7 Single/Dual Rail Conversion Circuits
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Single/Dual Rail Conversion Circuits
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ure 9.49. A straightforward approach to solving this problem is to use a differential amplifier circuit.
Differential Amplifier
The differential amplifier accepts two input voltages
depends upon the difference
and
and produces an output voltage that
This type of circuit can be used to convert a dual-rail logic signal back into a single-rail value that
can be interfaced with standard logic circuits.
A basic differential amplifier is shown in Figure 9.50. This uses a source-coupled pair of nFETs
Mn1 and Mn2 as the input devices. The pFETs Mp1 and Mp2 are used as active-load devices to
provide a pull-up path to the power supply voltage
A single output voltage is shown; it is a
single-rail variable corresponding to the value associated with the input voltage
The output
voltage is determined by the current
as it produces a voltage drop across Mp2. To analyze the
circuit, assume that the nFETs are both in saturation. The currents
and
are given by
where we will assume that Mn1 and Mn2 have the same aspect ratio so that
the source, the currents must sum to
applies to both. At
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due to the current source there. The behavior of the circuit revolves around finding
and
as
functions of the difference voltage
Let us calculate the currents by first noting that both nFETs have the same source voltage, so that
provides the important relationship between the input voltages
Next, we may rearrange the expression for
and
to give
and
and the current equations.
so that
This may be used to find the desired relations. For example,
Squaring both sides gives
may be eliminated by writing
Single/Dual Rail Conversion Circuits
471
Expanding and simplifying,
which, upon squaring and rearrangement, gives
This is a quadratic equation for
with a solution of
where we have chosen the positive root to insure that
is found to be
It is seen by inspection that
increases as
Also note that when
increases. Similarly, the
corresponding to
This corresponds to equal inputs giving balanced current flow.
The general behavior of both currents
and
are shown in Figure 9.51. As
increases
from a negative number
to a positive value
increases from 0 to the maximum value of
while
has the opposite behavior. The value of the difference voltage
needed to obtain
can be calculated by setting
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which is a quartic for
Solving for
and then taking the square root gives the value as
The same approach can be used to show that
gives
is
so that the total voltage change needed to divert the current from one FET to the other
This shows that the width of the input transition is set by the ratio of
for the circuit design since
This provides a basis
can be chosen according to the value of
9.7.3 A Basic Current Source
The above analysis shows that the sensitivity of the differential amplifier depends upon the value of
the current source
Although several types of current source circuits have been published in the
literature, the simple one illustrated in Figure 9.52 illustrates the important points. This circuit uses
a FET MnC to provide the current. It is biased with a gate-source voltage of
where
is a reference voltage supplied by the voltage divider circuit made up of MpR and MnR. Assuming that
MnC is biased into saturation, we can estimate
Problems
473
Channel length modulation effects may be included by multiplying this expression by the factor
To determine the value of
first note that MpR is defined by the terminal voltages
while the nFET MnR has
This shows that both transistors are saturated, so equating drain currents gives
Rearranging gives
This shows that the ratio
can be used to set
which in turn biases MnC to provide
The alert reader will have noticed that this identical to the formula for the inverter threshold voltage
this is due to the fact that the voltage divider circuit made up of MnR and MpR is simply an
inverter with the input shorted to the output!
9.8 Problems
[9-1] Design the CVSL logic gate for the function
and its complement using the AOI/OAI logic network design approach.
[9-2] Design the CVSL logic gate for the function
and its complement using the AOI/OAI logic network design approach.
[9-3] Design the CVSL logic gate for the function
using AOI/OAI design approach.
[9-4] Create the CVSL logic tree network for the 2-input function described by the table shown in
Figure P9.1.
[9-5] Design the CVSL gate by using the function in Figure P9.2 to construct the logic tree.
[9-6] Design the CVSL gate by using the information provided in the truth table of Figure P9.3 to
construct the logic tree for the function G and
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[9-7] Consider the logic function
(a) Construct the function table for this function using a horizontal format where the output F
and the input variables are in one column with the top-to-bottom order of F, A, B, C.
(b) Construct the CVSL logic tree as discussed in the text.
[9-8] Use 2-input CPL arrays to implement the NAND4
[9-9] Use CPL gates to construct the circuit for the logic function
and its complement.
[9-10] Create the DPL circuit for the odd function
and its complement using basic DPL cascades. Then compare your circuit with the CPL equivalent
by looking at device count and electrical operation.
References
475
9.9 References
[1] K.M. Chu and D.L. Pulfrey, “A Comparison of CMOS Circuit Techniques: Differential Cascode Voltage Switch Logic Versus Conventional Logic,” IEEE J. Solid-State Circuits, vol. SC-22,
no. 4, pp. 528-532, August, 1987.
[2] K.M. Chu and D.L, Pulfrey, “Design Procedures for Differential Cascode Voltage Switch Logic
Circuits,” IEEE J. Solid-State Circuits, vol. SC-21, no. 4, pp. 1082-1087, December, 1986.
[3] T. A. Grotjohn and B. Hoefflinger, “Sample-Set Differential Logic (SSDL) for Complex HighSpeed VLSI,” IEEE J. Solid-State Circuits, vol. SC-21, no. 2, pp. 367-369, April, 1986.
[4] L.G. Heller, et al., “Cascode voltage switch logic: a differential CMOS logic family, ISSCC84
Digest, pp. 16-17, February, 1984.
[5] N. Kanopoulos and N. Vasanthavada, “Testing of Differential Cascode Voltage Switch (DCVS)
Circuits,” IEEE J. Solid-State Circuits, vol. SC-25, no. 3, pp. 806-812, June, 1990.
