Electronics A - MIT OpenCourseWare...JV: 2.372J/6.777J Spring 2007, Lecture 6E - 1 Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical

JV: 2.372J/6.777J Spring 2007, Lecture 6E - 1

Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

Electronics A

Joel Voldman

Massachusetts Institute of Technology



Outline

> Elements of circuit analysis

> Elements of semiconductor physics• Semiconductor diodes and resistors• The MOSFET and the MOSFET inverter/amplifier

> Op-amps

TODAY



Elements of circuit analysis

> There are many ways to analyze circuits

> Here we’ll go over a few of them• Elements laws, connection laws and KVL/KCL• Nodal analysis• Intuitive approaches• Superposition



Lumped elements in circuits

> Circuit elements (R, L, C) are lumped approximations of complex devices

> The electrical capacitor• What is the relation

between Q and V?

0 ( , ) ( , )

( ) ( )

( ) ( )

closedsurface

b

a

d Q

EA Q

t V t

V b V a d

VV b V a V Eg E gAVQ A V CVg g

ACg

ε

ε

εε

ε

⋅ =

=

∇× = ⇒ = −∇

− = − ⋅

− = = ⇒ =

= = =

=

∫

∫

E a

E E r r

E l

+-V

Ig

A



Lumped elements in circuits

> The electrical capacitor• We can replace all of field theory with terminal relations• And introduce an element with an element law• As long as capacitor size << wavelength of electrical signal

» In general, MEMS are small e.g., λ=50 μm 600 GHz

CI

+ -V dtdVCI

CVdtd

dtdQI

CVQ

=

==

=

)(



Elements and element laws

> Do this with all three basic elements

> Resistor

> Capacitor

> Inductor

V RI=

RI

+ -VdIV Ldt

=

LI

+ -V

CI

+ -VdVI Cdt

=



Source elements

> We need elements to provide energy into the circuit

> Two common ones are voltage source and current source

I0

+

-V

I

+-V0

+

-V

I

0V V= 0I I=



KVL and KCL

> These are continuity laws that allow us to solve circuits

> Kirchhoff’s voltage law• The oriented sum of voltages around a loop is zero

> Kirchhoff’s voltage law• The sum of currents entering a node is zero



Complex impedances

> For LTI systems, use complex impedances instead

• Implicitly working in frequency domain

> Much easier circuit analysis

> All elements treated the same, like “resistors”

RI

+ -V

LI

+ -V

CI

+ -V

1CV IZ I

Cs= =

RV IZ IR= =

LV IZ ILs= =

+ -V

iZ ( ) ( ) ( )V s I s Z s=



Let’s analyze a circuit

1. Figure out what are you trying to determine

2. Replace elements with complex impedances

3. Assign across and through variables

4. Use KVL

5. Substitute in element laws

6. Solve

R

LC+

-V0

ZR

ZL

ZC+-V0

+ -VCiC

+- VR iR

+

-VL

iL

+

-VV

iV







4. Use KVL


6. Solve

0

00

V C L R

C C L L R R

C L R

V V V VV i Z i Z i Zi i i

− + − =− + − =

= − =

ZR

ZL

ZC+-V0

+ -VCiC

+- VR iR

+

-VL

iL

+

-VV

iV







4. Use KVL


6. Solve

0

0 0

02

0

1

1

R C R L R R

RC L R

R

V i Z i Z i ZV Vi

Z Z Z Ls RCsCsi V

LCs RCs

− − − =

= =+ + + +

=+ +

ZR

ZL

ZC+-V0

+ -VCiC

+- VR iR

+

-VL

iL

+

-VV

iV



Example #1

> Solve for VL

R

LC+

-V0VL

+

-



Nodal analysis

> Element law approach becomes tedious for circuits with multiple loops

> Nodal analysis is a KCL-based approach



Nodal analysis



3. Assign node voltages & ground node

4. Write KCL at each node

5. Solve for node voltages

6. Use node voltages to find what you care about

LC+

-V0 R

LC+

-V0 R

v2v1

+ -VC



Nodal analysis







LC+

-V0 R

i1 i2

i3

v2v1

1 0

1 2 3

1 2 2 2

00 0 0

C L R

v Vi i i

v v v vZ Z Z

=+ + =

− − −+ + =

Node 1:

Node 2:



Nodal analysis







LC+

-V0 R

i1 i2

i3

v2v1

( )

