Lecture 11 - University of California, Berkeleyee105/fa07/lectures/Lecture...Lecture 11...

Lecture 11ANNOUNCEMENTS

• Prob. 5 of Pre‐Lab #5 has been clarified. (Download new version.)Prob. 5 of Pre Lab #5 has been clarified. (Download new version.)• HW#6 has been posted.• Review session: 3‐5PM Friday (10/5) in 306 Soda (HP Auditorium)• Midterm #1 (Thursday 10/11):

• Material of Lectures 1‐10 (HW# 2‐6; Chapters 2, 4, 5)• 2 pgs of notes (double‐sided, 8.5”×11”), calculator allowed

OUTLINE

f l f

2 pgs of notes (double sided, 8.5 ×11 ), calculator allowed

• Review of BJT Amplifiers

• Cascode Stage

EE105 Fall 2007 Lecture 11, Slide 1 Prof. Liu, UC Berkeley

Reading: Chapter 9.1

Review: BJT Amplifier Design• A BJT amplifier circuit should be designed to

1. ensure that the BJT operates in the active mode, p ,2. allow the desired level of DC current to flow, and3. couple to a small‐signal input source and to an output “load”.

Proper “DC biasing” is required!(DC analysis using large‐signal BJT model)

• Key amplifier parameters: (AC analysis using small‐signal BJT model)

Voltage gain A ≡ v /v– Voltage gain Av ≡ vout/vin– Input resistance Rin ≡ resistance seen between the input

node and ground (with output terminal floating)O t t i t R i t b t th t t


– Output resistance Rout ≡ resistance seen between the output node and ground (with input terminal grounded)

Large‐Signal vs. Small‐Signal Models• The large‐signal model is used to determine the DC operating point (VBE, VCE, IB, IC) of the BJT.p g p ( BE, CE, B, C)

• The small‐signal model is used to determine how the output responds to an input signal.


Small‐Signal Models for Independent Sources

• The voltage across an independent voltage source does not vary with time. (Its small‐signal voltage is always zero.)y ( g g y )

It is regarded as a short circuit for small‐signal analysis.

Large-Signal Small-SignalLarge-SignalModel

Small-SignalModel

• The current through an independent current source does not vary with time. (Its small‐signal current is always zero.)

It i d d i it f ll i l l i


It is regarded as an open circuit for small‐signal analysis.

Comparison of Amplifier TopologiesCommon Emitter

L A 0

Common Base

L A 0

Emitter Follower

0 A 1• Large Av < 0‐ Degraded by RE‐ Degraded by RB/(β+1)

• Large Av > 0‐Degraded by RE and RS‐ Degraded by RB/(β+1)

• 0 < Av ≤ 1‐ Degraded by RB/(β+1)

• Large Ri• Moderate Rin

‐ Increased by RB‐ Increased by RE(β+1)

• Small Rin‐ Increased by RB/(β+1)‐ Decreased by RE

Large Rin(due to RE(β+1))

• Small Rout y E(β )

• Rout ≅ RC

• ro degrades A , Ro t

y E

• Rout ≅ RC

• ro degrades A , Ro t

‐ Effect of source impedance is reduced by β+1D d b Rro degrades Av, Rout

but impedance seenlooking into the collector can be “boosted” by

ro degrades Av, Routbut impedance seenlooking into the collector can be “boosted” by

‐ Decreased by RE

• ro decreases Av, Rin, and Rout


yemitter degeneration

yemitter degeneration

and Rout

Common Emitter Stage

Rv

∞=AV ∞<AV

BE

C

in

out

RRg

Rvv

+++

−≈

11

β

EBin

m

RRRrRR

g+++=+)1(1

ββ

π

[ ]( ))||(1|| RRR +EE105 Fall 2007 Lecture 11, Slide 6 Prof. Liu, UC Berkeley

Cout RR = [ ]( ))||(1|| πrRgrRR EmOCout +≈

Common Base Stage

∞=AV

ES

E

BES

C

in

out

RRR

RRR

Rvv

+⋅

++≈ 1

ESmg +1β

BR ⎞⎜⎛ 1 VE

B

min RR

gR ⎟

⎠

⎞⎜⎜⎝

⎛+

+≈1

1β

RR [ ]( ))||(1|| RRR +

∞<AV


Cout RR = [ ]( ))||(1|| πrRgrRR EmOCout +≈

Emitter Follower

∞=AV ∞<AV

1= Eout

RRv OEout

RrRv

1||

=

11

+++β

S

mE

inR

gRv

RR )1( β ( )( )

S

mOE

in

RR

Rg

rRv

||11

1||+

++

ββ

Ein RrR )1( βπ ++=

s RRR ||1 ⎞⎜⎜⎛

+=

( )( )s

OEin

rRRR

rRrR

||||1

||1

⎞⎜⎜⎛

+=

++= βπ


Em

out Rg

R ||1⎠⎜⎜

⎝ ++=β OE

mout rR

gR ||||

1 ⎠⎜⎜⎝

++

=β

Ideal Current Source

Equivalent CircuitCircuit Symbol I-V Characteristic

• An ideal current source has infinite output impedance• An ideal current source has infinite output impedance.

