on thi cuoi hk VLBD

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HBK TP HCMKHOA IN-IN T B MN IN T GV: H Trung M

Hng dn n thi LT mn Dng c bn dn HK 1 Nm hc: 2014-2015

Ch : thi trc nghim khong 50 cu S : 4 n 8 thi khng cho s dng ti liu

Trng tm n thi ca cc chng nh sau: Chng 4. Chuyn tip PN (phn cn li) in dung chuyn tip PN

o vi phn cc ngc: c in dung chuyn tip CJ [F] (J=junction=chuyn tip) ) . SJ dep

AC AC W

vi A: din tch mt ct ngang; S hng s in mi bn dn; W: b rng min ngho. o vi phn cc thun: c in dung khuch tn CD [F]: D TD

T

ICV

vi T l thi gian i qua diode (cn gi l thi gian chuyn tip [T=transit])

Hnh 4.6 in dung ca diode chuyn tip PN vi phn cc ngc VA = VR ( ch : V0=Vbi)

3. Cc loi diode khc

o Diode chnh lu: diode tip xc PN thng thng - chnh lu: cho dng in i qua 1 chiu (t anode sang cathode) - thng th c in p nh thng ln.

o Diode n p (cn gi l diode Zener) - S dng hiu ng nh thng Zener v/hoc hiu ng nh thng thc l. - Xem li nh hng ca nhit ? Vi in p nh thng VBR= VZ (vi VZ>0) th TCVZ < 0 vi nh thng Zener v TCVZ >0 vi nh thng thc l.

o Diode bin dung (varicap hay varactor) - ng dng in dung tip xc CJ = f(VR), khi VR tng th CJ gim (phn cc ngc VA = VR < 0) (xem hnh 4.6).

Hnh 4.7 M hnh diode Zener vi phn cc ngc VR > VZ0


o Diode Schottky - to t tip xc M-S (M=Metal=kim loi v S=Semiconductor=bn dn), th d M l platinum v S l bn dn loi N hnh thnh diode Schottky vi Anode bn M v cathode bn S. - Hy k thm cc kim loi khc ngoi Platinum? - c ro th nh ( ~ t 0.2V n 0.3V) - hot ng tt/dn tc chuyn mch cao.

4. Cc ng dng ca diode Mch chnh lu: bn k, ton sng, v chnh lu. S v cc cng thc Mch lc sng gn bng t. Cch tnh t vi cc mch chnh lu i vi mch lc. Mch xn dng diode thng v Zener. Mch n p dng diode Zener. . . .

Chng 5. BJT (TRANSISTOR TIP XC LNG CC) Ti sao c tn gi lng cc? tip xc (hay chuyn tip hay mi ni)? Cu to BJT loi NPN v loi PNP.K hiu jE, jC. Nng tp cht ca cc min?

Hnh 5.1 K hiu ca hai loi BJT: (a) PNP v (b) NPN.

k hiu BJT th mi tn cc E c ngha g? Cc dng in trong BJT ch tch cc [thun]:


Dng in r (r) ICBO (dng t C n B vi E h mch) v ICEO (dng t C n E vi B h mch) trong BJT (nhit tng dn n dng r tng)

o cu hnh CB: IC = IE + ICBO o cu hnh CE: IC = IB + ICEO vi ICEO = ICBO/(1-)

H s vn chuyn min nn B, hiu sut cc pht E? Chng ph thuc nh th no vi cc tham s ca BJT (nng tp cht, b rng min nn)? BJT tt c B, E tin gn ti 1.

o Hiu sut cc pht E

vi pe0 = ni2/NDE ; nb0 = ni2/NAB; WB = b rng min nn DE = h s khuch tn ca ht dn thiu s ti E = Dp DB = h s khuch tn ca ht dn thiu s ti B = Dn LE = chiu di khuch tn ca ht dn thiu s ti E= Lp Thay cc biu thc trn vo e , ta c dng biu din khc ca e nh sau:

1 pAB BeDE n p

DN WN D L

o H s vn chuyn min nn B

2112

Bn

B

WBL

vi WBn= b rng min nn phn trung ha WB

LB = chiu di khuch tn ca ht dn thiu s ti B = Ln li dng in cc nn chung = B.E = IC/IE li dng in cc pht chung = / (1 ) = IC/IB

o cao cn: tc ti hp thp min nn v thi gian chuyn tip (i qua) ngn min nn o ph thuc vo IC v nhit . o DC : dc = IC/IB vi IC, IB l dng DC o AC: ac = IC/IB vi IC, IB l s thay i ca IC, IB do dng tn hiu AC

Cc ch lm vic (ch hot ng) ca BJT v c im ca chng:

Ch :

o Vi BJT NPN Si th VON = 0.7V, VBEsat = 0.70.8V, v VCEsat 0.2V. o R l ch tch cc ngc. o SV t suy ra cch nhn bit vi BJT PNP.