02

2 0

2 0

1 1 1

C L R C

L R C R L C L R

L R

L R C R L C

VvZ Z Z Z

v Z Z Z Z Z Z V Z ZZ Zv V

Z Z Z Z Z Z

⎛ ⎞+ + =⎜ ⎟

⎝ ⎠+ + =

=+ +



Nodal analysis







LC+

-V0 R

i1 i2

i3

v2v1

2 0

2

2 0 2

2

1 2 0 0 2

1 1

C

LRsv VLRs R LsCs Cs

LRCsv VLRCs Ls R

LRCsV v v V VLRCs Ls R

=+ +

=+ +

= − = −+ +



Example #2

+

-L

C+-V0

VLR2

R1



Intuitive methods

> Instead of “solving” the circuit using equations, use series/parallel tricks to analyze the circuit by inspection

> Current divider & impedances in parallel

• Both elements have SAME voltage• Terminals connected together

+

-

V

ii1 i2

Z2

+

-

Vi

ZZ1

21

1 2

12

1 2

1 21 1

1 2

1 21 2

1 2

1 2

//

1 1 1

Zi iZ Z

Zi iZ Z

Z ZV i Z iZ Z

Z ZZ Z ZZ Z

Z Z Z

=+

=+

= =+

= =+

= +



Intuitive methods

> Voltage divider & impedances in series

• Both elements have SAME current1

11 2

22

1 2

11

1 1 2

1 2

ZV VZ Z

ZV VZ Z

V Vi iZ Z Z

Z Z Z

=+

=+

= = =+

= +

+

+

--

V

+

-V1

V2

i

Z2

+

-

Vi

Z

Z1



Intuitive methods

> Examples of elements NOT in series OR parallel

Z2Z1

Z4Z3

Z1

Z4Z3

Z1 and Z3 in seriesZ2 and Z4 in seriesZ1 and Z2 NOT in parallelZ3 and Z4 NOT in parallel

Z3 and Z4 in parallelZ1 and Z3 NOT in series



Intuitive methods

> Let’s use this approach to solve a circuit



3. Collapse circuit in terms of series/parallel relations till circuit is trivial

4. Re-expand to find signal of interest

//a R LZ Z Z=

iL

C+-V0

LC+

-V0 R

+

-

VaZa

0a

aa C

aL

L

ZV VZ ZViZ

=+

=



Intuitive methods

> Let’s use this approach to solve a circuit



3. Collapse circuit in terms of series/parallel relations till circuit is trivial

4. Re-expand to find signal of interest

( )

( )

0 0

0

0

0

0 2

//1 1//

1

1

11

a R LL

a C L R L C L

R L

R L

R L LC

R L

R L

R L R L C L

L

Z Z Zi V VZ Z Z Z Z Z Z

Z ZZ ZV Z Z ZZ

Z ZZ ZV

Z Z Z Z Z ZRLsV

LsRLs R Ls CsRCsi V

RLCs Ls R

= =+ +

+=

++

=+ +

=+ +

=+ +



Example #3

+

- L

C+-V0

VRR2

R1



Superposition

> These equivalent circuits are linear and obey the principles of superposition

• This can be useful

> For circuits with multiple sources, • Turn off all independent sources

except one• Solve circuit• Repeat for all sources, then add

responses

> Turning off a voltage source gives a short circuit

> Turning off a current source gives an open circuit

+-V0

I0

short

open


2 ZR CCite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

Superposition

> For circuits with multiple sources,

• Turn off all independent sources except one

• Solve circuit• Repeat for all sources, then add

responses

C+-V0

I0R

v2v1

CI0R

v2v1

Find v

2 0 //R Cv I Z Z=

C+-V0 R

v2v1

2 0RZv VZ

=+


2Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

Superposition

> For circuits with multiple sources,

• Turn off all independent sources except one

• Solve circuit• Repeat for all sources, then add

responses

C+-V0

I0R

v2v1

Find v

2 0 0

2 0 0

0 02

0 02

//

1

1

1

RR C

R C

R C R

R C R C

Zv I Z Z VZ Z

Z Z Zv I VZ Z Z Z

I R V RCsvR Cs

I R V RCsvRCs

= ++

= ++ +

+=

+

+=

+



Conclusions

> There are many ways to analyze equivalent circuits

> Use the simplest method at hand

> Element laws & connection laws are OK for simple ckts

> Nodal analysis works for most any circuit, but will be tedious for complicated circuits

> Try to use intuitive approaches whenever possible

Documents

Electronics A - MIT OpenCourseWare...JV: 2.372J/6.777J Spring 2007, Lecture 6E - 1 Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical