Problem for Problems for Discrete Transistor...

Problem for Problems for Discrete Transistor Amplifiers

ECE 65, Winter2013, F. Najmabadi

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (2/42)

Exercise 1: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).

This is a common collector amplifier (emitter follower) . o Input at the base, output at the emitter.

It has a emitter-degeneration bias with a voltage divider.

When Rsig & vsig are not given, it implies that vsig = vi and Rsig = 0

Bias circuit Thevenin form of the Voltage divider

V 95.49k 18k 22

k 90.9k 18||k 22

Real circuit

Exercise 1 (cont’d): β = 200, VA = 150 V.

Bias Calculations

A 3.20/ mA 05.4

107.0)1/(109.9 4.95

4.95 :KVLBE33

==≈=

+++×=

++=−

EEBEBB

IIRIVRI

V 7.0 and 0 V, 7.0: Activein is BJT Assume

≥>= CECBE VIV

V 0.7 V 5 10104 9

9 :KVLCE

=>=××+=

+=−−

V 95.4k 9.9 == BBB VR

k 28.1 r

k 0.3710x 4

mA/V 156 1026104.05

gIVIVVIg

Thevenin form of the Voltage divider

15115196510156)||||(

965k100||k1||k8.38)||||()||||(1

)||||(

=××=

RRrgRRr

RRrgRRrg

k 28.1 k, 8.38 mA/V, 156 === πom rrg

[ ][ ] k 9.9k194k3.1 ||k9.9

)||||( || ≈+=

RRRrrRR βπ

= 4.6||||

||ββππ rR

RRrRR E

sigBEo

Signal circuit (IVC = 0)

Amplifier Parameters Emitter Follower

1==×+

4.6k9.9

k 28.1k 8.38mA/V 156

Hz 39.31047.0)10100(6.4 2

Hz 2.341047.0)0109.9( 2

=×××+

=××+×

csigip

Hz 6.374.32.34 21 =+=+≈ ppp fff

Amplifier Cut-off frequency 2 caps: 2 poles

This is a common emitter amplifier with RE . o Input at the base, output at the collector.

It has a emitter-degeneration bias with a voltage divider.

Bias calculations Thevenin form of the Voltage divider

A 2.14/ mA 84.2

5107.0)1/(100.5 2.22

2.22 :KVLBE3

==≈=

+++×=

++=−

EEBEBB

IIRIVRI

≥>= CECBE VIV

V 0.7 V 5.10 )51010(1084.215

15 :KVLCE

=>=+××+=

++=−−

EECECC

k 83.1 r

k 8.5210 84.2

mA/V 10.9 1026102.84

gIVIVVIg

Exercise 2 (cont’d): β = 200, VA = 150 V

Bias circuit Caps open

V 22.215k 9.5k 34

k 0.5k 9.5||k 34

k 83.1 k, 8.52 mA/V, 10.9

Amplifier Parameters CE amplifier with RE

k 8.4104k ||k 0.5)/1](/)||[(1

RrRrRR

RrRRπ

0.990kk 100||k 1||)/1](/)||[(1

EoLCEm

RRrRrRRRg

||1 ||

+++≈

RRRrRrRR

59.1)64.1(100800,4

800,4−=−×

k 0.1 k 8.4 == oi RR

Amplifier Cut-off frequency (2 caps: 2 poles)

Hz 8.1510100)1010010( 2

Hz 91.6107.4)100108.4( 2

=×××+

=××+×

csigip

Hz 7.228.159.6 21 =+=+≈ ppp fff

k 83.1r k 8.52rmA/V 10.9

This is a PNP common emitter amplifier (no RE ). o Input at the base, output at the collector (RE is shorted

out for signal because of the by-pass capacitor)

It has a emitter-degeneration bias with two voltage sources (RB = ∞)

Exercise 3 (cont’d). β = 200, VA = 150 V.