0

0

1 1 eEP E BeEN b B E

pI D WI n D L


M hnh tn hiu ln ca BJT (TD: xt BJT Si loi NPN)

a) Tch cc (hay Tch cc thun)

b) Bo ha (VBE,sat=0.7V 0.8V)

c) Tt Hnh 5.2 M hnh tn hiu ln ca BJT SI loi NPN trong cc ch hot ng khc nhau

Cc cu hnh mc BJT: (Nh 1 mng 4 cc vi: Bn tri l mch vo v Bn phi l mch ra)

a) CB = Common Base =B chung b) CE=Common Emitter=E chung c) CC=Common collector=C chung

Hnh 5.3 Cc cch mc BJT NPN trong mch Hy nu c im ca cc cch mc vi ng dng kha in t v mch khuch i?

Phng trnh cc dng in trong BJT NPN ch tch cc thun: Dng [in cc] thu IC Dng [in cc] nn IB Dng [in cc] pht IE

1BE BE

T T

V VV V

C S S B EI I e I e I I

1

BE

T

VVC S E

BI I II e 1

BE

T

VVC S

E BI II e I

vi IS l dng bo ha: 2

E n iS

A Bn

qA D nIN W

trong AE l din tch mt ct ngang ti min pht, NA l nng tp cht Acceptor ti min nn v WBn l b rng phn trung ha trong min nn. Cc c tuyn Volt-Ampere ca BJT (cn gi l cc c tuyn I-V)

a) c tuyn vo b) c tuyn ra

Hnh 5.4 Cc c tuyn vo v ra ca BJT NPN mc CE


nh hng ca nhit n cc c tuyn ca BJT (TD vi BJT NPN)

iu ch min nn: Xt BJT NPN phn cc ch tch cc [thun] (khuch i), nu VCE tng b

rng hiu dng ca min nn gim dng IC tng. Ngha l b rng min nn b thay i (iu ch) khi in p VCE thay i.

in p Early VA: gi tr in p ti im nm trn trc honh m mi ng cong IC theo VCE ( phn khuch i) u i qua im ny. dc ti im lm vic Q:

C CQ CQ

CE CEQ A A

dI I IdV V V V

(nu VA >> VCEQ) Khi dng IC:

1BE

T

VV CE

C SA

VI I eV

in p nh thng BVCBO , BVCEO o CB: BVCBO khng b nh hng bi IE. o CE: BVCEO b nh hng bi IB (IB tng th BVCEO gim)

Hnh 5.5 Mt th d v nh thng BJT

Kha in t dng BJT : Kha m vi BJT tt (OFF) Kha ng vi BJT bo ha (ON)

a) Kha in t dng BJT NPN b) Kha in t dng BJT PNP


Do in tch cha ti JC khi bo ha nn gim tc chuyn mch ca BJT khi chuyn t bo ha

sang tt. VP Transistor Schottky:

Cu to Mch tng ng K hiu c im

Gim in tch cha ti JC khi BJT bo ha v diode Schottky c VON nh hn VON ca chuyn tip PN. Tng tc chuyn mch

M hnh tn hiu nh ca BJT (tn s trung bnh)

(a) M hnh (mc CE)

(b) M hnh T (nu khng b qua ro c th s c in tr ro ni t C n E)

M hnh tn s cao ca BJT ch tch cc: (C=Cbe, C=Cbc)

Tn s ct fT (khi ac =1)

01

2 2m

Tec

gf f fC C

Vi ec l thi gian in t i t E n C vi BJT NPN mc CB ch tch cc, f v f l cc tn s m ac v ac gim i 2 so vi tr s tn s thp, v 0 (dc=hFE) l gi tr ca ac tn s thp. M hnh tham s h. Cc tham s h cho BJT cu hnh CE:

'

