CHAPTER 16
Memory Circuits
Introduction
 The 2 major logic classifications are
 Combinational circuits: Their output depends only on the
present value of the input. These circuits do not have memory.
 Sequential circuits: Logic circuits that incorporate memory
are called sequential circuits. Their output depends not only on
the present of the input but also on the input’s previous values.
 2 approaches to provide memory to a digital circuits:
 Static sequential circuits: It employs positive feedback to
provide a circuit with 2 stable states (bistable state).
 Dynamic sequential circuits: It utilize a capacitor to store the
charge.
2
The Latch
 The basic memory element, the latch, is shown below.
 The latch memorizes the
external action by staying
indefinitely in the acquired state.
The state will be changed when
it is triggered.
 Point B is an unstable operating point.
By considering the noise that causes
positive (negative) small increment at W,
the regenerative process will shift the point
B upward (downward) to point C (A).
Figure 16.1 (a) Basic latch. (b) The latch with the feedback loop opened. (c) Determining the operating point(s) of the latch.
3
Figure 16.2
The SR Flip-Flop
 The latch together with the triggering circuitry forms a flip-flop.
 When S/R are raised to 1 simultaneously, Q/Q become 0.
However, when S/R return to rest state (S=R=0), the state of flipflop will be undefined.
Figure 16.3 (a) The set/reset (SR) flip-flop and (b) its truth table.
5
CMOS Implementation of
SR Flip-Flops (1)
 The clocked version of an SR flip-flop has the 2 cross-coupled
inverters as the heart and only NMOS transistors are used in the setrest circuitry (no conducting path between VDD and ground).
 If the flip-flop stores “0” (Q=0 and Q=1)
initially, we wish to set it.
1
 Arrange VDD on S input while R is held
at 0V.
0
ψ goes high  Q5/Q6 conduct and pull
the VQ down.
 If VQ goes below the VTH of Q3/Q4
inverter, it will change state  VQ rises.
1
0
 The increase of VQ is fed to Q1/Q2
inverter, causing the VQ go down further.
Figure 16. 4 CMOS implementation of a clocked SR flip
flop. The clock signal is denoted by Φ
6
CMOS Implementation of
SR Flip-Flops (2)
1
 The normal operation of the flip-flop is based on 2 important
assumptions.
 Transistors Q5/Q6 supply sufficient
current to pull the node Q down to a
voltage at least slightly below the VTH of
Q3/Q4 inverter.
0
 The set signal remains high for an
interval
long
enough
to
cause
regeneration to take over the switching
process..
0
1
7
Example 16.1 , p.1310
IDeq >= ID2
VGD=(1/2)VDD>Vt
Example 16.1
iC= iDeq - iD2
Figure 16.7 A simpler CMOS implementation of the clocked SR flip-flop. This circuit is popular as the basic cell in the design of static
random-access memory (SRAM) chips.
D Flip-Flop Circuits (1)
 The important type of flip-flop, the data (D) flip flop is shown below.
 When the clock is low, the flip-flop is in the memory (rest) state.
 As the clock goes high, the flip-flop acquires the logic level that
existed on the D-line just before the rising edge of the clock. Such a
flip-flop is said to be edge-triggered.
Figure 16.8 A block diagram representation of the D flip-flop.
11
D Flip-Flop Circuits (2)
 A simple implementation of the D flip-flop is shown below and its
operation can be described as follows.
Non-overlapping clock
 When ψ is high, the loop is opened, and the input D is connected to
input of inverter G1. The capacitance at input of G1 (G2) is charged to
the value of D (D).
 When ψ is low, the input line is isolated and the closed feedback loop
makes the latch acquire the state corresponding to the value of D just
before ψ went down.
Figure 16.9 A simple implementation of the D flip-flop. The circuit in (a) utilizes the two-phase non-overlapping clock whose
waveforms are shown in (b).
12
D Flip-Flop Circuits (3)
 The following is a flip-flop with a master-slave configuration.
 When ψ1 is high and ψ2 is low, feedback loop of slave is closed and
Q remains the value previously stored. Input capacitance of the master
latch is charged to the value of D.
 When ψ1 is low and ψ2 is high, master latch locks the value of D and
the node capacitances in the salve are charged to D.
 When ψ1 goes high again, the slave latch locks the new value of D
and output Q=D.
Figure 16.10 (a) A master–slave D flip-flop. The switches can be, and usually are, implemented with CMOS transmission gates. (b) Waveforms
of the two-phase nonoverlapping clock required.
13
16.2 Types of Semiconductor Memories (1)
 Computer memory can be divided into 2 types: main memory and
mass storage memory.
