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Dynamic Logic Circuits
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Transcript of Dynamic Logic Circuits
9 9 -- 11
CMOS Digital Integrated Circuits
Chapter 9 Dynamic Logic Circuits
馮武雄 教授長庚大學 電子系
9 9 -- 22
Dynamic Logic Circuits
1. Pass transistor circuits2. Voltage bootstrapping3. Synchronous dynamic circuit techniques4. Dynamic CMOS circuit techniques5. High-performance dynamic CMOS
circuits
9 9 -- 33
Static v.s. DynamicStatic Logic Gates
• Valid logic levels are steady-state operating points• Outputs are generated in response to input voltage levels after
a certain time delay, and it can preserve its output levels as long as there is power.
• All gate output nodes have a conducting path to VDD or GND, except when input changes are occurring.
Dynamic Logic Gates• The operation depends on temporary storage of charge in
parasitic node capacitances.• The stored charge does not remain indefinitely, so must be
updated or refreshed. This requires establishment of an update or recharge path to the capacitance frequently enough to preserve valid voltage levels.
9.1 Introduction
9 9 -- 44
Static v.s. Dynamic (Continued)Advantages of Dynamic Logic Gates
• Allow implementation of simple sequential circuits with memory functions.
• Use of common clock signals throughout the system enables the synchronization of various circuit blocks.
• Implementation of complex circuits requires a smaller silicon area than static circuits.
• Often consumes less dynamic power than static designs, due to smaller parasitic capacitances.
9 9 -- 55
Pass-Transistor Latch: Circuit and Operation
Operation• CK = H, D=H or L : CX is charged up or down through MP,
and X becomes H or L (depends on D input) since MP is on ⇒D and X are connected.
• CK = L: X is unchanged since MP is off and CX is isolated from D, and the charge is stored on capacitances CX.
• For X = H, Q = L and Q = H• For X = L, Q = H and Q = L
Cost: 3 to 5 devices (very low)
MP
CKCx
D VxML
MD
Q Q
Soft note
X
9.2 Basic Principles of Pass Transistor Circuits
9 9 -- 66
Pass-Transistor Latch: Soft Node Concept• During CK = 1: Let D = 1, i.e. VD = VOH = VDD MP is
conducting and charges CX to a “weak 1” (VX = VDD – VTD) ⇒Q = L (VQ<VTD) and Q = H(VQ=VDD).
• During CK = 0: Logic-level VX is preserved through charge storage on CX. However, VX starts to drop due to leakage.
• What value does VX have to deteriorate to no longer like a stored ?Example (see p359~359, Kang and Leblebici): For an inverter with VDD = 5V, VT,n = 0.8V , VOL = 2.9V and VIH = 2.9V, initial VX =4.2 V. But due to leakage currents, this will decline over time. When it declines below VIH(2.9V), then a logic 0 out of the inverter can no longer guaranteed.Thus, to avoid an erroneous output, the charge stored in CX
must be restored or refreshed to its original level before VX
declines below 2.9 V.
9 9 -- 77
Basic Principles of Pass Transistor CircuitsLogic “1” Transfer
Logic “1” Transfer: VX(t=0)=0V, Vin=VOH=VDD, CK=0 →VDD
• VGS = VDD - VX, VDS = VDD - VX = VGS. • Therefore, VDS> VGS – VT,MP⇒ MP is in saturation.
• Note that the VT,MP is subject to substrate bias effect and therefore, depends on the voltage level VX. We will neglect the substrate bias effect for simplicity.
MP
CKCx
Vx
Soft note
X MP
CKCx
Vin=VDDVx
X
IDVin ⇒ D S
( )VVVkdt
dVC MPTXDDnX
X ,2
2−−=
Fig. 9.2Fig. 9.1
9 9 -- 88
Basic Principles of Pass Transistor CircuitsLogic “1” Transfer (Cont.)
• Integrating the above equation with t from 0 → t and VX
from 0 → VX, we have
• Therefore,
• and,
( )V
MPTXDDn
X
V
MPTXDD
X
n
Xt
X
X
VVVkC
VVVdV
kCdt
0,
0 ,2
0
12
2
−−=
−−= ∫∫
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−−−
=VVVVVk
CtMPTDDMPTXDDn
X
,,
112
( )( )
( ) tC
VVk
tC
VVk
VVtV
X
MPTDDn
X
MPTDDn
MPTDDX
21
2)(,
,
, −+
−
−=
9 9 -- 99
Basic Principles of Pass Transistor CircuitsLogic “1” Transfer (Cont.)