0

BE

BCE

be T Tie be ac e

CE b EQ CQ

v V Vvh r r ri i I IV v


0

C

BCE

cfe ac m ie

CE b

iih g hi iV v

'

1CQm

T e

Ig

V r

0

1 CEC

B

A CEQce Ac O

Boe c CQ CQ

V Vv Vvr rh i i I II i

(nu VA >> VCEQ) Ch : dc = hFE = IC/IB ; ac = hfe = ic/ib ( tn s thp v trung bnh: ac dc=) Gng dng in (Current mirror)

Mch iu kin l ngun dng Phng trnh

1. Q1 v Q2 c c tnh ging nhau 2. Q1 (c mc nh diode) v Q2

lun ch tch cc thun (dn n c gii hn vi in tr ti RL)

Ch : VCC v VEE > 0

Dng hng qua ti:

21R

OUTII

vi dng chun IR: CC EE BE

RV V VI

R

Gii hn ca ti RL l ,0 CC EE CE satL

OUT

V V VR

I

Thyristor: l dng c cng sut quan trng, c thit k x l in p cao v dng in ln.

o Diode 4 lp p-n-p-n Cu to c tuyn dng-p ca diode p-n-p-n

o SCR (Silicon Controlled Rectifier)

Cu to ca SCR K hiu c tuyn dng-p ca SCR


Chng 7. MOSFET MOSFET c cch l gia cng v knh dn bng lp cch in, thnh phn c bn l kim loi (M=Metal), lp cch in SiO2 (O=Oxide), v bn dn (S=semiconductor). Cc tn gi khc ca MOSFET l MISFET (Metal-Insulator-Semiconductor), IGFET (Insulated Gate FET). Nguyn tc hot ng ca FET l dng ht dn t ngun in mng c iu khin bng in p cng hay in trng cng. in trng ny lm cm ng in tch trong bn dn giao tip bn dn-oxide. Cu trc ca MOSFET

MOSFET loi giu (cn gi l MOSFET knh dn cha lp sn)

MOSFET loi ngho (cn gi l MOSFET knh dn lp sn)

K hiu ca EMOS:

N-EMOS P-EMOS

M t nh tnh hot ng ca N-EMOS


Cc ch phn cc cho t MOS trong N-EMOS C 3 ch phn cc quan trng cho t MOS:

o Tch ly l (Hole Accumulation): khi phn cc m gia kim loi v bn dn (VGS < VFB < 0, VFB l in p di phng), ti giao tip gia bn dn v cch in s c tch ly l. o Ngho (Depletion): khi phn cc dng gia kim loi v bn dn (VFB < VGS < VTN, VTN > 0), ti giao tip gia bn dn v cch in s cc l b y xung di hnh thnh min ngho. o o ngc (Inversion): khi phn cc dng gi tr ln gia kim loi v bn dn (VGS > VTN), cc in t c ht vo min gn giao tip gia bn dn v cht cch in, do hnh thnh nn knh dn in t (knh N) trong bn dn P.

Cu trc N-EMOS

Vt liu dng cho bn cc dn in thng dng Silicon a tinh th c pha tp cht rt nhiu (cn c gi l polysilicon hay polySi hay poly). Vt liu cch in thng thng l SiO2. ti thiu ha dng in gia min thn v min S(source)/D(drain) ngi ta thng ni min thn vi cc ngun. S to thnh knh dn trong N-EMOS

S to thnh knh dn N S nh hng ca chiu di knh dn L v chiu rng knh dn W

MOSFET c gi l MOSFET knh ngn khi L < 1m, trong IC ngi ta thng dng MOSFET knh ngn.

MOSFET c gi l MOSFET knh di khi L > 1m

Cc min hot ng ca N-EMOS vi VGS > VTN

Min tuyn tnh (min Ohm hay min triode) (VDS < VDS,sat=VGSVTN)

Min bo ha (hay min tch cc) (VDS VDS,sat=VGSVTN)

cnh min bo ha VDS=VDS,sat

min bo ha VDS VDS,sat


Khi VDS nh (c th hon i D v S) th c th xem nh in tr c iu khin bng p (VGS3> VGS2> VGS1>VTN)

