πΆπ1π£π 1 (7)
ππ + π πΆππ 2 π£π β π£π 1 + ππ + π πΆππ 3 π£π β π£π 2 + π πΆπ2π£π = 0 (8)
ππ + π πΆππ 3 π£π = π£π 2(ππ + π (πΆππ 3 + πΆπ3)) (9)
From (9), find vs2 & Put in (8):
π£π 1 =
π£π1 2ππ + π πΆππ 2 + πΆππ 3 + πΆπ2 βππ + π πΆππ 3
2
ππ + π (πΆππ 3 + πΆπ3)
(ππ+π πΆππ 2)
Put vs1 in (7): π»ππ =π£π1
π£ππ=
(π πΆππ 1)
(ππ+π πΆππ 2)β 2ππ1+π πΆπ1+πΆππ 1+πΆππ 2
ππ+π πΆππ 22ππ+π πΆππ 2+πΆππ 3+πΆπ2 β
ππ+π πΆππ 32
ππ+π πΆππ 3+πΆπ3
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)
13-Jun-16 Snehasish Roychowdhury 17
Fig17: Proposed Attenuator structure
Fig18: ac equivalent model of Half
Circuit for Common mode inputs
πππππ = ππππππ + πππππππ
πππππ = ππππππ + πππππππ
πππππ = ππππππ + πππππππ
πΆ1 = πΆππ 1 + πΆππ₯π‘
Calculation for CMTF (Hcm)
Applying KCL at A,B,C,D,E,X nodes, we get six equations.
By solving these equations, we get:
π―ππ(β) =ππ
ππ=
πΆπΙ£πβπΆπΙ£π
πΆππ·πβπΆππ·π
Here, Ξ±1, Ξ±2 ,Ξ²1, Ξ²2 ,Ι£1, Ι£2 are constant values at very high
Frequency, coming from parasitic capacitances & C1.
By making , Hcm(β)=0, we find a quadratic equation of C1.
By solving the equation , (C1=Cgs1+Cext ) we find:
πͺπππ = ππππ
Adding this amount of external capacitance will make Hcm= 0 at very high frequency.
13-Jun-16 Snehasish Roychowdhury 18
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(Οt). ie.
π£ππ2 =
π΄2
21 β cos Οt .This DC
Offset is high when amplitude
is high as seen in figure.
But, by making Hcm(β)=0,
we are reducing non-linearity
of the filter at high frequency.
This is why at π β« πΊπ΅π
(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 weβll choose C1
KCL at node C(10),E(11),D(12):
π πΆππ + ππ1 π£1 + π£πΈ β 2π£π΄ β 2π£ππ 1 = π πΆπ1π£π΄ + π (πΆππ 1 + πΆπ1)π£ππ 1 .(10)
2π£πΈ β π£π· β π£πΆ π πΆππ + ππ1 + π πΆπ2π£πΈ = 0 .(11)
π£πΈ β π£π΅ β π£ππ 2 π πΆππ + ππ1 = π πΆππ 2+πΆπ3
2π£ππ 2 + π
πΆπ3
2π£π΅ .(12)
Solving (11) & (12) to omit vE:
ππ + π πΆππ +πΆπ π2+πΆπ3
22ππ + π 2πΆππ 2 + πΆπ2 β ππ + π πΆππ
2β π£ππ 2 + π£π΅ = (π£π΄ + π£ππ 1) ππ + π πΆππ
2 (13)
Put vE value from(12) in (10):
π πΆππ + ππ1 π£1 + π£ππ 2 ππ + π πΆππ +πΆπ π2+πΆπ3
2+ π£π΅ ππ + π πΆππ +
πΆπ3
2=