• VX rises from 0V and approaches a limit value Vmax = VX(t)|t=∞= VDD-VT,MP, but it can not exceed this value, since the pass transistor will turn off at this point (VGS=VT,MP). Therefore, it transfers a “weak logic 1”.
• The actual Vmax by taking the body effect into account is,
• and tcharge = time to VX = 0.9Vmax,
• Body Effect: Reduce VX, and Increase tcharge
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−−−
=VVVVVk
CMPTDDMPTmaxDDn
Xcharget
,,
19.0
12
( )φφγ 22,0 FmaxFMPTDDmax VVVV −+−−=
VX
VmaxVmax=VDD-VT,MP
t0Fig. 9.3
9 9 -- 1010
9 9 -- 1111
9 9 -- 1212
Basic Principles of Pass Transistor Circuits: Logic “0” Transfer
Logic “0” Transfer: VX(t=0)=Vmax= VDD – VT,MP, Vin=VOL=0V, CK= 0 → VDD
•VGS = VDD, VDS = Vmax = VDD – VT,MP. •Therefore, VDS≤VGS – VT,MP⇒ MP is in linear region.
•Note that the VSB=0. Hence, there is no body effect for MP(VT,MP= VT0,MP). But the initial condition VX(t=0)=VDD –VT,MP contains the threshold voltage with body effect. To simplify the expressions, we will use VT,MP in the following.
MP
CKCx
Vx
Soft note
X MP
CKCx
Vin=0 Vx
X
IDVin ⇒ DS
( )[ ]VVVVkdt
dVC XXMPTDDnX
X2
,22
−−=−
Fig. 9.6
9 9 -- 1313
Basic Principles of Pass Transistor CircuitsLogic “0” Transfer (Cont.)
•Integrating the above equation with t from 0 → t and VX from VT,MP → VX, we have
•Therefore,
•and,
( )[ ]
( )( ) V
VVX
XMPTDD
MPTDDn
X
V
VV XXMPTDD
X
n
Xt
X
MPTDD
X
MPTDD
VVVV
VVkC
VVVVdV
kCdt
,
,
,
,
2,0
2ln
22
−
−
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛ −−−
=
−−= ∫∫
( )( )
⎟⎠
⎞⎜⎝
⎛ −−−
=V
VVVVVk
CtX
XMPTDD
MPTDDn
X ,
,
2ln
( )( )e
VVtV CVVtkMPTDD
XXMPTDDn+
−=
−12)(
/,
,
9 9 -- 1414
Basic Principles of Pass Transistor CircuitsLogic “0” Transfer (Cont.)
• VX drops from Vmax = VDD-VT,MP, to 0V. Hence, unlike the charge-up case, it transfers a “strong logic 0”.•τfall = time of VX drops from 0.9Vmax to 0.1Vmax,
• where,
( )[ ]
( )VVkC2.74
VVkCtt
MPTDDn
X
MPTDDn
X
fall
,
,
%10%90
)22.1ln()19ln(
−=
−−
=
−=τ
( )( )( )
( )
( ) ( )
( ) ⎟⎠⎞
⎜⎝⎛
−=
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−−
−=
0.11.9
VVkCt
1.22VVk
CVV0.9
VV0.92VVk
Ct
MPTDDn
X
MPTDDn
X
MPTDD
MPTDD
MPTDDn
X
ln
ln
ln
,%10
,
,
,
,%90
VX
VmaxVmax=VDD-VT,MP
t0
Fig. 9.7
9 9 -- 1515
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage
• At t = 0, CK=0, VX= Vmax, Vin =0. The charge stored in CX
will gradually leak away, primarily due to the leakage currents associated with the pass transistor. The gate current of the inverter driver transistor is negligible.
n+ n+
VCK=0
Vin=0Ileakage VX
CX
Ireverse
Isubthreshold
p-type Si
MP
CK=0Cx
VxVin =0Ileakage Igate=0
Fig. 9.9
Fig. 9.8
9 9 -- 1616
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage (Cont.)
• Isubthreshold is the subthreshold current for the pass transistor with CK=0.• Ireverse is the reverse current for the source/drain pn junction at node X• Cj (VX) : due to the reverse biased drain-substrate junction, a function of
VX, Cin: due to oxide-related parasitics, can be considered constants.
n+ n+
VCK=0Vin=0
Ileakage VX
CX
Ireverse
Isubthreshold
p-type Si
Cin
VxIleakage
Cj(VX)
Ileakage= Isubthreshold + Ireverse
Ireverse
Cin= Cgb + Cpoly + Cmetal
CX= Cin + Cj
Isubthreshold
Drain-substrate pn-junction Fig. 9.10
9 9 -- 1717
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage (Cont.)