How can we increase the output impedance of a BJT that d ?


is used as a current source?

Boosting the Output Impedance• Recall that emitter degeneration boosts the impedance seen looking into the collector.g– This improves the gain of the CE or CB amplifier. However, headroom is reduced.

( )[ ] rRrrRgR ||||1 ++=


( )[ ] ππ rRrrRgR EOEmout ||||1 ++=

Cascode Stage• In order to relax the trade‐off between output impedance and voltage headroom we can use aimpedance and voltage headroom, we can use a transistor instead of a degeneration resistor:

||)]||(1[ rrrrrgR ++=( )1211

12112

||||)]||(1[

π

ππ

rrrgRrrrrrgR

OOmout

OOOmout

≈++=

• V for Q can be as low as ~0 4V (“soft saturation”)

1 if 1212 >>≅= βCEC III


• VCE for Q2 can be as low as ~0.4V (“soft saturation”)

Maximum Cascode Output Impedance• The maximum output impedance of a cascode is limited by rlimited by rπ1.

rrrgR β≈

: If 12 πrrO >>

1111max, 1 OOmout rrrgR βπ=≈


PNP Cascode Stage

( )121121

||||)]||(1[ ππ

RrrrrrgR OOOmout ++=


( )1211 || πrrrgR OOmout ≈

False Cascodes• When the emitter of Q1 is connected to the emitter of Q it’s not a cascode since Q is a diode‐connectedQ2, it s not a cascode since Q2 is a diode connected device instead of a current source.

11 ⎤⎡ ⎞⎜⎛

1

122

1122

1

1

||||1||||11 Om

OOm

mout

g

rrg

rrrg

gR

⎞⎜⎛

+⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎜⎝

⎛+= ππ


12

12

1 211 Om

Om

mout r

gr

ggR ≈+⎟

⎠

⎞⎜⎜⎝

⎛+≈

Short‐Circuit Transconductance

• The short‐circuit transconductance of a circuit is a measure of its strength in converting an inputmeasure of its strength in converting an input voltage signal into an output current signal.

i

0=

≡outvin

outm v

iG


Voltage Gain of a Linear Circuit• By representing a linear circuit with its Norton equivalent the relationship between V and V canequivalent, the relationship between Vout and Vin can be expressed by the product of Gm and Rout.

Norton Equivalent Circuit

RvGRiv ==

outminout

outinmoutoutout

RGvvRvGRiv

−=−=−=Computation of

short-circuit output current:


Example: Determination of Voltage Gain Determination of Gm Determination of Rout

1out giG =≡ 1

x rvR =≡

rgA =

10

mvin

m gv

Gout

≡=

1ox

out ri

R ≡


11 Omv rgA −=

Comparison of CE and Cascode Stages• Since the output impedance of the cascode is higher than that of a CE stage its voltage gain is also higherthan that of a CE stage, its voltage gain is also higher.

11 Oinmout rvgv −=

AOmv

VrgA −=−= 11 ( )21221 rrgrgA OO−≈


TOmv V

g 11 ( )21221 πrrgrgA OmOmv

Voltage Gain of Cascode Amplifier• Since rO is much larger than 1/gm, most of IC,Q1 flows into diode‐connected Q2. Using Rout as before, AV is Q2 g out , V easily calculated.

11 it gGvgi =⇒= 11 mminmout gGvgi ⇒

outmv RGA −=( )( )[ ]

( )[ ]22121

2122121

||||||1

OOmm

OOOmm

rrrggrrrrrgg

π

ππ

−≈++−=


( )

Practical Cascode Stage• No current source is ideal; the output impedance is finite.

)||(|| rrrgrR ≈


)||(|| 21223 πrrrgrR OOmOout ≈

Improved Cascode Stage• In order to preserve the high output impedance, a cascode PNP current source is usedcascode PNP current source is used.

[ ] [ ][ ] [ ]outmv

OOmOOmout

RgArrrgrrrgR

1

21223433 )||(||)||(−=≈ ππ


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