π£π΄ 2ππ + π 2πΆππ + πΆπ1 + π£ππ 1 2ππ + π 2πΆππ + πΆπ1 + πΆπ π1 (14)
13-Jun-16 Snehasish Roychowdhury 24
Fig12: ac equivalent of the CM Half circuit
Now, KCL at A(15), B(16), X(17):
π πΆ1 + ππ π£1 + π£π₯ + 2πππ + π πΆππ 1 π£ππ 1 = 2ππ + π 2πΆ1 + πΆπ1 π£π΄ β¦.β¦. (15)
π πΆ1 + ππ π£π₯ + πππ1 +π πΆπ π2
2π£π π2 = ππ + π πΆ1 +
πΆπ3
2π£π΅ β¦β¦. (16)
2ππ + π 2πΆ1 + πΆπ2 π£π₯ + πππ π£ππ 1 + π£ππ 2 = (π£π΄ + π£π΅)(ππ + π πΆ1) β¦β¦. (17)
For minimizing,we assume:
π π£ππ 2 + π£π΅ = π π£π΄ + π£ππ 1 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..β¦.β¦β¦ (13.a) , π = ππ + π πΆππ +πΆπ π2+πΆπ3
22ππ + π 2πΆππ 2 + πΆπ2 β ππ + π πΆππ
2 ,
π = ππ + π πΆππ 2
ππ£1 + ππ£ππ 2 + π π£π΅ = ππ£π΄ + ππ£ππ 1 β¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦β¦. (14.a) π = π πΆππ + ππ1 , π = ππ + π πΆππ +πΆπ π2+πΆπ3
2, π = ππ + π πΆππ +
πΆπ3
2
π΄ π£1 + π£π + π΅π£ππ 1 = πΆπ£π΄β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦... (15.a) π = 2ππ + π 2πΆππ + πΆπ1 , π = 2ππ + π 2πΆππ + πΆπ1 + πΆπ π1 , π΄ = π πΆ1 + ππ , π΅ = 2πππ + π πΆππ 1
π·π£π + πΈπ£ππ 2 = πΉπ£π΅ β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.... (16.a) πΆ = 2ππ + π 2πΆ1 + πΆπ1 , π· = π πΆ1 + ππ , πΈ = πππ1 +π πΆπ π2
2,
πΉ = ππ + π πΆ1 +πΆπ3
2
πΊπ£π + π» π£ππ 1 + π£ππ 2 = πΌ(π£π΄ + π£π΅) β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (17.a) πΊ = 2ππ + π 2πΆ1 + πΆπ2 , π» = πππ, πΌ = (ππ + π πΆ1)
Solving (13.a) & (14.a):
π£ππ 2 = π2π£1 + π2π£π΅ + π2π£π΄ where, π2 =ππ
(ππβππ), π2 =
β(πβπ)
(ππβππ), π2 =
β(ππβππ )
(ππβππ) β¦β¦β¦(18)
π£ππ 1 = π1π£1 + π1π£π΅ + π1π£π΄ where , π1 =ππ
(ππβππ), π1 =
π(π βπ)
(ππβππ), π1 = β 1 +
π πβπ
ππβππ β¦.β¦.. (19)
13-Jun-16 Snehasish Roychowdhury 25
At high frequency (sβ), we find the following:
π2 =ππ
(ππβππ)=
πΆππ 3
2πΆππ +πΆπ1+πΆπ π1 πΆππ +πΆπ π2+πΆπ3
22πΆππ +πΆπ2 βπΆππ
2 βπΆππ 2 (πΆππ +
πΆπ π2+πΆπ32
)
π2 =βπ(πβπ)
(ππβππ)=
πΆπ π1βπΆππ 2
2πΆππ +πΆπ1+πΆπ π1 πΆππ +πΆπ π2+πΆπ3
22πΆππ +πΆπ2 βπΆππ
2 βπΆππ 2 (πΆππ +
πΆπ π2+πΆπ32
)
π2 =β ππβππ
(ππβππ)=
πΆππ 2 (πΆππ +
πΆπ32
)β 2πΆππ +πΆπ1+πΆπ π1 πΆππ +πΆπ π2+πΆπ3
22πΆππ +πΆπ2 βπΆππ
2
2πΆππ +πΆπ1+πΆπ π1 πΆππ +πΆπ π2+πΆπ3
22πΆππ +πΆπ2 βπΆππ