Bias circuit (Signal = 0, Caps open)

A 0.5/ mA 0.1

100103.2 3 :KVLBE 3

µβ ==≈=

++×=−

≥>= ECCEB VIV

V 0.7 V 4.1 10106.46

3103.2103.23 :KVLCE

=>=××−=

−×++×=−−

k 26.5 r

k 15010150r

mA/V 38.5 1026

gIVIVVIg

3.85)k 100||k 3.2||k 150(1038.5

)||||(

3 −=×−=

RRrgvv

k 26.5 k, 150 mA/V, 38.5 === πom rrg

Amplifier Parameters 1) CE amplifier with no RE 2) No voltage divider biasing: no RB

RB is open circuit, RB = ∞

k 27.2k 3.2||k 150 ||

k 26.5 || === ππ rrRR Bi

3.85−=≈×+

isig RR <<

k 26.5r k 150r

mA/V 38.5

Hz 1451298.150321 =++=++≈ pppp ffff

k 27.2 k 26.5

Amplifier Cut-off frequency 1) 2 caps: 2 poles 2) no coupling capacitors at the input

csigip CRR

Hz 1291047)5.26||10(2.3 2

]/)||(/1[|| 21

=×××

EsigBmEp

CRRgRf

Hz 8.1510100)1010(2.3 2

=××+×

This is a MOS common source amplifier (no RS ). o Input at the gate, output at the drain (RS is shorted

out for the signal because of the by-pass capacitor)

It has a source-degeneration bias with a voltage divider.

Exercise 4: Find the amplifier parameters of this circuit. (µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1. Assume capacitors are large and ignore channel width modulation in biasing. )

V 353.18.110033

100V 0 G =+

=⇒=kk

Assume Saturation

232 10x3 5.0 OVOVoxnD VVL

WCI −== µ

V 803.0=+= tOVGS VVV

Saturation ⇒> OVDS VV

mA 275.010x3 23 == −OVD VI

V 550.0=−= GSGS VVV

V 249.18.1 =−= DDD IRV

V 699.0=−= SDDS VVV

V 303.0=OVV

DStOVDSGSG IRVVIRVV ++=+=

233 103 102 5.0353.1 OVOV VV −×××++=

0853.06 2 =−+ OVOV VV

GS-KVL:

Exercise 4 (cont’d): µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1.

k 3.3610 752.01.0

mA/V 1.82 0.303

10 0.275 2 2

32.3)k 50||k 2||k 3.36(101.82

)||||(

3 −=×−=

RRrgvv

Amplifier Parameters This is a CS amplifier with no RS

k 90.1k 2||k 3.36 ||

k 8.24== Gi RR

k 8.24k33||k100 ==GR

Exercise 4 (cont’d): µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1.

19.3k 1k 8.24

k 8.24−=×

Exercise 4 (cont’d):

Hz 1684.757.307.61321 =++=++≈ pppp ffff

k 9.1 k 8.24

Amplifier Cut-off frequency 1) 3 caps: 3 poles

Hz 7.6110100)1010(24.8 2

=××+×

csigip

Hz 4.75107.4]579||10[2 2

1)]/()||[(|| 2

=×××

SomLDoSp

CrgRRrRf

Hz 7.3010100)105010(1.9 2

=×××+×

k 3.36mA/V 1.82

This is a common collector amplifier (emitter follower). o Input at the base, output at the emitter.

It is biased with a current source! o No emitter degeneration: no RE (RE = ∞) and no RB (RB = ∞)

V 72.0100 3

−=++=

≥>= CECBE VIV

k 21.1 r

k 9.34104.3

mA/V 165 1026104.3

gIVIVVIg

A 5.21/mA 3.4

µβ ==≈=

V 0.7 V 72.44 0 =>=−= DECE VVV

k 21.1 k, 9.34 mA/V, 165 === πom rrg

1104.281

104.28

104.28)||||(

k9.25k100||||k9.34||||)||||(1

)||||(

≈×+

=∞=+

RRrgRRr

RRrgRRrg

≈ 111||

π sigBEo

[ ]M 2.5k9.25201k2.1

)||||)(1( || =×+=

RRRrrRR βπ

Signal circuit (IVC short, ICS open))

Amplifier Parameters 1) Emitter Follower 2) Biased with an ICS (RE = ∞ ) 3) No voltage divider biasing:

RB is open circuit, RB = ∞

1=≈×+

Signal circuit

k 21.1rk 9.34rmA/V 165

Hz 34.31047.0)11110100( 2

=××+×

111 M 2.5

Hz 34.32 == pp ff

Amplifier Cut-off frequency 1) 1 cap: 1 pole 2) no coupling capacitors at the input

csigip CRR

Exercise 6: Find the bias point and the amplifier parameters of the circuit below. (µpCox (W/L) = 400 µA/V2, Vtp = − 4 V, λ = 0.01 V-1. Ignore channel width modulation in biasing).

