EECS 105 Fall 2003, Lecture 1
Lecture 1: Course Overview and
Introduction to Phasors
Prof. Niknejad
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
EECS 105: Course Overview
Prof. A. Niknejad









Phasors and Frequency Domain (2 weeks)
Integrated Passives (R, C, L) (2 weeks)
MOSFET Physics/Model (1 week)
PN Junction / BJT Physics/Model (1.5 weeks)
Single Stage Amplifiers (2 weeks)
Feedback and Diff Amps (1 week)
Freq Resp of Single Stage Amps (1 week)
Multistage Amps (2.5 weeks)
Freq Resp of Multistage Amps (1 week)
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
EECS 105 in the Grand Scheme
Prof. A. Niknejad

Example: Cell Phone
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
Transistors are Bricks

Transistors are the building blocks (bricks) of the modern electronic world:
MOS Cap
Digital
Gate
Analog
“Amp”
Variable
Capacitor

PN Junction
Focus of course:
–
–
Understand device physics
Build analog circuits
– Learn electronic prototyping and measurement
– Learn simulations tools such as SPICE
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
SPICE stimulus
* Example netlist
Q1 1 2 0 npnmod
R1 1 3 1k
Vdd 3 0 3v
.tran 1u 100u netlist
SPICE response





SPICE = Simulation Program with IC Emphasis
Invented at Berkeley (released in 1972)
.DC: Find the DC operating point of a circuit
.TRAN: Solve the tran sient response of a circuit (solve a system of generally non-linear ordinary differential equations via adaptive timestep solver)
.AC: Find steady-state response of circuit to a sinusoidal excitation
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
BSIM






Transistors are complicated. Accurate sim requires 2D or
3D numerical sim (TCAD) to solve coupled PDEs (quantum effects, electromagnetics, etc)
This is slow
… a circuit with one transistor will take hours to simulation
How do you simulate large circuits (100s-1000s of transistors)?
Use compact models. In EECS 105 we will derive the so called “level 1” model for a MOSFET.
The BSIM family of models are the industry standard models for circuit simulation of advanced process transistors.
BSIM = Berkeley Short Channel IGFET Model
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
Berkeley…

A great place to study circuits, devices, and CAD!
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
Review of LTI Systems

Since most periodic (non-periodic) signals can be decomposed into a summation (integration) of sinusoids via Fourier Series (Transform), the response of a LTI system to virtually any input is characterized by the frequency response of the system:
Phase Shift
Any linear circuit
With L,C,R,M and dep. sources
Amp
Scale
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
Example: Low Pass Filter (LPF)
Prof. A. Niknejad


Input signal:
We know that: v s
( t ) v o
( t )


V s
K cos(
 s
 t ) cos(
 t
 
)
Phase shift
V
0
Amp shift v
0
( t )
 v s
( t )
 i ( t ) R i ( t )

C dv
0 dt v
0
( t )
 v s
( t )

RC dv
0 dt v s
( t )
 v
0
( t )
  dv
0 dt
Department of EECS University of California, Berkeley

Prof. A. Niknejad EECS 105 Fall 2003, Lecture 1
LPF the “hard way” (cont.)

Plug the known form of the output into the equation and see if it can satisfy KVL and KCL
V s cos
 t cos( x
 y )

V
0
 cos(
 t cos x cos
 
) y
 
 sin

V
0 sin(
 t x sin y
 
) sin(
V s x
 cos
 t y )
 sin

V
0 x cos y
 cos x sin cos
 t (cos
    sin y

)

V
0 sin
 t (sin
    cos

)

Since sine and cosine are linearly independent functions: a
1 sin
 t
 a
2 cos
 t

0
IFF a
1
 a
2

0
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
LPF: Solving for response…
Prof. A. Niknejad

Applying linear independence

V
0 sin
 
V
0
  cos

V
0 tan cos

 

 
V
0


 sin



V s
0

0
Phase Response:
   tan

1

V
0
(cos
  

 sin

)