• The total charge stored in the soft node can be expressed as,Q = Qj (VX) + Qin where Qin = Cin•VX
• The total leakage current can be expressed as the time derivative of the total soft-node charge Q
Cin
VxIleakage
Cj
Ileakage= Isubthreshold + Ireverse
Ireverse
Cin= Cgb + Cpoly + Cmetal
CX= Cin + Cj
Isubthreshold
Drain-substrate pn-junction
dtdVCdt
dVdV
VdQdt
dQdt
VdQdt
dQI
Xin
X
X
Xj
inXj
leakage
+=
+=
=
)(
)(
9 9 -- 1818
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage (Cont.)
• Where
• Therefore,
• We have to solve the above differential equation to estimate the actual charge leakage time from the soft node.
⎟⎠
⎞⎜⎝
⎛=
nNN
qkT
i
ASWDSW 20 lnφ⎟
⎠
⎞⎜⎝
⎛=
nNN
qkT
i
AD20 lnφ
φφ SW
X
SWj
X
j
XjX
Xj
VAC
VAC
VCdV
VdQ
0
0
0
0
11
)()(
++
+=
=
dtdVC
VPC
VAC
I Xin
SW
X
SWj
X
jleakage
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
++
++
=
φφ 0
0
0
0
11
9 9 -- 1919
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage (Cont.)
A quick estimate of the worst-case leakage behavior• Assume that the minimum combined soft-node
capacitance isCX,min = Cgb + Cpoly + Cmental + Cdb,min
Cdb,min is the minimum junction capacitance, obtained when VX=Vmax
• The worst-case holding time (thold) is the shortest time for VX to drop from its initial logic-high value to the logic threshold voltage due to leakage.
thold = ΔQcritical,min/Ileakage,max
• whereΔQcritical,min =CX,min (Vmax-VDD/2)
Vth
9 9 -- 2020
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage (Cont.)
Example 9.2: Consider the soft-node structure shown below, which consists of the drain (or source, depending on current direction) terminal of the pass transistor, connected to the polysilicon gate of an nMOS driver transistor via a metal interconnect.
Question: is to estimate thold if VDD=5V and the soft-node is initially charged to Vmax.
MP
Cx
Vx
CK
M1
2
31
41 1
3
6 65
52
2 4
soft node
M1MP
diffusion metal polysiliconCK
9 9 -- 2121
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage (Cont.)
• Material parameters:VTO = 0.8V COX = 0.065 fF/μm2
γ = 0.4V1/2 C’metal = 0.036 fF/ μm2
|2φF| = 0.6V C’poly = 0.055 fF/ μm2
φ0 = 0.88V Cj0 = 0.095 fF/ μm2
φ0SW = 0.95V Cj0SW = 0.2 fF/μmIleakage,max = 0.85 pA
Soft-node Capacitance Calculation• Oxide-related (constant) parasitic capacitances
» Cgb = COX·W·Lmask = 0.065 fF/μm2· (4 μm×2 μm) = 0.52 fF» Cmetal = C’metal·W·Lmetal = 0.036 fF/μm2· (5 μm×5 μm) = 0.90 fF» Cploy = C’poly·W·Lpoly = 0.055 fF/μm2· (36+6+2 μm2) = 2.42 fF
2
31
41 1
3
6 65
52
2 4M1MP
diffusion metalpolysiliconCK
9 9 -- 2222
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage (Cont.)
•Parasitic junction capacitanceBy zero-bias unit capacitance values in the previous slide, we have» Cbottom = Abottom·Cj0 = 0.095 fF/μm2· (36 μm2 + 12 μm2 ) = 4.56 fF» Csidewall = Cj0SW·Psidewall = 0.2 fF/μm2· (30 μm) = 6.00 fFTherefore» Cdb,max = Cbottom + Csidewall = 4.56 fF + 6.00 fF = 10.56 fFThe minimum drain junction capacitance is achieved as the
junction is biased with Vmax. We need to find Vmax to determine Cdb,min
» Vmax = 5.0 - 8.0 - 0.4 ( 0.6+ Vmax - 0.6 )⇒ Vmax = 3.68 V
Therefore,
fF
VC
VCC
SW
maxX
sidewall
maxX
bottommindb
71.4
95.068.31
0.6
88.068.31
56.4
110
,
0
,,
=+
++
=
++
+=
φφ
9 9 -- 2323
Basic Principles of Pass Transistor CircuitsCharge Storage and Charge Leakage (Cont.)