 The main memory is usually of the random-access type. A randomaccess memory (RAM) is one in which the time required for writing and
reading information is independent of the physical location in which the
information is stored.
 RAMs should be contrasted with serial or sequential memories.
Sequential access means that the accessing a piece of information will
take a varying amount of time, depending on which piece of information
was accessed last. The device may need to seek (e.g. to position the
read/write head correctly), or cycle (e.g. to wait for the correct location
in a revolving medium to appear below the read/write head). Floppy
disk and hard disk are of this type.
14
Types of Semiconductor Memories (2)
Source: G. Chou at ProMOS
15
Memory-Chip Organization
 The bits on a memory chip are addressable either individually or in
groups of 4 to 16.
 64M-bit chip in which all bits individually addressable is said to be
organized as 64M words x 1 bit (64M x 1, it needs a 26-bit address). It
can also be organized as 16M words x 4 bits (16M x 4, 24-bit address is
required).
 The bulk of the memory circuit consists of the cells in which the bits
are stored. Each memory cell is an electronic circuit capable of storing
one bit.
Source: ProMOS
16
Innovation in Organization
 As the number of cells in the array increases, the physical lengths of
the word lines and bit lines increase. This has occurred even though for
each generation of memory chips, the transistor size has decreased
(currently 90 nm feature size are utilized).
 The net increase in word-line and bit-line length increases their total
resistance and capacitance, and slows down the transistor response.
 This problem has been solved by partitioning the memory chip
into a number of blocks.
17
Memory Structure (2M rows x 2N columns)
Figure 16.11 A 2M+N -bit memory chip organized as an array of 2M rows × 2N columns.
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Memory Chip Timing
 The
memory access time is the time between the
initialization of a read operation and the appearance
of the output data.
The memory cycle time is the minimum time
allowed between 2 consecutive memory operation.
 MOS memories have access and cycle times in the
range of a few to few hundred nano-seconds.
 More information about memory chip timing will be
discussed in DRAM part.
19
16.3 Random-Access Memory Cells
 There are basically 2 types of MOS RAM: static and dynamic.
 Static RAMs (called SRAMs for short) utilize static latches as
the storage cells. SRAMs can hold their stored data indefinitely,
provided the power supply remains on.
 Dynamic RAMs (called DRAMs), store the binary data on
capacitors, resulting in further reduction in cell area. It requires
periodic refreshing to regenerate the data stored on the
capacitors.
 Both static and dynamic RAMs are volatile, that is, they
require continuous presence of a power supply.
20
Static Memory Cell Structure
 A CMOS SRAM memory cell is a flip-flop comprising 2 cross-coupled
inverters and 2 access transistors, Q5 and Q6.
Figure 16.12 A CMOS SRAM memory cell.
21
Static Memory Cell - Read Operation (1)
 Relevant parts of the SRAM cell circuit during a read operation when
the cell is storing a logic 1.
1
 The cell must be designed so that the changes in vQ and vQ are small
enough to prevent the flip-flop from changing state during readout.
22
Static Memory Cell - Read Operation (2)
 Assume the cell is storing a 1 (Q=1/Q=0).
 B and B lines are precharged to an intermediate voltage, between
the low and high values, say VDD (usually VDD/2).
 When word line is selected and Q5/Q6 are turned on
 current flow from VDD through Q4/Q6 and onto line B, charging CB.
 current flow from precharged B line through Q5/Q1 to ground,
discharging CB.
 Voltage across CB will rise and that across C B will fall.
 A differential voltage develops between line B and line B. Usually
only 0.2 V is required for the sense amplifier to detect the presence of a
1 in the cell.
 The read operation in an SRAM is non-destructive.
23
(W/L)5 ↓, ensure
VQ < Vtn of latch
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∆t = CB *∆V/I5 (16.7)
Large I5→ small ∆t
Figure 16.15 Voltage waveforms at various nodes in the SRAM cell during a read-1 operation.
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0
1
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Static Memory Cell - Write Operation (1)
 Assume the cell is storing a 1 (Q=1/Q=0) and we wish to write a 0.
 The B line is lowered to 0 and the B line is raised to VDD. The cell is
selected by raising the word line to VDD.
 Q is being pulled up from 0 toward VDD/2 and Q is being pulled down
from VDD toward VDD/2.
 CQ will discharge faster than CQ will charge because
 The current (ICQ=I5-I1) that charges the CQ will decrease when
current of Q1 increases due to the raised VQ.
Body effect of Q5 makes the I5 decreases as VQ increases.
* There are some corrections above !!
27
Static Memory Cell - Write Operation (2)
 The write delay is much smaller than the corresponding component in
the read operation.
