CIRCUITS AND ELECTRONICS. Filters

(1)

6.002 Fall 2000 Lecture

18

Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

6.002 CIRCUITS

AND

ELECTRONICS

(2)

6.002 Fall 2000 Lecture

18 Review

+

–

C

v

I

v

+

–

R

C

Reading:

Section 14.5, 14.6, 15.3 from A & L.

+

–

c

V

i

V

+

–

R

Z

C

Z

i

R

C

c

V

Z

V

⋅

+

=

RC

j

1

1 R

C

j

1 C

j

1 V

V

i

c

ω

+

=

+

=

(3)

6.002 Fall 2000 Lecture

18 A Filter

RC

j

1

1 V

Z

V

_i

R

C

c

=

₊

⋅

=

₊

_ω

“Low Pass Filter”

ω

1 ( )

i

c

V

H

ω

=

Demo

with audio

+

–

c

V

i

V

+

–

R

Z

C

Z

(4)

6.002 Fall 2000 Lecture

18 Quick Review of

Impedances-Just as

2

1 ab

ab

AB

R

I

V

R

=

+

L

j

R

I

V

Z

₁

ab

AB

=

+

ω

1 R

ab

I

+

–

ab

V

2 R

A

B

1 R

ab

I

+

–

ab

V

L

j

ω

A

B

(5)

6.002 Fall 2000 Lecture

18 Quick Review of Impedances

Similarly

L

2 C

1 AB

R

Z

||

R

Z

=

+

L

2 C

1 Z

R

Z

R

Z

R

+

=

L

j

CR

j

1 R

R

2

1 +

₊

_ω

+

ω

=

1 R

L

A

B

2 R

C

(6)

6.002 Fall 2000 Lecture

18 We can build other filters by

combining impedances

( )

ω

Z

L

R

C

_ω

Z

(7)

6.002 Fall 2000 Lecture

18 We can build other filters by

combining impedances

HPF

High Pass Filter

ω

( )

ω

H

ω

( )

ω

H

LPF

Low Pass Filter

ω

( )

ω

H

HPF

( )

ω

Z

L

R

C

ω

Z

+

–

+

–

+

–

(8)

6.002 Fall 2000 Lecture

18 Check out:

R

C

j

1 L

j

R

V

i

r

+

=

ω

RC

j

LC

1 RC

j

2 ω

ω

+

−

=

(

₂

)

2 (

)

2 i

r

RC

LC

1 RC

V

ω

+

−

=

ω

+

–

L

C

R

+

–

r

V

i

V

LC

1 o

=

ω

At resonance,

ω

=

ω

_o

and

Z

_L

+

Z

_C

= 0

,

so

V

_i

sees

only

R

!

More later…

Intuitively:

i

r

V

1 L blocks high

_freq

C blo

cks l

(9)

6.002 Fall 2000 Lecture

18 What about:

+

–

L

_C

R

+

V

lc

–

i

V

Band Stop Filter

i

lc

V

ω

1 _{C open}

L open

(10)

6.002 Fall 2000 Lecture

18 Another example:

+

–

+

–

L

R

i

V

C

V

o

i

o

V

o

ω

BPF

C sho

_rt

L sho

rt

(11)

6.002 Fall 2000 Lecture

18 AM Radio Receiver

crystal radio demo

Thévenin

antenna

model

+

–

L

R

i

V

C

demodulator

amplifier

antenna

(12)

6.002 Fall 2000 Lecture

18 AM Receiver

“Selectivity” important —

relates to a parameter “

Q

” for the filter. Next…

+

–

L

R

i

V

C

demodulator

amplifier

f

signal

strength

540 …1000 1010 1020 1030 … 1600 KHz

10 KHz

filter

_WBZ

News

Radio

(13)

6.002 Fall 2000 Lecture

18 Recall,

Selectivity:

Look at series RLC in more detail

+

–

L

C

R

+

–

r

V

i

V

C

j

1 L

j

R

V

i

r

ω

+

=

i

r

V

ω

o

ω

2

1 higher

Q

1 Define

quality factor

Δ

=

Q

o

ω

Δ

bandwidth

⇒

Q

(14)

6.002 Fall 2000 Lecture

18 ω

ω

Δ

=

o

Q

LC

1 o

=

ω

Quality Factor Q

⎟

⎠

⎞

⎜

⎝

⎛

₋

+

=

+

=

CR

1 R

L

j

1

1 C

j

1 L

j

R

i

V

r

V

ω

?

ω

Δ

at

ω

_ο

=0

ω

_ο

:

(15)

6.002 Fall 2000 Lecture

18 Note that abs magnitude is

1 ₂

when

₁

1 _j

₁

CR

1 R

L

j

1

1 V

V

i

r

±

=

⎟

⎠

⎞

⎜

⎝

⎛

₋

+

=

ω

i.e. when

1 CR

1 R

L

₋

₌

_±

ω

0 LC

1 L

R

2 ω

₋

₌

ω

∓

:

ω

Δ

ω

Δ

=

o

Q

Quality Factor Q

Looking at the roots of both equations,

LC

4 L

R

2

1 L

2 R

2

1 =

+

ω

LC

4 L

R

2

1 L

2 R

2

2 =

−

+

ω

L

R

=

−

=

Δ

ω

₁

ω

₂

(16)

6.002 Fall 2000 Lecture

18 R

L

R

Q

=

ω

o

=

ω

o

The lower the

R

(for series

R

),

the sharper the peak

ω

Δ

=

o

Q

Quality Factor Q

LC

1 o

=

ω

(17)

6.002 Fall 2000 Lecture

18 Another way of looking at

Q

:

cycle

per

lost

energy

stored

energy

2 π

=

Q

0

2 r

2 R

I

2

1 I

L

2

1

2 ω

π

=

R

L

Q

=

ω

o

Quality Factor Q