•Combining the Oxide-related (constant) parasitic capacitances with the parasitic junction capacitance, CX,min
can be got asCX,min = Cgb + Cpoly + Cmental + Cdb,min
= 0.52 + 2.42 + 0.90 +4.71 = 8.55 fF•The amount of the critical charge drop is
ΔQcritical = CX,min(VX,min-VDD/2)=8.55 (3.68-2.5)=10.09 fC
•Finally, thold = Δ Qcritical /Ileakage,max=11.87ms
•The worst-case hold time for this structure is relatively long, even with a very small soft-node capacitance of 8.55fF. It means that the logic gate can be preserved in a soft node for a long time period when the leakage current is small.
9 9 -- 2424
9.3 Voltage Bootstrapping• The Voltage bootstrapping is a technique to overcome
the threshold voltage drops of the output voltage levelsin pass transistor gates or enhancement-load inverters and logic gates.
• Consider the following circuit with VX≤VDD ⇒ M2 is in saturation. If Vin is low, the maximum output voltage is limited as
Vout(max) = VX – VT2(Vout)
M1 Cout
Vx
Vin
M2
VDD
Vout
Fig. 9.11
9 9 -- 2525
Voltage Bootstrapping (Cont.)• To overcome the voltage drop, the voltage VX must be increased. This
can be achieved by adding a third transistor M3 into the circuit.» CS and Cboot represent the capacitances which dynamically couple
VX to the ground and to the output.» The goal of the above circuit is to provide a high enough voltage VX
to let Vout go to VDD instead of VDD-VT2(Vout).
• Initially, let Vin=H⇒ M1 and M2 are on, and Vout=L.• Now Vin goes to L ⇒ M1 turns off, and Vout starts to rise. This change
will be coupled to VX through the bootstrap capacitor, Cboot.
M1 CoutVin
M2
VDD
Vout
Vx
M3
CbootCS
Fig. 9.12
9 9 -- 2626
Voltage Bootstrapping (Cont.)
• Let iCboot be the transient current through Cboot during the charge-up event, and let iCS be the current through CS. Assume iCS ≈ iCboot, we have
iCS ≈ iCboot⇔ CS·dVX/dt ≈ Cboot·d(Vout-VX)/dt⇒ (CS+Cboot)·dVX/dt ≈ Cboot·dVout/dt⇒ dVX/dt ≈ Cboot /(CS+Cboot) ·dVout/dt
• This expression can be integrated to give VX such that Vout will rise to VDD.
• If Cboot >> CS, then for Vout rising to VDD, VX(max) ≈ 2VDD – VT3 – VOL > VDD – VT2.
for realistic values of the voltages. Thus, it is feasible to use the circuit to obtain Vout =VDD.
( ) ( )VVCC
CVVV
dVCC
CdV
OLDDbootS
bootTDDX
V
Vout
bootS
bootV
VVX
DD
OL
X
TDD
−+
+−=⇒
+= ∫∫ −
3 .
3
9 9 -- 2727
Voltage Bootstrapping (Cont.)
• To overcome the threshold voltage drop at Vout, the minimum VX is
VX(min) = VDD + VT2|Vout = VDD
= [VDD-VT3(VX)]+Cboot /(CS+Cboot) ·(VDD-VOL)• Therefore, the required capacitance ratio Cboot /(CS+Cboot) is
• CS is the sum of the parasitic source-to-substrate capacitance of M3 and the gate-to-substrate capacitance of M2.
XDDout
XDDout
XDDout
VTVVTOLDD
VTVVT
S
boot
OLDD
VTVVT
bootS
boot
VVVV
VV
CC
VV
VV
CCC
32
32
32
.−−−
+=
−
+=
+
=
=
=
9 9 -- 2828
Voltage Bootstrapping (Cont.)• Cboot can be specifically constructed to control its value by
using a transistor with the source and drain connected together at Vout and the gate attached to VX. Since its drain and source tied together, it simply acts as an MOS capacitor between VX and Vout.
• See Kang and Leblebici at pp. 373 for a SPICE example.
M1Vin
M2
VDD
Vx
M3
Vout
Cboot
Fig. 9.13
9 9 -- 2929
9.4 Synchronous Dynamic Circuit Techniques –Dynamic Pass Transistor Circuits
• The multi-stage synchronous circuit is shown below. The circuit consists of cascaded combinational logic stages interconnected through nMOS pass transistors. Its operation depends on temporary charge storage in the parasitic input capacitances.
• Logic levels are stored on input capacitances during the inactive clock phase.
AB
φ1φ1 φ2C D
F1
F2
Comb.Logic
1
Comb.Logic
2
Comb.Logic
3
φ1
phase1 phase2
t
tφ2
φ1,φ2 non-overlapping clocks
Fig. 9.14
Fig. 9.15
9 9 -- 3030
Dynamic Pass Transistor CircuitsTwo-Phase Clock Dynamic Shift Register
Depletion-Load Dynamic Shift Register• The max clock frequency is determined by signal
propagation delay through one inverter stage.• One half-period of the clock signal must be long enough to
allow Cin to charge up or down, and Cout to charge to the new value.
