ECE 342 – Jose Schutt-Aine 1 ECE 342 Solid-State Devices & Circuits 16. Active Loads Jose E....

ECE 342 – Jose Schutt-Aine 1

ECE 342Solid-State Devices & Circuits

16. Active Loads

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]


Ideal MOS Common Source CKT

, ,i vo m ds o dsR A g r R r

m dsIntrinsic gain is g r


PMOS Implementation of Active Load

1 1 1ds oLet r r for Q

2 2 2ds or r for Q

1 2, ||out o othen R r r

1MB m outA g R

1 11

1o ds

o

Let g gr

2 22

1o ds

o

g gr


'1

1

42 2 200 100 0.63 /

0.4m n REF

Wg k I mA V

L

Example

Assume VDD=3 V, Vtn = |Vtp| = 0.6 V, mnCox = 200 mA/V2, mpCox=65 mA/V2, L = 0.4 mm, W = 4 mm, VAn = 20 V, |VAP| = 10 V, IREF= 100 mA. Find small-signal gain.

11

20200

0.1An

oD

V Vr k

I mA

22

10100

0.1Ap

oD

V Vr k

I mA

1 1 2|| 0.63 200 ||100 42v m o oA g r r


IC BJT CE

, ,i vo m o out oR r A g r R r


1 2 1 2|| ||out o o ce ceR r r r r

1 2 1 2out s sC C C C C

Cs1 & Cs2 are the collector-to-substrate capacitances of Q1 and Q2 respectively

IC BJT Common Emitter


IC Common Emitter – High Frequency Model

• High Frequency Calculations 1. Upper corner frequencies more difficult to

evaluate than for discrete amp2. Miller effect will be larger corner

frequency lower


1

2in highin eq

fC R

Where CM1 is the Miller capacitance associated with Cm1

IC-CE: High-Frequency Analysis

1

2out highout out

fC R

1 1in MC C C

The total input capacitance in parallel with rp1 is

The input and output corner frequencies are


ExampleThe CE circuit (see next page) is biased so that the collector currents of Q1 and Q2 are 1.14 mA. The parameters for Q1 are: b=160, rx1= 10 W, rce1= 68 kW, Cp1 = 20 pF, and Cm1 = 2.1 pF. For device Q2, the parameters are rce2= 21 kW and Cm2 = 3.1 pF. Each device has a value of Ccs1 = Ccs2 = 2.5 pF. In this circuit, the power supply is 10 V and R1 = 10 kW. Find the midband gain and the upper corner frequency.

1 2 1 2|| || 68 || 21 16out o o ce ceR r r r r k

Evaluate rp1 and Rout

1 1

261 161 3672

1.14er r


The midband gain is:

1 1

outMB

x

RA

r r

160 16,000695 /

10 3672MBA V V

1 1 1 1

1 1

2 2in highin eq M x

fC R C C r r

The corner frequency of the input circuit is

Example (cont’)


Example (cont’)

The corner frequency of the output circuit is

1

2out highout out

fC R

12

1

2 2.1 3.1 2.5 2.5 10 16,000out highf

975out highf kHz

12

1 110.7

2 2 1481 10 10in highin eq

f MHzC R

For overall corner frequency, use SPICE


Source Follower

' 1|| ||L L o

mb

R R rg

'o m gs Lv g v R

Source Follower

gs i ov v v

'

'1o m L

vi m L

v g RA

v g R

1m o

vom mb o

g rA

g g r

1

1m

vom mb

gA

g g


Source Follower – Output Resistance

1||o o

m mb

R rg g

1/ 1o mR g


Source Follower with Active Load

• Characteristics – Provides a buffer stage– M1 is amplifying stage– M2 is active load


Source Follower – Incremental Model

1

1 1 1 2

mMB

m mb ds ds

gA

g g g g


(current mirror)

Emitter Follower with Active Load

Emitter follower can be used to drive a low-impedance load


1 1 2

1 1 1 1 2

(1 )

(1 )

m ce

MBs x m ce

g r rA

R r r g r r

Midband gain:




• AC Properties1. Gain is less than 1 and near 1 for typical

element values2. Frequency response has one zero and two

poles3. Exact frequency response is difficult Use

SPICE4. Output stage of NPN current mirror serves

as high impedance load at emitter of Q1

IC - Common Gate Amplifier

Substrate is not connected to the source must account for body effect

Drain signal current becomes

D m gs mb bsi g v g v

And since gs bsv v

Body effect is fully accounted for by using m m mbg g g


i m mb i roi g g v i

m mb ioi o i i L

ro io o L

o

1g g v

rv v v i Ri i

r r R1

r

with i sv v



i o L

ini m mb o

v r RR

i 1 g g r

As o inm mb

1r , R

g g

If L iR , R

o ro i m mb o i iv i v g g r v v

vo m mb oA 1 g g r



whereo L Lin o m mb o

vo m mb o

r R R1R , A g g r

A g g A

Taking ro into account adds a component (RL/Ao) to the input resistance.