 This is because in the write operation, only the small capacitance CQ
needs to be charged (discharged), whereas in the read operation,
charge (discharge) the much larger capacitances CB is required.
28
To ensure VQ < Vtn,
(W/L)4 / (W/L)6 ↓,
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Dynamic Memory Cell (1)
 Where can we find DRAMs?
30
Dynamic Memory Cell (2)
The memory controller acts like a traffic signal that
directs the movement of data across the memory
channel. For example, data arriving to the Memory
Controller is first stored in the memory modules (2),
then is re-read (3) and finally transferred to the
processor (4).
31
Dynamic Memory Cell (3)
 The DRAM cell consists of a single N-MOSFET (access transistor)
and a storage capacitor Cs.
 When the cell is storing a 1, the capacitor is charged to VDD-Vt; when
a 0 is stored, the capacitor is discharged to 0V.
 Because of the leakage effect, the cell must be periodically refreshed.
During refresh, the cell content is read and the data bit is re-written,
thus restoring the capacitor voltage to a proper value.
The refresh operation must
be performed every 5-10 ms
(currently 64 ms). During
refresh, the chip is not
available
for
read/write
operation. The interval required
to refresh the entire chip is less
than 2% between refresh
cycles.
Figure 16.18 The one-transistor dynamic RAM (DRAM) cell.
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Dynamic Memory Cell (4)
Transistor
(電晶體)
Capacitor
(電容器)
Cell (記憶胞) = 1 bit
Transistor
(電晶體)
+
Capacitor
(電容器)
=
Cell (記憶胞)
33
Dynamic Memory Cell – Read Operation
 The bit lines are precharged to VDD/2.
 The word line is raised to VDD and the access transistors on the
selected row will conduct.
 Small voltage difference will appear on the bit lines.
 Assume the initial voltage on the cell capacitor be VCS (it is VDD-Vt
for a “1” and 0 V for a “0”).
 For CB is 30 CS, VDD=5V and Vt=1.5V, V (0) = -83 mV and V (1) =
33 mV. VDD is currently 1.8 V for DDR2 DRAM chip.
 The readout process of DRAM is destructive.
Figure 16.19 When the voltage of the selected word line is raised, the transistor conducts, thus connecting the storage capacitor CS to the
bit-line capacitance CB.
34
Dynamic Memory Cell – Read Operation
cycle time
access time
35
16.4 Sense amplifiers and address decoders
Peripheral Circuits of A RAM Chip
 Besides memory cell, other important circuit blocks in a memory chip
are peripheral circuits. It contains sense amplifiers and address
decoders.
 The
input
and
output
terminals are the same for a
sense amplifier.
The
sense
amplifier
is
required to detect a small signal
appearing between B and B
and to provide a full-swing
signal at B and B.
 The output of a sense
amplifier is directed to the chip
I/O pin and at the same time is
used to rewrite a data into cell.
36
Operation of A Sense Amplifier (1)
 Before read operation, B and B
line are equal voltage (VDD/2) by
turning on Q7/Q8/Q9 through the
signal ψp. Q7 helps speed the
process by equalizing the initial
voltages on the two lines.
 Initially the latch will be at its
unstable equilibrium point.
 ψs controls whether to turn on
the sense amplifier. This is
important to conserve power.
37
Operation of A Sense Amplifier (2)
 When read operation commences, the voltage between the 2 lines
develops and the latch will quickly move to one of its 2 equilibrium
points.
Figure 16.21
38
Operation of A Sense Amplifier (3)
 What happens to the B line when a particular word line is selected?
Dummy cell has a storage
capacitor CD = CS.
Figure 16.22 An arrangement for obtaining differential operation from the single-ended DRAM cell. Note the dummy cells at the far right
and far left.
39
Figure 16.23
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Figure 16.24 The active-loaded MOS differential amplifier as a sense amplifier.
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The Row-Address Decoder
 The
row-address
decoder is required to
select on of 2M word
lines in response to an
M-bit address input.
 For M=3, 1 out of 8
lines is selected using
3 bit address. Each row
connects
to
3
transistors.
 During precharge (ψp
is low) and all word
lines at at VDD.
For A2A1A0=011, row
3 is selected and kept
at high voltage.
 precharge and then turn off QP.
off
0
off
off
1
1
 address bits
are applied.
Figure 16.25 A NOR address decoder in array form. One out of eight lines (row lines) is selected using a 3-bit address.
42
The Column-Address Decoder (1)
 A column decoder realized by a combination of a NOR decoder and
a pass-transistor multiplexer.
Figure 16.26 A column decoder realized by a combination of a NOR decoder and a pass-transistor multiplexer.