2 βπΆππ 2 (πΆππ +
πΆπ π2+πΆπ32
)
π1 =ππ
ππβππ=
πΆππ +πΆπ π2+πΆπ3
2 2πΆππ +πΆπ2 βπΆππ 2 πΆππ
2πΆππ +πΆπ1+πΆπ π1 πΆππ +πΆπ π2+πΆπ3
2 2πΆππ +πΆπ2 βπΆππ 2 βπΆππ
2 (πΆππ +πΆπ π2+πΆπ3
2 )
π1 =π(π βπ)
ππβππ=
β πΆππ +πΆπ π2+πΆπ3
2 2πΆππ +πΆπ2 βπΆππ 2 β
πΆπ32
2πΆππ +πΆπ1+πΆπ π1 πΆππ +πΆπ π2+πΆπ3
2 2πΆππ +πΆπ2 βπΆππ 2 βπΆππ
2 (πΆππ +πΆπ π2+πΆπ3
2 )
π1 =π(πβπ)
ππβππ=
πΆππ +πΆπ π2+πΆπ3
2 2πΆππ +πΆπ2 βπΆππ 2 βπΆπ π1
2πΆππ +πΆπ1+πΆπ π1 πΆππ +πΆπ π2+πΆπ3
2 2πΆππ +πΆπ2 βπΆππ 2 βπΆππ
2 πΆππ +πΆπ π2+πΆπ3
2
β 1
All terms are independent of C1
13-Jun-16 Snehasish Roychowdhury 26
Put equation(19) in (15.a): π£1 π΄ + π΅π1 + π΅π1π£π΅ + π΄π£π = πΆ β π΅π1 π£π΄
Put equation(18) in (16.a): π·π£π + πΈπ2π£1 + πΈπ2π£π΄ = πΉ β πΈπ2 π£π΅
Put value of vA from first equation in second equation: πΌ1π£π΅ = π½1π£π + Ι£1π£1 (20)
where, πΌ1 = πΉ β πΈπ2 πΆ β π΅π1 β πΈπ2π΅π1 , π½1 = π· πΆ β π΅π1 + πΈπ2π΄ , Ι£1 = πΈπ2 πΆ β π΅π1 + πΈπ2 π΄ + π΅π1
From (18) & (19), put values vbs1 & vbs2 in (17.a): πΌ2π£π΅ = π½2π£π + Ι£2π£1 (21)
where, πΌ2 = πΌ +πΌπ΅π1
πΆβπ΅π1β π» π1 + π2 β
π» π1+π2 π΅π1
πΆβπ΅π1 , π½2 = πΊ +
π΄π» π1+π2
πΆβπ΅π1β
π΄πΌ
πΆβπ΅π1 , Ι£2 = π» π1 + π2 +
π» π1+π2 π΄+π΅π1
πΆβπ΅π1β
π΄+π΅π1 πΌ
πΆβπ΅π1
From (20) & (21): πΌ2
πΌ1π½1π£π + Ι£1π£1 = π½2π£π + Ι£2π£1
π―ππ =πΆπΙ£π β πΆπΙ£π
πΆππ·π β πΆππ·π
For Hcm=0, we find: πΆπΙ£π β πΆπΙ£π=0
πΌ(πΆ β π΅π1) + πΌπ΅π1 β π» π1 + π2 (πΆ β π΅π1) β π» π1 + π2 π΅π1 β πΈπ2 πΆ β π΅π1 + πΈπ2 π΄ + π΅π1 = πΉ β πΈπ2 πΆ β π΅π1 β πΈπ2π΅π1 β π» π1 + π2 (πΆ β π΅π1) + π» π1 + π2 π΄ + π΅π1 β π΄ + π΅π1 πΌ
1st term: πΌ(πΆ β π΅π1) + πΌπ΅π1 β π» π1 + π2 (πΆ β π΅π1) β π» π1 + π2 π΅π1 divide numerator and denominator by [order of
max(num, den) = 2] , weβll find the term is simplified to: (πΆ1) 2πΆ1 + πΆπ1 β πΆππ 1π1 + πΆ1πΆππ 1π1
2nd term: πΈπ2 πΆ β π΅π1 + πΈπ2 π΄ + π΅π1 divide numerator and denominator by s2 gives:
πΆπ π2
2π2 2πΆ1 + πΆπ1 β πΆππ 1π1 +
πΆπ π2
2π2 C1 + Cbs1π1
13-Jun-16 Snehasish Roychowdhury 27
3rd term: πΉ β πΈπ2 πΆ β π΅π1 β πΈπ2π΅π1 is simplified by dividing num & den by s2 and put sβ
: πΆ1 +πΆπ3
2β
πΆπ π2
2β π2 2πΆ1 + πΆπ1 β πΆππ 1π1 β
πΆπ π2
2πΆππ 1 π2π1