vo m mb oA 1 g g r

The open-circuit voltage gain is:

The voltage gain of the loaded CG amplifier is:

Lv vo

L o vo s

RG A

R r A R



x x m mb ov i g g v r v with x sv i R

orout o m mb o s out o vo sR r 1 g g r R R r A R

CG Output Resistance

o Lin

o L

e e

r RR

r R1

r 1 r

vo m oA 1 g r 'out o m o eR r 1 g r R

'e eR R || re

rr

1

CB Amplifier


11

1o

o

Define gr

Common source amplifier, followed by common gate stage – G2 is an incremental ground

MOS Cascode Amplifier

22

1o

o

gr

LR current source impedance

1L

L

GR


• CS cacaded with CG CascodeVery popular configurationOften considered as a single stage

amplifier• Combine high input impedance and

large transconductance in CS with current buffering and superior high frequency response of CG

• Can be used to achieve equal gain but wider bandwidth than CS

• Can be used to achieve higher gain with same GBW as CS

MOS Cascode Amplifier


MOS Cascode Incremental Model

Lo

L

iv

G

2 2 2 2 2 2L mb s m s s o oi g v g v v v g

2 2 2 2 2 2L

L s mb m s o oL

ii v g g v g g

G


22 2 2 21 o

L s mb m oL

gi v g g g

G

22

2 2 2

1 /L o Ls

mb m o

i g Gv

g g g

KCL at vs2

1 2 1 2 2 2 2 2 2( ) 0m gs s o m s mb s o s og v v g g v g v g v v

MOS Cascode Analysis


1 21 1

2 2 2

1 /1 0o o L

m gs Lo m mb

g g Gg v i

g g g

1 1

1 2

2 2 2

1 /1

m gsL

o o L

o m mb

g vi

g g G

g g g

1

1 2

2 2 2

o m

in o L oL

o m mb

v g

v g G gG

g g g

Two cases



1 2 2 2 1 2 om o m mb o o

in

vg g g g r r

v

1 2 1 2 1 2 1 2 MB m o m m m mb o oA g g g g g g r r

1 1 2 2MB m o m oA g r g r

CASE 1

The voltage gain becomes

1: 0LCase If G



1 2

2 2 2

2 : o L oL

o m mb

g G gCase If G

g g g

1 o m

MBin L

v gA

v G

CASE 2

The voltage gain becomes



Cascode Example


Cascode Example

The cascode circuit has a dc drain current of 50 mA for all transistors supplied by current mirror M3. Parameters are gm1=181 mA/V, gm2=195 mA/V, gds1= 5.87 mA/V, gmb2=57.1 mA/V, gds2= 0.939 mA, gds3= 3.76 mA/V, Cdb2 = 9.8 fF, Cgd2 = 1.5 fF, Cdb3=40.9 fF, Cgd3= 4.5 fF. Find midband gain and approximate upper corner frequency

1 2

2 2 2

5.87 3.76 0.9390.109 /

0.939 195 57.1o L o

Lo m mb

g G gA V G

g g g

3 3.76 / L dsG g A V

Therefore, we use Case 2 to compute the gain

The internal conductance of the current source is:


Cascode Example

1 1

3

18148.1 /

3.76 m m

MBL ds

g gA V V

G g

Gain can be approximated by

Upper corner frequency is approximated by

2 15 33 2

1 110.6

2 2 56.7 10 266 10ods

f MHzr C


Common emitter amplifier, followed by common base stage – Base of Q2 is an incremental ground

BJT Cascode Amplifier


BJT Cascode Incremental Model

2 2o m Lv g R v

2 22 2

2 2 2

1x xx

x

v r rv v v

r r r


BJT Cascode Analysis

1 1 2 22 2

xm m

x

vg v g v

r r

Ignoring rx2

21 1 2 2

2m m

vg v g v

r

1 1 2 22

1m mg v v g

r



11

1 1in

s x

rv v

R r r

1 12

1 12

2

1

1m in

s xm

g r vv

R r rg

r

2 1 1

1 12

2

1

1m L m in

os x

m

g R g r vv

R r rg

r



2 2 1 1

1 1 2 21o m m L

in s x m

v g r g r R

v R r r g r

1 2

1 1 2 1o L

in s x

v R

v R r r

1 2v m LA g R

If Rs << rx1+rp1, the voltage gain can be approximated by

ECE 342 – Jose Schutt-Aine 1 ECE 342 Solid-State Devices & Circuits 16. Active Loads Jose E....

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Transcript of ECE 342 – Jose Schutt-Aine 1 ECE 342 Solid-State Devices & Circuits 16. Active Loads Jose E....