This is a PMOS common drain amplifier. o Input at the gate, output at the source.

It has a source-degeneration bias with two voltage sources.

Exercise 6 (Cont’d): µpCox (W/L) = 400 µA/V2, Vtp = − 4 V,λ = 0.01 V-1

Assume Saturation:

09104005.010

||101013

=−+×××

++×=+×=−

tpOVDSGD

VVIVISG-KVL:

5.0 2OVoxpD V

LWCI µ=

V 9.5=+= tOVSG VVV

mA 71.0104005.0 26 =××= −OVD VI

V 9.5 0 =→−= SSSG VVV

V 9.10)5( =−−= SSD VV

V 9.1=OVV

k 141100.7101.0

mA/V 0.747 0.303

100.71 2 2

Signal circuit

87.038.61

38.66.38)||||(

k54.8k100||k10||k 141||||

88.0)||||(1

)||||(

RRrgRRr

RRrgRRrg

Amplifier Parameters Common collector Amp.

k 2.1k 34.1||k 101 ||

87.0==×+

k 141 mA/V, 0.747 == om rg

Amplifier Cut-off frequency 1) 1 cap: 1 pole 2) no coupling capacitors at the input

Signal circuit

k 141mA/V 0.747

Hz 39.31047.0)102.110( 2

=×××+

Exercise 7: Find the bias point and the amplifier parameters of the circuit below. (µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1. Ignore channel width modulation in biasing).

This is a NMOS common gate amplifier. o Input at the source, output at the drain.

It has a source-degeneration bias with a voltage divider.

Bias circuit

05108005.010

=−+×××

×++=×+==−

DtOVDGSG

IVVIVVGS-KVL:

Assume Saturation: 5.0 2OVoxnD V

LWCI µ=

V 0.2=+= tOVGS VVV

mA 40.0108005.0 26 =××= −OVD VI

V 74415101015 44

=−−=×++×=

V 0.1=OVV

Exercise 7 (Cont’d): µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1.

V 615M8.1M2.1

M2.1=×

DS-KVL:

k 250100.401.0

mA/V 0.800 1

100.4 2 2

69.7)||k 10||k 250(108.0

)||||(

3 =∞×=

RRrgvv

Signal circuit (IVC short)

Amplifier Parameters 1) CG amplifier 2) Biased with two voltage sources (RG = 0)

11.469.7k 1k 15.1

k 15.1=×

k 250mA/V 0.800

k 15.1k 1.30||k 10

/)||(1 ||

k 77.9k 432||k 10

)]||(1([ ||

sigSmoDo

RRRgrRR

Hz 74021 =+≈ ppp fffk 250mA/V 0.80

Signal circuit (IVC short) Amplifier Cut-off frequency

2 caps: 2 poles

Hz 74010100)1010(1.15 2

=××+×

csigip

010100)10(9.77 2

=××∞+×

k 77.9 k 15.1

Exercise 8: Design a common emitter amplifier with RE with vo / vi = 5 when driving a 100 kΩ load and a bias point of IC = 4 mA and VCE = 6 V. The circuit should have a cut-off frequency of 50~Hz for Rsig = 100 Ω. A 16 V power supply is available. Use a Si BJT with β = 200, β min = 100, VA = 150 V, ignore Early effect in bias calculations.

Prototype of the circuit is shown.

Problem specification gives

We should find RC , RE , RB1 , RB2 , Cc1 , and Cc2

k 100 , 100 V, 16 =Ω== LsigCC RRV

* See Exercise 4 of BJT lecture slides for bias design

mA, 4 V, 6 k, 100 , 100 V, 16 ===Ω== CCELsigCC IVRRV

Exercise 8 (Cont’d):

k 30.1

k 5.37104

mA/V 154 1026104

1. Bias point (IC = 4 mA & VCE = 6 V) gives the SSM parameters as well as a relationship between RC and RE through CE-KVL:

k 5.2104

+≈−++=

ECCCECC

EECECCCC

RRIVVIRVIRV

2. Amp gain of vo / vi = 5 also gives another relationship between RC and RE:

CE-KVL:

5)/1](/)||[(1

)||( =+++

=πrRrRRRg

EoLCEm

Solution can be simplified by noting RC + RE = 2.5k and thus RC < 2.5k: 1) Since RC << RL = 100 k, then RC || RL ≈ RC 2) Since RC << ro = 37.5 k, we can drop the 3rd term in the denominator of the gain formula:

5.32555

k 09.2 411

k 5.25.325

A 0.20/ µβ == CB II

3102.73.631316

)/1](/)||[(1)||( −×++

=πrRrRRRg

EoLCEm

In fact, because gm terms are large:

Note that the (open-loop gain) is independent of BJT parameters!

