radio frequency electronics - university of...

A. Kruger 1 Radio Frequency Electronics The University of Iowa

Radio Frequency Electronics

Image from Wikipedia

• Born in 1791 in Massachusetts

• Fairly accomplished painter

• After witnessing various electrical experiments,

got intrigued by electricity

• Designed the first single-wire telegraph

• Invented the concept “relay” what we now call

repeaters

• Created Morse Code (digital communications?)

• Held several patents related to the telegraph

• Dies in 1872

Active Components IV

Samuel Morse


CE Amplifier (CS is Similar)

Inverting amplifier

High-gain because of CE

)]([2

1

eqSB

HCRRr

f

Use time constant technique:

p

Hf2

1

eqC


SPICE Results for Common Emitter


Common Base Amplifier

Notice where the input signal goes


Common Base Amplifier

Remember, it is 𝑣𝐵𝐸 that controls collector current in BJTs

𝐼𝐶 = 𝐼𝑆𝑒𝑉𝐵𝐸𝑉𝑇

Common Emitter Common Collector Common Base

𝑣𝑏𝑒 𝑣𝑏𝑒 𝑣𝑏𝑒


Designing a Common-Base Amplifer

Design a CE amplifier Ground the base

𝑅1, 𝑅2, 𝐼𝐶 , 𝑅𝐸 , 𝑅𝐶 are determined to

achieve a desired operating point

𝐼𝐶

𝐴𝑣 = −𝑔𝑚𝑅𝐶

𝐴𝑣 = − 40𝐼𝐶 𝑅𝐶 = ? ?

𝑅𝑖

𝑅𝑖 = 𝑟𝜋 =𝛽

𝑔𝑚=

100

40𝐼𝐶= ? ?

Feed signal into emitter

𝐴𝑣 = Same as for CE

𝑅𝑖 =𝑟𝜋𝛽= =

1

𝑔𝑚= ? ?

𝛽 = 100 𝛽 = 100

𝑅𝑖

𝐼𝐶


CB Amplifier

This is not an inverting amplifier

Thus, Miller no multiplication effect.


CB Amplifier (CG is Similar)

These are NOT inverting

amplifiers.

Thus, Miller no multiplication

effect.


CB Amplifier

CRR

rf

SE

H

12

1

CRR

fLC

H)(2

1

Equivalent input circuit

Equivalent output circuit

Either one could

determine bandwidth

(normally Cμ)

Regardless, higher bandwidth than CE


Cascode Circuit

1 𝑔𝑚

Cascode amplifiers are composite amplifiers where

a CE amplifier feeds a CB amplifier.

One can view the CB amplifier as the load of the CE

amplifier.

The CB has input impedance 1 𝑔𝑚 .

If we were to “glue” the CB to the CE, the CE

would see a load of 𝑅𝐶 = 1 𝑔𝑚 .


Cascode Circuit

1 𝑔𝑚

Cascode amplifiers are composite amplifiers where

a CE amplifier feeds a CB amplifier.

One can view the CB amplifier as the load of the CE

amplifier.

The CB has input impedance 1 𝑔𝑚 .

If we were to “glue” the CB to the CE, the CE

would see a load of 𝑅𝐶 = 1 𝑔𝑚 . 1 𝑔𝑚

The gain of the CE would be −𝑔𝑚𝑅𝑐 = −1

This means small Miller Effect

The gain of the CB stage is 𝑔𝑚𝑅𝑐 and there is no

Miller effect

𝑅𝐶


BJT Cascode Amplifier

36 dB Voltage Gain at 6 MHz

12 dB Voltage Gain at 70 MHz

2N2222 BJTs Common Base

Common

Emitter


BJT Cascode Circuit

CE is an inverting amplifier => Miller effect present

CE voltage gain ~ 1 => low Miller effect


BJT Cascode Circuit

)(2

1

1111 MBS

HCCrRR

f

2)(2

1

CRR

fLC

H

Either one could

determine bandwidth

(normally Cμ) Wide bandwidth


SPICE Results for Cascode


FET Cascode

Common Gate

Common Source


FET Cascode

1 𝑔𝑚


Dual Gate FETs

The cascode amplifier is very popular in RF and the “stacked” configuration has other

useful applications.