4th term: π» π1 + π2 (πΆ β π΅π1) + π» π1 + π2 π΄ + π΅π1 β π΄ + π΅π1 πΌ is simplifed by dividing num & den by s2 and put sβ : β C1 + πΆππ 1π1 (πΆ1)
πͺππ ππΏπ + πππ β π +
πͺπ ππͺπ»ππΏπ β ππͺππππππΏπ + ππππ¬ππΏπππ + πͺπ»π β πͺπππππ + πͺπππππ ππΏπ + ππ β πͺπ»π β πͺπππππ β πͺπππππππ + ππͺππππΏπ +πͺπ»π β πͺπππππ + πͺπππππ πͺπ»ππΏπ β πͺππππππΏπ + πππ¬ππΏπππ β πͺππππΏπ(πͺπ»π β πͺπππππ β πͺπππππππ) = 0
This follows the form: ππΆ12 + ππΆ1 + π = 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, π = β4.03, π = β2.53 β 10β15 & π = 4.12 β 10β26
πͺπ = βπ. ππ β ππβππ β βπ. ππ β ππβππ π + π β π. ππ β π. ππ β ππβππ
π β π. ππ= πππ. ππ ππ
πΆ1 = πΆππ + πΆππ₯π‘ , Here, Cgs=7.05fF , πΆππ₯π‘ = 93.74ππΉ
We consider, πͺπππ = ππππ
13-Jun-16 Snehasish Roychowdhury 28
RETURN
Infinite frequency for 1st term:
1st term:
πΌ(πΆ β π΅π1) + πΌπ΅π1 β π» π1 + π2 (πΆ β π΅π1) β π» π1 + π2 π΅π1
= ππ + π πΆ1 2ππ + π 2πΆ1 + πΆπ1 β 2πππ + π πΆππ 1 π1 + 2πππ + π πΆππ 1 π1
β πππ π1 + π2 2ππ + π 2πΆ1 + πΆπ1 β ,πππ π1 + π2 2πππ + π πΆππ 1 π1-
The critical frequencies above which (at least 10 times) we are considering the approximation of sβ is valid:
π ππ
πΆ1=
42.2π
100.79π = 67.48 πΎπ»π§
ππ2ππ
2πΆ1 + πΆπ1= 64.27 πΎπ»π§
πππ2πππ
πΆππ 1= 74.9 πΎπ»π§
13-Jun-16 Snehasish Roychowdhury 29
Infinite frequency for 2nd term:
2nd term:
πΈπ2 πΆ β π΅π1 + πΈπ2 π΄ + π΅π1
= πππ1 +π πΆππ 2
2,π2 2ππ + π 2πΆ1 + πΆπ1 β 2πππ + π πΆππ 1 π1 + π2* ππ + π πΆ1 + 2πππ + π πΆππ 1 π1+-
The critical frequencies above which (at least 10 times) we are considering the approximation of sβ is valid:
π2πππ2 β 2ππππ1
2πΆ1 + πΆπ1 π2 β πΆππ 1π1= 71.33 πΎπ»π§
ππ(ππ+2ππππ1)
πΆ1 + πΆππ 1π1= 67.16 πΎπ»π§
13-Jun-16 Snehasish Roychowdhury 30
Infinite frequency for 3rd term:
3rd term:
πΉ β πΈπ2 πΆ β π΅π1 β πΈπ2π΅π1
= ππ + ππππ2 + π πΆ1 +πΆπ3
2β
πΆππ 2
2π2 2ππ β 2πππ + π 2πΆ1 + πΆπ1 β πΆππ 1π1 β π2π1(πππ
+π πΆππ 2
2)(2πππ + π πΆππ 1)
The critical frequencies above which (at least 10 times) we are considering the approximation of sβ is valid:
πππ + ππππ2
πΆ1 +πΆπ32 β
πΆππ 22 π2
= 64.18 πΎπ»π§
ππ2(ππ β πππ)
2πΆ1 + πΆπ1 β πΆππ 1π1= 54.6 πΎπ»π§
13-Jun-16 Snehasish Roychowdhury 31
Infinite frequency for 4th term:
4th term:
π» π1 + π2 (πΆ β π΅π1) + π» π1 + π2 π΄ + π΅π1 β π΄ + π΅π1 πΌ
= πππ π1 + π2 2ππ + π (2πΆ1+πΆπ1 β 2πππ + π πΆππ 1 π1- + πππ(π1 + π2),ππ + π πΆ1 + 2πππ + π πΆππ 1 π1-
β (ππ + π πΆ1),ππ + π πΆ1 + 2πππ + π πΆππ 1 π1-
The critical frequencies above which (at least 10 times) we are considering the approximation of sβ is valid:
πππ + 2ππππ1
πΆ1 + πΆππ 1π1= 67.16 πΎπ»π§
ππ2(ππ β ππππ1)
2πΆ1 + πΆπ1 β πΆππ 1π1= 64.37 πΎπ»π§
13-Jun-16 Snehasish Roychowdhury 32
π = ππ + π πΆππ +πΆπ π2+πΆπ3
22ππ + π 2πΆππ 2 + πΆπ2 β ππ + π πΆππ
2 ,
π = ππ + π πΆππ 2, π = π πΆππ + ππ1 ,
π = ππ + π πΆππ +πΆπ π2+πΆπ3
2, π = ππ + π πΆππ +
πΆπ3
2
π = 2ππ + π 2πΆππ + πΆπ1 , π = 2ππ + π 2πΆππ + πΆπ1 + πΆπ π1 ,
π΄ = π πΆ1 + ππ , π΅ = 2πππ + π πΆππ 1
πΆ = 2ππ + π 2πΆ1 + πΆπ1 , π· = π πΆ1 + ππ ,
πΈ = πππ1 +π πΆπ π2
2, πΉ = ππ + π πΆ1 +
πΆπ3
2
πΊ = 2ππ + π 2πΆ1 + πΆπ2 , π» = πππ, πΌ = (ππ + π πΆ1)
π2 =ππ
(ππβππ), π2 =
βπ(πβπ)
(ππβππ), π2 =
β(ππβππ )
(ππβππ)
π1 =ππ
(ππβππ), π1 =
π(π βπ)
(ππβππ), π1 = β 1 +
π πβπ
ππβππ
13-Jun-16 Snehasish Roychowdhury 33
For X1,2 , Y1,2, Z1,2 , the frequencies above which we can considerπ β , are given as following:
M,Q: (i) ππ
πΆππ +πΆππ 2+πΆπ3
2
= 324.5πΎπ»π§
(ii) 2ππ
2πΆππ +πΆπ2= 549.3πΎπ»π§
N,P : ππ
πΆππ = 957.2πΎπ»π§
R: 2ππ
2πΆππ +πΆπ3= 549.3πΎπ»π§
T: ππ
πΆππ +πΆπ1+πΆππ 1
2
< 549πΎπ»π§
S: 2ππ
2πΆππ +πΆπ1= 549.3πΎπ»π§
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(β)=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
π£π1 = π»πππ£ππ + π»πππ£ππ
2 &
π£π2 = π»πππ£ππ β π»πππ£ππ
2
And, ππ1 = π£π1 + ππ·πΆ & ππ2 = π£π2 + ππ·πΆ
CASE1: If Common mode interference
comes at attenuator output, then from
equations above: |π£π1| β |π£π2| Unequal capacitive division occurs
across gate-source of M1 & M2 due to
CT,tail and as a result, |vgs,M1|β |vgs,M2| ie
finite offset current IOS at output.
CASE2: Instead, if we make Hcm=0
somehow, ie no CM Interference at
attenuator output, π£π1 = |π£π2|
In this case, P acts as virtual ground as second stage doesnβt 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.
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