For discrete BJT amplifiers (RC << ro ), dropping the third term is a very good approximation because gm terms are actually large:

k 09.2 , 411 =Ω= CE RR

k 30.1 k, 5.37 mA/V, 154 === πom rrg

A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ====Ω== BCCELsigCC IIVRRV

Exercise 8 (Cont’d): A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ====Ω== BCCELsigCC IIVRRV

1. Bias point (IC = 4 mA & VCE = 6 V) and requirements of good bias give RB1 and RB2 through BE-KVL. However, it is easier to find RB and VBB (Thevenin equivalent) first.

2. Resistors RB1 and RB2 are found from:

BE-KVL:

Dividing the above equations gives RB1 :

k 89.4k 4.27

k 09.2 , 411 =Ω= CE RR

k 4.154111011.04111011.0)1( 0.1

=××=××=+≤

V 43.2114 10 4 0.710 4.15 1020 336

=××++×××=

EEBEBBBB

RIVRIV

152.016

k 15.4||

RRRRRRR

The 2nd condition for stable bias is RE IE ≥ 1 V:

V 1 64.1114 10 4 3 >=××= -EE RI

Coupling capacitors Cc1 and Cc2 are found from specification of cut-off frequency (50 Hz). We need to compute Ri and Ro first.

Can choose either capacitor and compute the other one. A good design guideline is to keep capacitors small.

k 15.4 ,k 89.4 ,k 4.27 ,k 09.2 , 411 21 ====Ω= BBBCE RRRRR

k 95.3k 81.6 ||k 15.4)/1](/)||[(1

RrRrRR

RrRRπ

k 09.2M 1.74 ||

||1 ||

RRRrRrRR

k 30.1 k, 5.37 mA/V, 154 === πom rrg

Amplifier Cut-off frequency (2 caps: 2 poles)

10102 21

1005.4 21

ccsigip

5010102 2

11005.4 2

We can choose either capacitor and compute the other one. A good design guideline is to keep capacitors small. Since the coefficient of CC1 is 25 times smaller than that CC1, a reasonable choice is to assign 80% of fp to CC1 :

k 15.4 ,k 89.4 ,k 4.27 ,k 09.2 , 411 21 ====Ω= BBBCE RRRRRk 30.1 k, 5.37 mA/V, 154 === πom rrg

nF 156 502.010102 2

F 982.0 508.01005.4 2

=→×=××

Design Commercial Value Value (5%)

nF 156F 982.0

411k 09.2k 89.4k 4.27

Ω====

CCRRRR

µnF 160

F 1 390

k 2k 7.4

Ω====

CCRRRR

This is a two-stage amplifier. o Stage 1 is a common emitter amplifier with RE o Stage 2 is a common collector amplifier (emitter follower).

Both stages have source-degeneration bias with a voltage divider. o Each stage is independently biased because of the 470 nF

coupling capacitor between stages.

Stage 1 Bias Thevenin form of the Voltage divider

V 37.215k 2.6k 33

k 22.5k 2.6||k 33

Exercise 9 (cont’d) : β = 200, VA = 150 V

A 9.15/ mA 18.3

5007.0)1/(1022.5 2.37

:KVLBE

111111

==≈=

+++×=

++=−

EEBEBBBB

IIRIVRIV

111 ≥>= CECBE VIV

V 0.7 V 05.7 )500102(1018.315

15 :KVLCE

=>=+×××+=

++=−−

EECECC

k 64.1

k 2.4710 18.3

mA/V 122 1026103.18

Stage 2 Bias Thevenin form of the Voltage divider

V 25.815k 18k 22

k 90.9k 22||k 18

A 36/ mA 20.7

107.0)1/(109.9 8.25

8.25 :KVLBE

==≈=

+++×=

++=−

EEBEBB

IIRIVRI

222 ≥>= CECBE VIV

V 0.7 V 80.7 101020.715

15 :KVLCE

=>=××+=

+=−−

k 722.0

k 8.2010 20.7

mA/V 277 1026107.20

26226294510277)||||(

)||||(1)||||(

=××=

RRrgRRrg

RRrgvv

[ ][ ] k 41.9k 189722 ||k 90.9

)||||( ||

2222222

RRRrrRR βπ

Stage 2: Emitter Follower For Gain & Ri , start from the load side because we need to know RL