• The logic-high input value is one VT0 lower than VDD.VDD
φ1 φ2 φ1
Vin
Vout
Cin1 Cin2 Cin3Cout1 Cout2
VDD VDD
Cout3
Fig. 9.16
9 9 -- 3333
Dynamic Pass Transistor CircuitsEnhancement-Load Dynamic Shift Register
Enhancement-Load Dynamic Shift Register 1• Instead of biasing load transistors with a constant gate voltage, a
clock signal is applied to the gate of the load transistor ⇒ power dissipation and silicon area are reduced.
• The power supply current flows only when the load devices are activated by the clock signal, the power consumption is lower than the depletion-load nMOS logic.
VDDφ1 φ2 φ1
Vin
Vout
Cin1 Cin2 Cin3Cout1 Cout2
VDD VDD
Cout3
φ2
Fig. 9.19
9 9 -- 3434
Enhancement-Load Dynamic Shift Register 1 (Cont.)General Structure
BC
φ1
VDD
nMOSLogic
Stage 1
nMOSLogic
Stage 2
A
D
VDD
φ2 φ1
Z
General Circuit Structure of Ratioed Synchronous Dynamic Circuit
Fig. 9.20
9 9 -- 3535
9 9 -- 3636
Enhancement-Load Dynamic Shift Register 1 (Cont.) VDD
φ1
φ2 φ1
Vin
Vout3
Cin1 Cin2 Cin3Cout1 Cout2
VDD VDD
Cout3
φ2Vout1 Vout2
Vout2⇒VOLVDD
φ1
φ2 φ1
VinCin1 Cin2 Cin3Cout1 Cout2
VDD VDD
Cout3
φ2
φ1=H
φ2=H
Vout1 Vout2 Vout3
Vout1⇒VOL Vout3⇒VOL
• VOL → kdriver/kload ⇒Ratioed Dynamic Logic.• Cout1, Cin2 & Cout2, Cin3 interact ⇒ Charge Sharing
9 9 -- 3737
Enhancement-Load Dynamic Shift Register 2
Enhancement-Load Dynamic Shift Register 2• The input pass transistor and the load transistor are driven by
the same clock phase.• The valid low-output voltage level VOL=0V can be achieved
regardless of the driver-to-load ratio, this circuit is a ratiolessdynamic logic.
VDDφ1 φ2 φ1
Vin
Vout
Cin1 Cin2 Cin3Cout1 Cout2
VDD VDD
Cout3
Fig. 9.21
9 9 -- 3838
Enhancement-Load Dynamic Shift Register 2(Cont.)General Structure
General Circuit Structure of Ratioless Synchronous Dynamic Circuit
BC
φ1
VDD
nMOSLogic
Stage 1
nMOSLogic
Stage 2
A
D
VDD
φ2
Z
Fig. 9.22
9 9 -- 3939
Enhancement-Load Dynamic Shift Register 2 (Cont.)
• VOL → 0V ⇒Ratioless Dynamic Logic.
VDD
φ1φ2 φ1
Vin
Vout3
Cin1 Cin2 Cin3Cout1 Cout2
VDD VDD
Cout3
Vout1 Vout2
Vout2⇒ 0V
φ1=H
Vout1⇒VOL Vout3⇒VOH
VOH ⇒0VOH
VDDφ1 φ2 φ1
VinCin1 Cin2 Cin3Cout1 Cout2
VDD VDD
Cout3
φ2=H
Vout1 Vout2 Vout3
Vout1⇒0V Vout2⇒VOH
VOH⇒0VOH 0
Vout3=VOH
9 9 -- 4040
Enhancement-Load Dynamic Shift Register 2 (Cont.)
• Cini << Couti-1 for i=2,3 ⇒ Minimum Charge Sharing
VDD
φ1φ2 φ1
Vin
Vout3
Cin1 Cin2 Cin3Cout1 Cout2
VDD VDD
Cout3
Vout1 Vout2
Vout2= VOH
φ1=H
Vout1=0V Vout3=?
0V 0VOH
Charge Sharing
VOH
9 9 -- 4141
Enhancement-Load Dynamic Shift Register 2 (Cont.)Charge Sharing
• φ2 = 0: Qout2 = Cout2Vb and Qin3 = Cin3Va
• φ2 = 1: Qtotal = Cout2Vb + Cin3Va and Ctotal = Cout2 + Cin3
The resulting voltage across Ctotal is VR = Qtotal / Ctotal = (Cout2Vb + Cin3Va )/ (Cout2 + Cin3)
• If Vb = VOH and Va << Vb ⇒ VR ≈ Cout2VOH /(Cout2 + Cin3)VR ≈ VOH if Cin3 << Cout2
φ2=1
Cin3Cout2
Va
Charge SharingCin1
Vi1=0
φ2=1
Cin2Cout1
Vb=VOL →0 Va
Cin1
Vin1=1
Vb=VOH
9 9 -- 4242
9.5 Dynamic CMOS Circuit TechniquesCMOS Transmission Gate Logic
• Each transmission gate is controlled by the clock signal and itscomplement. Therefore, the two-phase clocking need four clock signals.
• As in the nMOS structures, the CMOS dynamic circuit relies on charge storage in parasitic input capacitances during the inactive clock cycles.
D
φ1 φ2
A
B
CStage 1 Stage 2
φ1
φ1
φ1
φ2
F1
Fig. 9.23
9 9 -- 4343
Dynamic CMOS Transmission Gate LogicShift Register
• The basic building block of the shift register consists of a CMOS inverter, which is driven by a TG.
• CK=1⇒Vin is transferred onto the parasitic input capacitance CX.
• The low on-resistance of TG results in»A smaller transfer time compared to nMOS-only switches.»No threshold voltage drop across TG
CK
VDD
VoutVX
CX CyCK
Vin
soft node
Fig. 9.24
9 9 -- 4444
Dynamic CMOS Transmission Gate LogicShift Register (Cont.)
• The single-phase CMOS shift register is built by» Cascading identical inverter units» Driving each stage alternately with the CK and CK.
• Ideally: The odd-numbered stages are on as CK=1, while the even-numbered stages are off ⇒ the cascaded inverter stages are alternately isolated.
• Practically:» The CK and CK are not a truly nonoverlapping signal pair,
since their waveforms have finite rise and fall times.» One of the signals is generated by inverting the other ⇒ the
clock skew is unavoidable.» True two-phase clocking is preferred over single-phase
clocking.CK
CK
V1
CK
CK
CK
CK
V2 V3 V4
Fig. 9.225
9 9 -- 4545
Dynamic CMOS Precharge-Evaluate LogicReduced Transistor Count
•φ=0 ⇒ Mp on and Me off⇒ Cprecharges to VDD (output is not available during precharge)
•φ =1 ⇒ Mp off and Me on ⇒C selectively discharges to 0 (output is only available after discharge is complete)
φ
VDD
nMOSLogicinputs
C
Vout
Me
Mp
Internal capacitance
φ
t
t
Voutprecharge precharge
evaluate
9 9 -- 4646
Dynamic CMOS Precharge-Evaluate LogicAn Example
φ
VDD
Vout
Me
Mp
A1
A2
A3
B1
B2
Z is high when φ=0
Z=(A1 A2A3 +B1B2)
Fig. 9.26
9 9 -- 4747
Dynamic CMOS Precharge-Evaluate LogicAdvantages/Disadvantages
Advantages• Need only N+2 transistors to implement a N-input gate.• Low static power dissipation• No DC current paths to place constraints on device sizing• Input capacitance is same as pseudo nMOS gate.• Pull-up time is improved by active switch to VDD.
Disadvantages• The available time of output is less than 50 % of the time.• Pull-down time is degraded due to series active switch to 0.• Logic output value can be degraded due to charge sharing with other
gate capacitances connected to the output.• Minimum clock rate determined by leakage on C.• Maximum clock rate determined by circuit delays.• Input can only change during the precharge phase. Inputs must be
stable during evaluation; otherwise an incorrect value on an input could erroneously discharge the output node. (single phase P-E logic gates can not be cascaded)
• Outputs must be stored during precharge, if they are required during the next evaluate phase.
9 9 -- 4848
Dynamic CMOS Precharge-Evaluate LogicCascading Problem
• Evaluate:» Me1, Me2 ⇒ ON» Mp1, Me2 ⇒ OFF
• Problem: All stages must evaluate simultaneously one clock does not permit pipelining of stages.