V s
V
0 cos

( 1
   tan

)
V
0
V
0

1 cos


(

( 1


)
2
(


1 /

2 
)
2
)
V s

V s

V s
Amplitude Response:
V
0
V s

1
1

(
 
)
2
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
LPF Magnitude Response

1
1

0 .
707
0.8
0.6
V
0
V s

1
1

(
 
)
2
0.4
0.2
Prof. A. Niknejad

0 .
1
2 4
 

1
Passband of filter
6 8 10
 
10 /

Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
-40

V
0
V s
  tan

1

-60
-80
0
-20
LPF Phase Response

45 o
 

1
2 4 6 8
 
10
10 /

Prof. A. Niknejad
 
90 o
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad dB: Honor the inventor of the phone…



The LPF response quickly decays to zero
We can expand range by taking the log of the magnitude response dB = deciBel (deci = 10)
0
Department of EECS
-40
-10
-20


V
V s
0

 dB

20 log


V
0
V s


-30
0.1
1 10 100
University of California, Berkeley

Prof. A. Niknejad EECS 105 Fall 2003, Lecture 1
Why 20? Power!


Why multiply log by “20” rather than “10”?
Power is proportional to voltage squared: dB

10 log


V
0
V s


2

20 log


V
0
V s




At breakpoint:
 
100 /

 
1000 /




1 /


V
0
V s


 dB



V
0
V s
 
40

 dB dB
 
3 dB



V
V s
0

 dB
 
60 dB
Observe: slope of signal attenuation is 20 dB/decade in frequency
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
Why introduce complex numbers?



They actually make things easier
One insightful derivation of e ix
Consider a second order homogeneous DE y
''  y

0 y
 sin
 cos x x

Since sine and cosine are linearly independent, any solution is a linear combination of the
“fundamental” solutions
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
Insight into Complex Exponential



 e
That means: e ix  a
1 sin x
 a
2 cos x
To find the constants of prop, take derivative of this equation: i e ix   a
2 sin x
 a
1 cos x
Now solve for the constants using both equations:

 sin cos x x
 cos sin x x



 a a
2
1




 i e ix e ix


A

 a a
2
1


 b det A
 
1

0
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
The Rotating Complex Exponential

So the complex exponential is nothing but a point tracing out a unit circle on the complex plane: e ix  cos x
 i sin x
Department of EECS e i
 t e
 i
 t e i
 t  e
 i
 t
2
University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
Magic: Turn Diff Eq into Algebraic Eq
Prof. A. Niknejad

Integration and differentiation are trivial with complex numbers:

 d dt e i
 t  i
 e i
 t
 e i
 d
 
1 i
 e i
 t
Any ODE is now trivial algebraic manipulations … in fact, we’ll show that you don’t even need to directly derive the ODE by using phasors
The key is to observe that the current/voltage relation for any element can be derived for complex exponential excitation
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
Complex Exponential is Powerful

To find steady state response we can excite the system with a complex exponential
Mag Response
 e i
 t
LTI System
H
H (

) e i (
 t
 
)
Phase Response
At any frequency, the system response is characterized by a single complex number H :
H (

)
   H (

)


This is not surprising since a sinusoid is a sum of complex exponentials (and because of linearity!) sin
 t
 e i
 t
 e
 i
 t
2 i cos
 t
 e i
 t
 e
 i
 t
2
From this perspective, the complex exponential is even more fundamental
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
LPF Example: The “soft way”
Prof. A. Niknejad

Let’s excite the system with a complex exp: v s
( t )
 v
0
( t )
  dv v s
( t )

V s e j
 t v o
( t )

V
0 e j (
 t
 
) dt
0

V
0 e j
 t use j to avoid confusion real complex
V s e j
 t 
V
0 e j
 t
V s
  

V
0

1
 j
 
V
0 e j
 t j
   
V
0
V s


1

1 j
    Easy!!!
University of California, Berkeley Department of EECS

EECS 105 Fall 2003, Lecture 1
Magnitude and Phase Response
Prof. A. Niknejad

The system is characterized by the complex function
H (

)

V
V s
0


1

1 j
   

The magnitude and phase response match our previous calculation:
H (

)

V
0
V s

1
1

(

)
2

 H (

)
  tan

1


Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
Why did it work?