43
The Column-Address Decoder (2)
 A tree column decoder. Note that the colored path shows the
transistors that are conducting when A0 = 1, A1 = 0, and A2 = 1, the
address that results in connecting B5 to the data line.
Small number of
transistors
but
with slower speed
(relatively
large
number of series
transistors in the
signal path).
Figure 16.27 A tree column decoder. Note that the colored path shows the transistors that are conducting when A0 = 1, A1 = 0, and A2 = 1,
the address that results in connecting B5 to the data line.
44
The Ring Oscillator
 The ring oscillator is formed by connecting an odd number of
inverters in a loop.
 Note that the rising edge at node 1 propagates through gate 1, 2,
and 3 to return inverted after a delay of 3 tp.
 The circuit oscillates with a period of 6 tp.
Figure 16.28 (a) A ring oscillator formed by connecting three inverters in cascade. (Normally at least five inverters are used.) (b) The resulting
waveform. Observe that the circuit oscillates with frequency 1/6tP.
45
Figure 16.29 (a) A one-shot or monostable circuit. Utilizing a delay circuit with a delay T and an XOR gate, this circuit provides an output
pulse of width T. (b) The delay circuit can be implemented as the cascade of N inverters where N is even, in which case T = NtP .
Microelectronic Circuits, International Sixth Edition
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How to Measure Propagation Delay (1)
 Ring oscillators propagation delays are used to benchmark the
performance of CMOS fabrication process.
 The simplest ring oscillator contains an odd number of cascaded
inverters, with the output of the cascade fed back to the input of the
inverter chain. In this configuration, the oscillator frequency is
completely dependent on the inherent inverter time delay and is
therefore not externally controllable.
47
How to Measure Propagation Delay (2)
 For a ring oscillator, odd-number n (usually n > 5) of inverters are
connected in a chain and it will oscillate with period T.
Source: K. Banerjee at UC Santa Barbara
48
How to Measure Propagation Delay (3)
 If the gate delay (CV/I) of a
transistor is less than 1 ps,
then we call it a TeraHertz
transistor.
 To further improve transistor
speed for TeraHertz operation,
the following structure will be
used.
Source: 2001 Si nanotechnology
workshop from Intel
49
A MOS ROM
 The following is a 32 bit (8 words x 4 bit) MOS ROM. The major
disadvantage is the static power dissipation. The data stored in the
ROMs is determined at the time of fabrication.
Figure 16.30 A simple MOS ROM organized as 8 words ×4 bits.
50
Mask-Programmable ROMs and PROMs
Mask-programmable ROMs
To avoid having to custom-design each ROM from scratch, ROMs are
manufactured using a process known as mask programming.
 MOSFET can be included at all bit locations, but only the gates of
those transistors where 0s are to be stored are connected to the word
lines.
Programmable ROMs
 PROMs are ROMs that can be programmed by the user, but only
once (They are frequently seen in computer games, or such products
as electronic dictionaries).
 A typical PROM comes with all bits reading as 1, burning a fuse
during programming causes its bit to read as 0. The memory can be
programmed just once after manufacturing by "blowing" the fuses
(using a PROM blower), which is an irreversible process.
51
EPROMs (1)
Erasable-Programmable ROMs
 It is a type of computer memory chip that retains its data when its
power supply is switched off. Once programmed, an EPROM can be
erased only by exposing it to strong ultraviolet light (wavelength=2537A).
 EPROMs are easily recognizable by the transparent window in the top
of the package, through which the silicon chip can be seen, and which
permits UV light during erasing.
52
EPROMs (2)
 Its structure and programming principle are illustrated below.
Figure 16.31 (a) Cross section and (b) circuit symbol of the floating-gate transistor used as an EPROM cell.
Microelectronic Circuits, International Sixth Edition
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Figure 16.32
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Figure 16.33 The floating-gate transistor during programming.
Microelectronic Circuits, International Sixth Edition
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EEPROMs
 A more versatile PROM is the electrically erasable PROM
(EEPROM). As the name implies, an EEPROM can be erased and
reprogrammed electrically without the need for ultraviolet illumination.
 The difference between EPROM and EEPROM lies in the thickness
of the insulating layer — in an EPROM it is about 25 nm thick whereas
in an EEPROM it is much thinner — typically around 10 nm. This
thinner insulating layer allows for lower voltages to be used for
programming.
56
Flash Memory
 Flash memory is a form of non-volatile computer memory that can be
electrically erased and reprogrammed.
 Unlike EEPROM, it is erased and programmed in blocks consisting
of multiple locations. Flash memory costs far less than EEPROM and
therefore has become the dominant technology wherever a significant
amount of non-volatile, solid-state storage is needed.
Source: Samsung Electronics
57