φ
VDD
nMOSLogic
Vout1
1st stage
1
Vout2
VDD
2nd Vout1 does not switch fromt
t
φ
Vout
t
Voutcorrect state
erroneous state
precharge evaluate
“1” to “0” fast enoughinputs
Me1Me2
Mp1 Mp2
9 9 -- 4949
9.6 High Performance Dynamic CMOS CircuitsDomino CMOS Logic
φ
VDD
nMOSLogic
Vout
VDD
inputs
X
φ precharge evaluate1
t
Static inverter serves to buffer the logic part of the circuit from its output load
• φ=0 » X precharges to VDD, and Vout
= 0.• φ=1
» X remains high, and Vout
remains low.» X discharges to 0, and Vout
changes from 0 to 1. Fig. 9.27
9 9 -- 5050
9 9 -- 5353
9 9 -- 5454
Domino CMOS Logic
φ
VDD
nMOSLogicinputs
VDD
nMOSLogic
X1
VDD
nMOSLogic
X2 X3
t
tX1
t
precharge
evaluate
t
X2
X3
φ evaluate
teval
Max number gates limited: total propagation delay < teval
9 9 -- 5555
Domino CMOS Logic (Cont.)
•The problem in cascading conventional dynamic CMOS occurs when one or more inputs make a 1 to 0 transition during evaluation.•Domino circuits can fix the above problem
»During the evaluation, each buffer output can make at most one transition (from 0 to 1), and thus each input of all subsequent logic stages can also make at most one (0 to 1) transition.
X3
φ
VDD
nMOSLogicinputs
VDD
nMOSLogic
X1
VDD
nMOSLogic
X2
9 9 -- 5656
Domino CMOS Logic The Limitations
•The static CMOS and domino gates can be used together, see Fig. 9.31. in Kang and Leblebici. The limitation: the number of inverting static logic stages in cascade must beeven, to let the inputs of next domino stage can have only 0 to 1 transitions during the evaluation.•Can implement only non-inverting logic•Due to precharge use, can suffer from charge sharingduring the evaluation which may cause erroneous outputs.
»The problem will be described in the next slide, and several solutions will be presented later.
9 9 -- 5757
Domino CMOS LogicCharge Sharing
• Assume that all inputs are low initially, and the voltage acrossC2=0V
• During the precharge, C1 is charged to VDD
• If transistor N switches from 0 to 1 during the evaluation phase, the charge initially stored in C1 will be shared by C2. Therefore, the value of VX will reduced.
φ
VDD
Vout
VDD
VX
C1
C2VX = VDDC1/(C1+C2)
Keep C2 << C1
N
0V(initially)
Fig. 9.32
9 9 -- 5858
Domino CMOS LogicReduce Charge Sharing Degradation of VX
φ
VDD
nMOSLogicinputs
VoutVX
Push VX to VDD unless there is a strong pull-down path between Vout and ground
weak pull-up pMOS
Fig. 9.33
9 9 -- 5959
Domino CMOS LogicReduce Charge Sharing Degradation of VX (Cont.)
Use separate pMOS transistors to precharge all intermediate nodes in nMOS pull-down tree which have a large parasitic capacitance.
Effectively eliminate all charge sharing problems during evaluation
Allow implementation of multiple-output domino structures.
Can cause additional delay since the nMOS tree need to drain a larger charge to pull down VX
VDD
φ
nMOSLogic
Vout1
VX2
VX1
nMOSLogic
Vout2
C1
C2
Another Way: Use a smaller threshold voltage⇒ the final stage output is not affected by lowering of VX ⇒trade off the pull-up speed (weaker pMOStransistor)
Fig. 9.34
9 9 -- 6060
Domino CMOS LogicAn Example of Using Separate pMOS Transistor
VDD
φ
VDD
Vout
VDD
VX1
C1
C2
VA
VB
VX2
• Let C1 = C2 = 0.05pF. VX1 = 0, and VX2 = 0 at t=0
• Without this extra pMOStransistor
» Precharge: VX1 ≠VX2
» Evaluation: VX1 = VDDC1/(C1+C2) = VDD/2
• With this extra pMOS transistor » Presharge: VX1 = VX2
» Evaluation: VX1 = VDD
• See pp.392~393 for the HSPICE simulation result
• Note that there is a speed penaltyfor adding this extra pMOSprecharge transistor.
9 9 -- 6161
Domino CMOS LogicAn Example of Multiple-Output Domino Circuits
C1=G1+P1C0
C2=G2+P2G1+P2P1C0
C3=G3+P3G2+P3P2G1+P3P2P1C0
C4=G4+P4G3+P4P3G2+P4P3P2G1+P4P3P2P1C0
φ
P1
C0
P2
P3
P4
G2
G3
G1
G4
C4
C3
C2
C1
VDD
Gi = Ai · Bi
Pi = Ai ⊕ Bi
Reduce transistor count
Fig. 9.35
9 9 -- 6262
FET Scaling in Domino CMOS GatesThe transient performance can be improved by adjusting the nMOS transistor sizes in the pull-down path to reduce the discharge time.