Consequently, semiconductor companies make dual-gate FETs where two gates

“squeeze” the same channel

Cascode amplifier with 2 separate FETs

Cascode amplifier with dual gate FET


Dual Gate FETs

𝑫 𝑫

𝑺 𝑺

𝑮𝟏

𝑮𝟐

𝑮𝟏

𝑮𝟐

Construction Schematic Symbols

n-Channel p-Channel

𝑮𝟏

𝑮𝟐

𝑫

𝑺


Dual Gate FETs

Note the two gates

These form back-to-back Zeners that

protect the FET against damage from

static electricity


Emitter-Follower Circuit (Source-Follower is Similar)


'

'

1)1(2

1

Lm

LmBS

H

Rg

CCrRgRR

f

'

'

1)1(

Lm

LmBSpRg

CCrRgRR

Wide bandwidth

'

''

'

11

1

1L

LL

b R

gmR

sC

gmRr

Z


SPICE Results for Emitter Follower


Single-Tuned Amplifier

Tuned amplifier using a depletion-mode MOSFET

Circuit for bias calculations

The equivalent ac circuit The small-signal model


𝑣𝑜 𝑠 − 𝑣𝑖(𝑠 )𝑠𝐶𝐺𝐷 + 𝑔𝑚𝑣 + 𝑣𝑜 𝑠1

𝑟𝑜+

1

𝑠𝐿+ 𝑠𝐶 +

1

𝑅𝐷 𝑅3 = 0

The small-signal model

Let 𝑅𝑝 = 1 𝐺𝑝 where 𝐺𝑝 = 1 𝑟𝑜 + 1 𝑅𝐷 + 1 𝑅3 (i.e., the parallel combination of the resistances

au the output. Solving for the voltage transfer function 𝑣𝑜(𝑠) 𝑣𝑖(𝑠) yeilds

𝐴𝑣 𝑠 = 𝑣𝑜 𝑠

𝑣𝑖(𝑠)= 𝑠𝐶𝐺𝐷 − 𝑔𝑚 𝑅𝑝

𝑠𝑅𝑝 𝐶 + 𝐶𝐺𝐷

𝑠2 +𝑠

𝑅𝑝 𝐶 + 𝐶𝐺𝐷+

1𝐿 𝐶 + 𝐶𝐺𝐷

𝐴𝑣 𝑠 ≅ 𝐴𝑚𝑖𝑑

𝑠𝜔𝑄

𝑠2 + 𝑠𝜔𝑄+ 𝜔0

2, 𝜔0 =

1

𝐿(𝐶 + 𝐶𝐺𝐷), 𝑄 = 𝜔0𝑅𝑝 𝐶 + 𝐶𝐺𝐷 , 𝐴𝑚𝑖𝑑 = −𝑔𝑚𝑅𝑃

Neglecting the right-half-plane zero ( “𝑠𝐶𝐺𝐷” in 𝑠𝐶𝐺𝐷 − 𝑔𝑚 ) then

Further, 𝑄 = 𝑅𝑝 𝜔𝐿 and 𝐵𝑊 = 𝜔𝑜 𝑄

KCL @ output


𝐴𝑣 𝑠 ≅ 𝐴𝑚𝑖𝑑

𝑠𝜔𝑄

𝑠2 + 𝑠𝜔𝑄+ 𝜔0

2,

𝜔0 =1

𝐿(𝐶 + 𝐶𝐺𝐷,

𝑄 = 𝜔0𝑅𝑝 𝐶 + 𝐶𝐺𝐷

𝐴𝑚𝑖𝑑 = −𝑔𝑚𝑅𝑃

𝐶𝐺𝐷

𝜔0 = 1 𝐿 𝐶 + 𝐶𝐺𝐷 = 1 10 × 10−6 120 × 10−12 = 28.6 × 106 rad s

𝑓0 = 𝜔0 2𝜋 = 4.59 MHz

𝑟𝑜 = 1 𝜆𝐼𝐷 = 1 0.02 3.2 × 10−3 = 15.6K

𝐶 + 𝐶𝐺𝐷 = 100 + 20 = 120 pF 𝑅𝑝 = 𝑟𝑜 100K 100K=11.9K

𝑔𝑚 = 2 𝐾𝑛𝐼𝐷 = 2 2.5 × 10−3 3.2 × 10−3 = 5.66 mA V2

𝐴𝑚𝑖𝑑 = −𝑔𝑚𝑅𝑝 = − 5.66 × 10−3 11.9 × 103 = −67

𝑄 = 𝜔0𝑅𝑝 𝐶 + 𝐶𝐺𝐷 = − 28.6 × 106 11.9 × 103 120 × 10−12 = 40.8

𝐵𝑊 = 𝑓0 𝑄 = 4.59 × 106 40.8 = 112 kHz

Assume 𝜆 = 0.02 𝑉−1, 𝐼𝐷 = 3.2 mA, 𝐶𝐺𝐷 = 20 pF

radio frequency electronics - university of...

Documents