The system is linear:
Re[ y ]

L (Re[ x ])

Re[ L ( x )]
If we excite system with a sinusoid: v s
( t )

V s cos
 t

V s
Re[ e j
 t
]

If we push the complex exp through the system first and take the real part of the output, then that’s the
“real” sinusoidal response v o
( t )

V o cos(
 t
 
)

V o
Re[ e j (
 t
 
)
]
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
And yet another perspective…
Prof. A. Niknejad

Again, the system is linear: y

L ( x
1
 x
2
)

L ( x
1
)

L ( x
2
)

To find the response to a sinusoid, we can find the e e e i
 t e i
 t e
 i
 t
 e
 i
 t
2
Department of EECS
LTI System
H
LTI System
H
LTI System
H
H (

) e i (
 t
 
1
)
H (
 
) e i (
  t
 
2
)
H (

) e i
 t 
H (
 
) e
 i
 t
2
University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
Another persepctive (cont.)
Prof. A. Niknejad


Since the input is real, the output has to be real:
H (

) e i
 t 
H (
 
) e
 i
 t y ( t )

2
That means the second term is the conjugate of the first:
H (
 
)

H (

)
 H (
 
)
   H (

)
  
( even function)
( odd function)

Therefore the output is: y ( t )

H

(

)  e i (
 t
 
)
2
H (

)
 cos(
 t e
 i (
 t
 
)
 
)


Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
“Proof” for Linear Systems
Prof. A. Niknejad y


For an arbitrary linear circuit ( L , C , R , M , and dependent sources), decompose it into linear suboperators, like multiplication by constants, time derivatives, or integrals: y

L ( x )
 ax
 b
1 d dt x
 b
2 d
2
2 x
    x
  x
  x
  dt
For a complex exponential input x this simplifies

L ( e j
 t
) y
 to:
 ae j
 t ae j
 t  b
1
 b
1 j
 e j
 t d dt e j
 t
 b
2
(
 b
2 j

)
2 e d
2 dt 2 j
 t e j
 t   
   c
1 c
1
 e j
 t j

 e j
 t c
2 y

Hx
 e j
 t

 a
 b
1 j
  b
2
( j

)
2    c
1 j


 c
2

( e j
 t j

)
2 e j
 t
 
 
( j c

2
)
2
 


Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
“Proof” (cont.)

Notice that the output is also a complex exp times a complex number: y

Hx
 e j
 t

 a
 b
1 j
  b
2
( j

)
2    c
1 j


( j c

2
)
2
 



The amplitude of the output is the magnitude of the complex number and the phase of the output is the phase of the complex number y

Hx
 e j
 t

 a
 b
1 j
  b
2
( j

)
2    c
1 j


( j c

2
)
2



 y
 e j
 t
H (

) e j  H (

)
Re[ y ]

H (

) cos(
 t
  H (

))
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1 Prof. A. Niknejad
Phasors




With our new confidence in complex numbers, we go full steam ahead and work directly with them … e i
 t we can even drop the time factor since it will cancel out of the equations.
Excite system with a phasor:
Response will also be phasor:
~
V
1

~
V
2

V
1 e
V
2 e j

1 j

2
For those with a Linear System background, we’re going to work in the frequency domain
– This is the Laplace domain with s
 j

Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
Capacitor I-V Phasor Relation
Prof. A. Niknejad

Find the Phasor relation for current and voltage in a cap:
+ i c
( t )

C dv
C
( t ) dt i c
( t ) v c
( t )


Re[ I c e j
 t
]
Re[ V c e j
 t
] v
C
( t ) i c
( t )
_
Re[ I c e j
 t
]

C d dt
Re[ V c e j
 t
]
Re[ CV c d e j
 t
] dt
I
I c e
 c j
 t  j
 j

C V c
CV c e
 j
 t
Re[ j

CV c e j
 t
]
Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 1
Inductor I-V Phasor Relation
Prof. A. Niknejad

Find the Phasor relation for current and voltage in an inductor: v ( t )

L di ( t ) dt i ( t )

Re[ Ie j
 t
] v ( t )

Re[ Ve j
 t ]
+ v ( t ) i ( t )
_
Re[ Ve j
 t
]
 d
L dt
Re[ Ie j
 t
]
Re[ LI
Ve j
 t  d dt j
 e j
 t
]
LIe
 j
 t
Re[
V
 j

L I j

LIe j
 t
]
Department of EECS University of California, Berkeley