φ
VDD
Vout
Me
Mp
A
B
C
D
CL
CLC1C0
R0 R10 1
ABCD
Fig. 9.36
9 9 -- 6363
The nMOS Scaling in Domino CMOS Gates
V1=V0=VDD⇒VDDe-1 after time T1T1 =R0(C0+C1+CL)+R1(C1+CL)
Let the last nMOS is increased by a fraction of ∆k thenC1⇒ C1(1+∆k); R1⇒ R1/(1+∆k)
T1 =R0(C0+C1+CL)+R1(C1+CL)+(C1-R1CL/R0)∆kIf
CL<(R0/R1)C1
T1 decreases by decreasing the size of the last nMOS. R0/R1 is the number of series-connected nMOS minus one, times a factor γ that takes the many effects that makes a real nMOS different from a linear resistor, into account. Using the approximation γ=1/2, we conclude
If CL<C1(N-1)/2 is satisfied, the overall delay can be reduced by decreasing the size of last nMOS.
The above result can be iteratively applied to the other transistors, which leads to graded sizing of all nMOS devices.
CLC1C0
R0 R10 1
9 9 -- 6464
NORA CMOS Logic (NP-Domino Logic)
•Advantages»An Inverter is not required at the output of stages»Allow pipelined system architecture
•Disadvantages: Also suffer from charge sharing and leakage
VDD VDD VDD
φ
nMOSLogic
pMOSLogic
nMOSLogic
φ φ
to nMOS stage to pMOS stagenMOS stageprechargepMOS stagepre-discharge
all stagesevaluate nMOS stageprechargepMOS stagepre-discharge
all stagesevaluateφ
Fig. 9.37
9 9 -- 6565
NORA CMOS Logic (NP-Domino Logic)Examples
VDD VDD VDD
φ φ φ
• φ=L: nMOS precharges to H, and pMOS pre-discharges to L.
• φ=L→H: All cascaded nMOS and pMOS logic stages evaluate one after the other.
Fig. 9.38
9 9 -- 6666
NORA CMOS Logic (NP-Domino Logic)Examples (Cont.)
• Pipelined System Architecture: See Fig. 9.39 – Use of CMOS2 latches (three state latches storing on logic inputs.)
• Zipper Logic: See Fig. 9.40 – Identical to NORA except for weird clock signals that keep precharge devices weakly on to handle charge leakage and charge sharing
9 9 -- 6969
Pipelined True Single-Phase Clock (TSPC) Dynamic CMOS
• φ=L: nMOS blocks precharge to VDD
pMOS blocks evaluate by selective pull-up to VDD
• φ=H: pMOS blocks pre-discharge to VDD
nMOS blocks evaluate by selective pull-down to 0V• φ is not used, no clock skew problem can arise.• Provide similar performance to NORA structure
Using tristate inverters between stages decouples the stages and enables pipelined operation
VDD VDD
φ
nMOSLogic
pMOSLogic
φ
VDD
φ
φ
VDD
N-block P-block
to next N-block
Fig. 9.41
9 9 -- 7070
TSPC-Based Rising Edge-triggered D-type Flip-Flop
• Need only 11 transistors. • Static Edge Triggered D Flip-flop (see Fig. 8.30) need 16
transistors.Common Advantages of dynamic Logic Styles
• Smaller area than fully static gates.• higher speed: smaller parasitic capacitances.• Glitch free operation if design carefully
VDD
φ
VDD
φ
VDD
Q
VDD
φD
Fig. 9.42
9 9 -- 7373
Summary •Full complementary static logic is best option in the majority of CMOS circuits.
»Noise-immunity is not sensitive to kn/kp
»Does not involve precharge of nodes»Dissipate no DC power»Layout can be automated»Large fan-in gates lead to complex circuit structures (2N transistors)
»Larger parasitics»Slower and higher dynamic power dissipation than alternatives
»No clock
9 9 -- 7474
Summary (Cont.)
•Pseudo-nMOS static logic finds widest utility in large fan-in NOR gates.
» Require only N+1 transistors for N fan-in» Smaller parasitics» Faster and lower dynamic power dissipation
than full CMOS» Noise immunity sensitive to kn/kp
» Dissipate DC power when pulled down» Not well suited for automated layout» No clock
9 9 -- 7575
Summary (Cont.)
• CMOS domino logic should be used for low-power, high speed applications
» Require only N+k transistors for N fan-in, size advantages of pseudo-nMOS.
» Dissipate no DC power» Noise immunity is not sensitive to kn/kp
» Use of clocks enables synchronous operation» Rely on storage on soft node» Require exhaustive simulation at all the process
corners to insure proper operation» Some of the speed advantage over static gates is
diminished by the required per-charge (pre-discharge) time.