[6] F-S. Lai and W. Hwang, “Design and Implementation of Differential Cascode Voltage Switch
with Pass-Gate (DCVSPG) Logic for High-Performance Digital Systems,” IEEE J. Solid-State Circuits, vol. 32, no. 4, pp. 563-573, April, 1997.
[7] S-H. Lu, “Implementation of Iterative Networks with CMOS Differential Logic,” IEEE J.
Solid-State Circuits, vol. 23, no. 4, pp. 1013-1017, August, 1988.
[8] D. Somasekhar and K. Roy, “Differential Current Switch Logic: A Low Power DCVS Logic
Family,” IEEE J. Solid-State Circuits, vol. 31, no. 7, pp. 981-991, July, 1996.
[9] M. Suzuki, et al., “A 1.5ns 32-b CMOS ALU in Double Pass-Transistor Logic,” IEEE J. SolidState Circuits, vol. 28, no. 11, pp. 1145-1151,November, 1993.
[10] K. Yano, et al., “A 2.8-ns CMOS 16xl6-b Multiplier Using Complementary Pass-Transistor
Logic,” IEEE J. Solid-State Circuits, vol. 25, no. 2, pp.388-395, April, 1990.
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Chapter 10
Issues in Chip Design
Designing a CMOS integrated circuit requires more than just understanding the
logic circuits. Items such as interconnect delay on the chip, and interfacing the circuit to the outside world require special considerations. In this chapter we will
study some important circuit problems that occur at the chip level and affect the
internal operations. The introduction here is designed to provide a solid background for more specialized studies.
10.1 On-Chip Interconnects
It is interesting to examine the evolution of MOS technology in recent years. The typical channel
length in a transistor has shrunk to a nominal value of less than 0.2 microns using the best manufacturing technology. With this type of resolution, the footprint area required for an FET has shrunk to
the point where it is almost insignificant when compared with the surface area needed for contacts,
vias, and interconnect routing. This leads to the conclusion that modern CMOS chip design is
interconnect-limited. In other words, we usually don’t worry about the number of FETs on a chip;
in most cases, the wiring complexity is much more important to the real estate consumption.
Modern CMOS process flows provide several metal layers for use as interconnect wiring.
Although 3 or 4 interconnect layers were sufficient for networks with a million or so FETs, the
high-density compact systems being designed at the start of the 21st century require the use of 7-to10 or more interconnect layers. Obviously, accurate modelling of on-chip wiring is important to the
circuit designer. Parasitic-induced delays and stray coupling may require “tweaking” or re-design at
the circuit level to make a chip operational. We will therefore direct our attention to this important
topic from the viewpoint of electrical modelling of the interconnect structures.
10.1.1 Line Parasitics
Let us examine the basic geometry shown in Figure. 10.1. This is representative of an interconnect
line that is described by a width w, and has a distance of d. The material layer itself has a height (or
thickness) h. Parasitic electrical elements include resistance and capacitance; although the wire also
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has inductance associated with it, we usually do not encounter magnetic effects at local circuit
level.1 Our program at this point is two-fold. First, we want to determine the values of the parasitic
elements that are introduced by the interconnect. Once these have been calculated, we can then proceed to evaluate their effects on the performance.
Line Resistance
The resistance of the line from one end to the other is computed by using the standard equation
where is the resistivity in units of
and A =hw is the cross-sectional area of the line. Every
material is characterized by a value of When choosing interconnect lines, metals dominate due to
their small values of resistivity. Although this expression can be used directly, a more useful formulation for use in chip design is based on the use of the sheet resistance which has units of ohms
for the layer. This is defined by
and is the end-to-end resistance of a square section of material wit d = w as seen from the top. The
sheet resistance is useful as it can be directly measured on a test structure in the laboratory. Once
is known, then the total resistance of a line that has a width w and spans a distance d is given by
where
1
This is due to the small current flow levels. One exception (among several) to this statement are the power
supply and ground lines, which can exhibit inductive effects. For example, if the current changes very quickly
in a power supply line then there is a temporary voltage drop of v=L(di/dt) on the line, which reduces the voltage reaching the circuit.
On-Chip Interconnects
479
is the number of squares of dimensions (w × w ) encountered by the current. This can be seen by
the top view of the interconnect shown in Figure 10.2. Owing to this observation,
is sometimes
labeled as having units of “ohms per square” in processing jargon.It is worthwhile to note that the
sheet resistance is quite sensitive to the height h of the interconnect. As the minimum linewidth has
decreased, the thickness of the material layer has remained fairly large. In fact, most of the interconnects in a state-of-the art process are thicker than they are wide.
Example 10-1
Doped polysilicon has a best-case sheet resistance of about 20-to-25 per square. Consider a poly
line that has a width of 0.4
and a length of 20
. The number of squares of size (0.4
x 0.4
is
so with
we have a line resistance of
To show the relative significance of this results, consider an nFET with
an aspect
ratio of 10, and a threshold voltage of 0.7 volts. With a 3.3v power supply, the linearized resistance
is
so that the line resistance is about five times larger than the FET drain-source resistance. If we used
this type of interconnect at the output of a logic gate, the parasitic resistance would dominate the
delay times.
Line Capacitance
The capacitance of an interconnect line can be the limiting factor in high-speed signal transmission.
Consider the cross-sectional geometry shown in Figure 10.3(a). Most formulations are based on the
capacitance per unit length c’ with units of farads per centimeter such that the total capacitance of
the line in farads is given by