Transcript Lecture 41

Lecture #41: Active devices
• This week we will be reviewing the
material learned during the course
• Today: review
– Active circuits
– Digital logic
– CMOS transistors
12/8/2004
EE 42 fall 2004 lecture 41
1
Example of the Load-Line Method
Lets hook our 2K resistor + 2V source circuit up to an LED (light-emitting
diode), which is a very nonlinear element with the IV graph shown below.
Again we draw the I-V graph of the 2V/2K circuit on the same axes as
the graph of the LED. Note that we have to get the sign of the voltage
and current correct!!
At the point where the two graphs intersect, the voltages and the currents
are equal, in other words we have the solution.
I (ma)
I
+
2V
-
LED
2K
+
4
LED
Solution: I = 0.7mA,
V = 1.4V
V
-
2
5
12/8/2004
EE 42 fall 2004 lecture 41
V (Volt)
2
Simplification for time behavior of RC Circuits
We call the time period
during which the output
changes the transient
We can predict a lot about the
transient behavior from the pre- and
post-transient dc solutions
12/8/2004
input
time
voltage
Long after the input change
occurs things “settle down” ….
Nothing is changing …. So
again we have a dc circuit
problem.
voltage
Before any input change occurs we have a dc circuit
problem (that is we can use dc circuit analysis to relate the
output to the input).
EE 42 fall 2004 lecture 41
output
time
3
RC RESPONSE
Example – Capacitor uncharged: Apply voltage step of 5 V
5
Vin
Input node
Output node
+
Vout
Vin
-
0
0
R
time
Vout
C
ground
• Clearly Vout starts out at 0V ( at t = 0+) and approaches 5V.
•
We know this because of the pre-transient dc solution (V=0) and
post-transient dc solution (V=5V).
So we know a lot about Vout during the transient - namely its initial value,
its final value , and we know the general shape .
12/8/2004
EE 42 fall 2004 lecture 41
4
LOGIC GATE DELAY D
Time delay D occurs between input and output: “computation” is not
instantaneous
Value of input at t = 0+ determines value of output at later time t = D
F
A
Logic State
B
Capacitance to Ground
Input (A and B tied together)
1
0
t
0
1
Output (Ideal delayed step-function)
F
0
12/8/2004
D
Actual exponential voltage versus time.
0
EE 42 fall 2004 lecture 41
t
5
SIGNAL DELAY: TIMING DIAGRAMS
Show transitions of variables vs time
Oscilloscope Probe
A
B
C
D
Logic state
A
Note B changes one gate delay
after A switches
Note that C changes two gate delays
after A switches.
1
0
B

t
 2
t
 2 3
t
C
Note that D changes three gate delays
D
after A switches.
12/8/2004
t
0
EE 42 fall 2004 lecture 41
6
EXAMPLE OF THE USE OF DEPENDENT SOURCE
IN THE MODEL FOR AN AMPLIFIER
AMPLIFIER SYMBOL
Differential Amplifier
V+
V
+
A

V0  A( V  V )
AMPLIFIER MODEL
Circuit Model in linear region
V0
Ri
+

V1
+

AV1
+

V0 depends only on input (V+  V-)
See the utility of this: this Model when used correctly
mimics the behavior of an amplifier but omits the
complication of the many many transistors and other
components.
12/8/2004
EE 42 fall 2004 lecture 41
7
V0
NODAL ANALYSIS WITH DEPENDENT SOURCES
Example circuit: Voltage controlled voltage source in a branch
R5
R1
R3
Va
Vb
Vc
R2
+

VAA
+

AvVc
R4
R6
ISS
Write down node equations for nodes a, b, and c.
(Note that the voltage at the bottom of R2 is “known” so current
flowing down from node a is (Va  AvVc)/R2.)
Va  VAA Va  A v Vc Va  Vb


0
R1
R2
R3
Vc  Vb Vc
Vb  Va Vb Vb  Vc

 ISS


0
R5
R6
R3
R4
R5
12/8/2004
EE 42 fall 2004 lecture 41
CONCLUSION:
Standard nodal
analysis works
8
NODAL ANALYSIS WITH DEPENDENT SOURCES
Finding Thévenin Equivalent Circuits with Dependent Sources Present
Method 1: Use Voc and Isc as usual to find VT and RT
(and IN as well)
Method 2: To find RT by the “ohmmeter method” turn
off only the independent sources; let the dependent
sources just do their thing.
12/8/2004
EE 42 fall 2004 lecture 41
9
NODAL ANALYSIS WITH DEPENDENT SOURCES
Example : Find Thévenin equivalent of stuff in red box.
R3
Va
Vc
R2
+