k 41.9 21 == iL RR

k 100 2 == LL RR

k 90.9k 722.0

k 8.20mA/V 277

k 95.497.2k ||k 22.5)/1](/)||[(1

RrRrRR

RrRRπ

25.30.62

201046.0500101221

1065.110122

046.0)/1](/)||[(k 1.65k 41.9||k 2||

)/1](/)||[(1)||(

1111111

−=−=+××+×××

+++−=

EoLCEm

rRrRRRR

rRrRRRgRRg

Stage 1: CE amplifier with RE k 41.9 21 == iL RR

k 22.5k 64.1k 2.47mA/V 122

Stage 2: Emitter Follower

For Ro , start from the source side because we need to know Rsig

k 0.2 12 == osig RR

100 1 Ω== sigsig RR

||1 ||

RRRrRrRR

Stage 1: CE amplifier with RE

= 8.119.11||k 1||

22222 β

π sigBEo

k 22.5k 64.1k 2.47mA/V 122

k 90.9k 722.0

k 8.20mA/V 277

k 0.2 k 95.4

25.3/ k 41.9

−===

sigsig

vvRRRR

k 22.5k 64.1k 2.47mA/V 122

k 90.9k 722.0

k 8.20mA/V 277

8.11 k 41.9

1/ k 100

vvRRRR

19.31)25.3(98.0

11.8 k, 4.95

−=×−×=××+

Ω====

k 0.2 k 41.9 11.8

k 4.95

Hz 3.527.299.1571.6321 =++=++≈ pppp ffff

Amplifier Cut-off frequency 3 caps: 3 poles

Hz 71.6)( 2

csigip CRR

Hz 9.15)( 2

p CRRf

Hz 7.2910470)(2

=××+

cjjojipj

Coupling cap at the input:

Coupling cap at the output:

Coupling cap between stages:

This is a two-stage amplifier. Both stages are common emitter amplifier with RE.

Both stages have source-degeneration bias with two power supplies (no voltage divider). o There are NO coupling capacitors and the second stage is

biased from VC1.

A 3.33/mA 65.6

510)1/(106.3 510106.351.3

3 5107.0106.31083.3106.3 15

3 510)(106.3 15

==≈=

++×=+×=

−++×+×××=

−+++×=−

IIIIII

A 2.19/mA 83.3

3 6007.0)1/( 100 03 600 100 0

==≈=

−+++=−++=

V 7.0 and 0 V, 7.0: Activein are sBJTboth Assume

111 ≥>= CECBE VIV

BE1-KVL:

BE2-KVL:

CE2-KVL:

0.7 V 63.43 510105.1 15

=>=−++×=

CE1-KVL:

0.7 V 79.13 600)(106.3 15

=>=−+++×=

VVIVII

k 36.1

k 2.3910 83.3

mA/V 147 1026103.83

V 63.4A 2.19

mA 83.3

µV 79.1

A 3.33mA 65.6

k 782.0

k 6.2210 65.6

mA/V 256 10261065.6

For Gain & Ri , start from the load side because we need to know RL

k 8.92 21 == iL RR

k 100 2 == LL RR

∞====

k 782.0k 6.22mA/V 256

k 8.9292.8k ||)/1](/)||[(1

RrRrRR

RrRRπ

88.2132379

108.05101056211048.110562

108.0)/1](/)||[(k 1.48k 100||k 5.1||

)/1](/)||[(1)||(

2222222

−=−=+××+×××

+++−=

EoLCEm

rRrRRRR

rRrRRRgRRg

k 108k 108 ||)/1](/)||[(1

RrRrRR

RrRRπ

71.53.89

510127.0600101471

1047.310147

127.0)/1](/)||[(k47.3k 8.92||k 6.3||

)/1](/)||[(1)||(

1111111

−=−=+××+×××

+++−=

EoLCEm

rRrRRRR

rRrRRRgRRg

Stage 1: CE amplifier with RE k 8.92 21 == iL RR

∞====

k 36.1k 2.39mA/V 147

∞====

k 782.0k 6.22mA/V 256

∞====

k 36.1k 2.39mA/V 147

For Ro , start from the source side because we need to know Rsig

k 6.3 12 == osig RR

100 1 Ω== sigsig RR

||1 ||

RRRrRrRR

||1 ||

RRRrRrRR

∞====

k 782.0k 6.22mA/V 256

∞====

k 36.1k 2.39mA/V 147

k 6.3 k 108

71.5/ k 8.92

−===

sigsig

vvRRRR

k 5.1 k 8.92

88.2/ k 100

−===

vvRRRR

4.16)88.2()71.5(1

k 5.1 k, 108

=−×−×=××+

Amplifier Cut-off frequency: only Coupling cap at the output:

Hz 7.15)( 2

p CRRf

Problem for Problems for Discrete Transistor...