A v Vcs
R6
ISS
With method 2 we first find open circuit voltage (VT) and then we
“measure” input resistance with source ISS turned off.
R 2 (R 6  R 3 )
You verify the solution: V  ISS R 6 (R 2  AR 3 ) R TH 
TH
R 2  R 3  R 6 (1 - A)
R 2  R 3  R 6 (1 - A)
12/8/2004
EE 42 fall 2004 lecture 41
10
EXAMPLE: AMPLIFIER ANALYSIS
USING THE AMPLIFIER MODEL WITH Ri = infinity:
Assume the voltage between the inputs is zero, and then figure out if
that is consistent, or if the amplifier will hit a rail.
RF
VIN RS
V+
V-

A
+
RF
VIN
RS
V-
-
V0
V+
+
V1 +

V0
AV1
Method: We substitute the amplifier model for the amplifier, and
perform standard nodal analysis
- AR F
solution: RIN = R F  (1  A)R S VO/VIN =
R F  (1  A)R S
1 A
12/8/2004
EE 42 fall 2004 lecture 41
11
OP-AMPS AND COMPARATORS
A very high-gain differential amplifier can function either in extremely
linear fashion as an operational amplifier (by using negative feedback)
or as a very nonlinear device – a comparator. Let’s see how!
Differential Amplifier
V+
V
+
A

“Differential”
V0  A( V  V )
Circuit Model in linear region
V0
Ri
+

V1
+

AV1
+

 V0 depends only on difference (V+  V-)
“Very high gain”  A  
But if A ~ , is the
output infinite?
The output cannot be larger than the supply voltages. It will limit or
“clip” if we attempt to go too far. We call the limits of the output the
“rails”.
12/8/2004
EE 42 fall 2004 lecture 41
12
V0
WHAT ARE I-V CHARACTERISTICS OF AN ACTUAL
HIGH-GAIN DIFFERENTIAL AMPLIFIER ?
• Circuit model gives the essential linear part
VIN +

+

V0
• But V0 cannot rise above some physical voltage related to
the positive power supply VCC (“ upper rail”)
V0 < V+RAIL
• And V0 cannot go below most negative power supply, VEE
i.e., limited by lower “rail”
V0 > V-RAIL
Example: Amplifier with gain of 105, with max V0 of 3V and min V0 of 3V.
(a)
V-V near V0 (V)
origin
(b)
V-V over wider V0 (V)
range
3
0.2
0.1
3 2 1
.2
12/8/2004
upper “rail”
2
1
1
2
3
VIN(V)
lower “rail”
30 20 10
EE 42 fall 2004 lecture 41
1
2
3
10 20 30
VIN(V)
13
THE RAILS
The output voltage of an amplifier is of course limited by whatever
voltages are supplied (the “power supplies”). Sometimes we show
them explicitly on the amplifier diagram, but often they are left off.
VDD=2V
Differential Amplifier
V+
V
+
A

V0= A( V  V )
V+
V
+
A

V0= A( V  V )
VSS=0
If the supplies are 2V and 0V, the output cannot swing beyond
these values. (You should try this experiment in the lab.) For
simplicity we will use the supply voltages as the rails.
So in this case we have upper rail = 2V, lower rail = 0V.
12/8/2004
EE 42 fall 2004 lecture 41
14
I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN
DIFFERENTIAL AMPLIFIER (cont.)
Example: Amplifier with gain of 105, with upper rail of 3V and lower rail
of 3V. We plot the V0 vs VIN characteristics on two different scales
(b)
V0 (V)
V-V over
wide range 3
upper
“rail”
(c)
Same V0 vs VIN over even
wider range
V0 (V)
3
2
1
2
1
lower
“rail”
12/8/2004
30 20 10
1
2
3
10 20 30
VIN(V)
3 2 1
EE 42 fall 2004 lecture 41
1
2
3
1
2
3
VIN(V)
15
Logic Gates
These are circuits that accomplish a given logic function such as “OR”. We will
shortly see how such circuits are constructed. Each of the basic logic gates has a
unique symbol, and there are several additional logic gates that are regarded as
important enough to have their own symbol. The set is: AND, OR, NOT, NAND,
NOR, and EXCLUSIVE OR.
A
A
AND
C=A·B
B
B
A
B
A
C=A+B B
OR
A
A
NOT
12/8/2004
A
B
NAND
C = A B
NOR
C = A B
C  AB
EXCLUSIVE OR
EE 42 fall 2004 lecture 41
16
Evaluation of Logical Expressions with “Truth Tables”
The Truth Table completely describes a logic expression
In fact, we will use the Truth Table as the fundamental meaning
of a logic expression.
Two logic expressions are equal if their truth tables are the
same
12/8/2004
EE 42 fall 2004 lecture 41
17
Some Useful Theorems
1) A  B  B  A
Defined from form
of truth tables
Communicative
2) A  B  B  A
3) A  B  C  C  B  A
Associative
4) A  B  C  C  B  A
5) A  A  0
6) A  A  1
7) A  B  A  C  A  (B  C)
Distributive
8) A  B  A  B
9) A  B  A  B
12/8/2004
Each of these can be
proved by writing out
truth tables
} de Morgan’s Laws
EE 42 fall 2004 lecture 41
18
Synthesis
Designing the combinatorial logic circuit, con’t
Method 3: NAND GATE SYNTHESIS. If we may use De Morgan’s theorem we
may turn the sum-of-products expression into a form directly implementable
entirely with NAND gates. (We also need the NOT function, but that is
accomplished by a one-input NAND gate). function.
Starting with any SUM-OF-PRODUCTS expression:
Y = ABC+DEF we can rewrite it by “inverting” with De Morgan:
Y  (ABC)  (DEF) Clearly this expression is realized with three NAND
gates: one three-input NAND for (ABC) , one for
(DEF) , and one two-input gate to combine them:
A
B
C
The NAND realization, while based on
Y
DeMorgan’s theorem, is in fact much
D
E
simpler: just look at the sum of products
F
expression and use one NAND for each term
and one to combine the terms.
12/8/2004
EE 42 fall 2004 lecture 41
19
Synthesis
Designing the combinatorial logic circuit, con’t
Method 3: NAND GATE SYNTHESIS (CONTINUED).
Two Examples of SUM-OF-PRODUCTS expressions:
X  AB  AB (X-OR function)
Y  ABC  AB C
A
B C
A
X
Y
B
(No connection)
12/8/2004
We could make the drawings simpler
by just using a circle for the NOT
function rather than showing a oneinput NAND gate
EE 42 fall 2004 lecture 41
20
Controlled Switch Model of Inverter
VDD = 3V
+
-
SP
SP is closed if
VIN < VDD by 2V
RP
VIN
+
RN
Input
Note top, type
P, switch is
“upside down”
VOUT
+
+SN is closed if Output
SN VIN > VSS by 2V -
VSS = 0V
The idea: If input is 3V then top switch open, bottom one closed. And
if input is 0V, bottom switch is open, and top switch closed. Thus we
connect the output (through one of the resistors RP or RN) to either
ground or VDD.
12/8/2004
EE 42 fall 2004 lecture 41
21
CMOS
Both NMOS and PMOS on a single silicon chip
NMOS needs a p-type substrate
PMOS needs an n-type substrate
But we can build in the same substrate by changing doping type
D
G
S
G
D
S
oxide
p
p
n-well
n
n
p-well
We can butt the p and n together, or even let, for example
the entire non n-wellEE
region
be lecture
p type.
12/8/2004
42 fall 2004
41
22
Basic CMOS Inverter
Inverter
CMOS
Inverter
IN
VDD
OUT
IN
VDD
p-ch
OUT
n-ch
Al “wires”
IN
VDD
PMOS Gate
Example layout of
CMOS Inverter
N-WELL
OUT
NMOS Gate
GROUND
12/8/2004
EE 42 fall 2004 lecture 41
23
Al “wires”
IN
VDD
PMOS Gate
N-WELL
OUT
NMOS Gate
GROUND
12/8/2004
EE 42 fall 2004 lecture 41
24
NMOS TransistorV
DS
- +
VGS
drain
gate
- +
source
ID
IG
metal
oxide insulator
metal
n-type
metal
n-type
p-type
metal
G
IG
ID
S
12/8/2004
-
VDS +
D
EE 42 fall 2004 lecture 41
25
NMOS I-V Characteristic
G
IG
ID
S
-
VDS +
D
• Since the transistor is a 3-terminal device, there
is no single I-V characteristic.
• Note that because of the insulator, IG = 0 A.
• We typically define the MOS I-V characteristic as
ID vs. VDS
for a fixed VGS.
• The I-V characteristic changes as VGS changes.
12/8/2004
EE 42 fall 2004 lecture 41
26
NMOS I-V Curves
ID triode mode
saturation mode
VGS = 3 V
VDS = VGS - VTH(n)
VGS = 2 V
VGS = 1 V
cutoff mode (when VGS < VTH(N))
12/8/2004
EE 42 fall 2004 lecture 41
VDS
27
Saturation in a MOS transistor
• At low Source to drain voltages, a MOS transistor looks
like a resistor which is “turned on” by the gate voltage
• If a more voltage is applied to the drain to pull more
current through, the amount of current which flows stops
increasing→ an effect called pinch-off.
• Think of water being sucked through a flexible wall tube.
Dropping the pressure at the end in order to try to get
more water to come through just collapses the tube.
• The current flow then just depends on the flow at the
input: VGS
• This is often the desired operating range for a MOS
transistor (in a linear circuit), as it gives a current source
at the drain as a function of the voltage from the gate to
the source.
12/8/2004
EE 42 fall 2004 lecture 41
28