Documents

Transcript of Problem for Problems for Discrete Transistor...

Measuring the common emitter current gain β in a …Measuring the common emitter current gain β in a BJT 9 DAC programming (frame #0 ) Measurement I/O -> DAQmx Data Acquisition ->

The role of quinazoline in ameliorating intervertebral disc ......The role of quinazoline in ameliorating intervertebral disc degeneration 2079 blocked with 5% bovine serum albumin/PBS

ECEN 326 LAB 1 Design of a Common-Emitter BJT Ampliﬁerhoyos/courses/326/lab326.pdfECEN 326 LAB 1 Design of a Common-Emitter BJT Ampliﬁer 1 Circuit Topology and Design Equations

TGF-β1 aggravates degenerative nucleus pulposus cells ... › wp › wp-content › uploads › 12025-12033.pdfcells degeneration. To determine the effect of TGF-β1 and ANGPTL2 on

Preclinical Amyloid-β and Axonal Degeneration Pathology in ... · brain β-amyloidosis, but it is not settled whether this represents preclinical Alzheimer’s disease (AD) or have

Lecture 19 Bipolar Junction Transistors (BJT): Part 3 …alan.ece.gatech.edu/ECE3040/Lectures/Lecture19-BJT Ebers-MollModel.pdf · E. I. C. Base. I. B. Collector . Base . Emitter

High-Precision Half-Life Measurements for the β Emitter C ......= constant Δ R = nucleus independent inner radiative correction: 2.361(38)% δ R = nucleus dependent radiative correction

980nm Multi-Emitter Laser Diode, 70W Fiber-Coupled Output

TNF-α antagonist attenuates systemic lipopolysaccharide-induced … · 1 day ago · etanercept of 5˙ mg/kg reduced neuronal degeneration after 1˙h, and in a study involving a

CHAPTER.3: Transistor at Low Frequenciesajaybolar.weebly.com/uploads/1/0/1/0/10106930/unit_3_transistor_at… · Common emitter Ib Ic Vbe Vce Common base Ie Ic Veb Vcb Common Collector

ΚΥΚΛΩΜΑΤΑ ECL (Emitter Coupled Logic)”ιάλεξη... · (Emitter Coupled Logic) ECL (Emitter Coupled Logic) 1. ΨηφιακάΟλοκληρωμέναΚυκλώματακαιΣυστήματα2008

EE105 – Fall 2014 Microelectronic Devices and Circuitsee105/fa14/lectures... · Lecture23-Amplifier Frequency Response 7 C-S Amplifier High Frequency Response Source Degeneration

Lucio Ludovici / INFN-Roma · Lucio Ludovici / INFN-Roma . ... 2885 (1995) 611 8. ... e lm ntc au io s. In models with neutrino mass degeneration ﬁ constraint on

EMITTER-COUPLED LOGIC INEL 4207. Differential Pair as basic element of ECL.

BIPOLAR JUNCTION TRANSISTORS - Home | University … · BIPOLAR JUNCTION TRANSISTORS - GENERAL USE ... by adding an emitter follower as a second stage. In the emitter follower circuit,

Infrared Emitter (940 nm) IR-Lumineszensdiode (940 nm ... · • Sensor technology • Sensorik • Discrete optocouplers • Diskrete Optokoppler • Discrete interrupters • Diskrete

The afterglow and host galaxy of GRB 090205: evidence for a Ly- α emitter at z = 4.65

Penguat EmittER

RESEARCH ARTICLE Open Access Impaired ossification coupled … · 2017. 8. 26. · RESEARCH ARTICLE Open Access Impaired ossification coupled with accelerated cartilage degeneration

Open Access Catabolic cytokine expression in degenerate ... · While several cytokines have been implicated in the process of IVD degeneration and herniation, investigations have