Khi VDS tng, im nght di chuyn v pha cc ngun

c tuyn truyn t v c tuyn ra ca N-EMOS

c tuyn truyn t c tuyn ra

BVDSS= in p nh thng gia DS khi ngn mch cng

Cc phng trnh dng in mng ID trong N-EMOS VGS VTN : min tt ID=0 VGS > VTN : (VDS,sat = VGS VTN )

o 0 < VDS < VDS,sat : min tuyn tnh (cn gi l min Ohm, min in tr hay min triode) 2

2DS

D n ox GS TN DSVWI C V V V

L

vi n l linh ng in t v Cox l in dung lp cch in trn 1 n v din tch. Nu 2DS GS TNV V V th ID l hm tuyn tnh theo VDS: (c th hon i D v S)

D n ox GS TN DSWI C V V VL Khi MOSFET tng ng vi in tr RDS (RON hay RDS,ON):

1DS

n ox GS TN

R WC V VL

o VDS VDS,sat = VGS VTN : min bo ha (cn gi l min tch cc) vi 212D n ox GS TNWI C V VL

Ngi ta thng ng dng min tt v tuyn tnh cho MOSFET lm kha in t, v min bo ha cho MOSFET lm phn t khuch i tn hiu hoc lm ngun dng. Mt s c tnh khng l tng ca MOSFET (Xt N-EMOS min bo ha)

o iu ch chiu di knh dn: tng t hiu ng Early trong BJT, khi tng VDS th im nght dch chuyn v min ngun, dn n chiu di knh dn hiu dng nh hn hay dng ID tng ln. Khi phng trnh dng in mng c dng

21 12D n ox GS TN DS

WI C V V VL

vi 1AV

v VA l in p Early o Hiu ng thn: khi tng VSB lm in p ngng VTN tng nh hng c tuyn I-V.


o nh hng ca nhit : khi T tng VTN v linh ng gim dng ID gim o S bo ha vn tc: khi kch thc transistor gim, dy lm oxide mng hn vn tc in t

bo ha v lc phng trnh dng ID: 12D n ox GS TNWI C V VL

vi =1 2, ty theo cng ngh. M hnh tn hiu ln ca N-EMOS (dng phn tch tng qut hay tnh im tnh)

M hnh tn hiu nh ca N-EMOS (khi N-EMOS lm vic min bo ha v 0.2gs GS TNv V V )

M hnh M hnh T Tn s cao

H dn gm:

dDm n ox GS TNGS gsQ

idI Wg C V VdV v L

in tr ra ro:

A DSQDS ds Ao

D d DQ DQQ

V VV v VrI i I I

Tn s ct 2 mT gs gdgf

C C

o Cc cch mc MOSFET: CS, CD v CG. o Cc ng dng tiu biu ca MOSFET l kha analog, in tr c iu khin bng p, ngun dng

v phn t khuch i tn hiu trong mch khuch i. Tm tt quan h dng-p ca MOSFET

NMOS PMOS Min tt (VGS VTN): ID = 0 Min triode (VGS > VTN v 0 VDS < VDS,sat)

22DS

D n GS TN DSVI K V V V

vi n n ox WK C L

Min bo ha (VGS > VTN v VDS VDS,sat) 2

2n

D GS TNKI V V

2m n GSQ TN n DQg K V V K I im chuyn tip VDS,sat = VGS VTN Loi giu: VTN > 0 Loi ngho: VTN < 0

Min tt (VGS VTP): ID = 0 Min triode (VGS < VTP v 0 VDS > VDS,sat)

22DS

D p GS TP DSVI K V V V

vi p p ox WK C L

Min bo ha (VGS < VTP v VDS VDS,sat) 2

2p

D GS TP

KI V V

2m p GSQ TP p DQg K V V K I im chuyn tip VDS,sat = VGS VTP Loi giu: VTP < 0 Loi ngho: VTP > 0

BT v BJT1

HQG Tp HCM HBK Khoa in-TB mn in T Mn hc: Dng c bn dn GVPT: H Trung M

Chng 5 Bi tp gii sn v BJTAY1112-S1

(Gi s cc BJT Si c in p JE dn l 0.7V, c in p bo ha[NPN]: VBEsat=0.8V v VCEsat=0.2V)

1. Vi cc trng hp sau BJT hot ng min no? a) NPN: VBE = 0.8 V, VCE = 0.4 V d) PNP: VCB = 0.9 V, VCE = 0.4 V b) NPN: VCB = 1.4 V, VCE = 2.1 V e) PNP: VEB = 0.6 V, VCE = 0.4 V c) NPN: VBE = 1.2 V, VCB = 0.6 V f) PNP: VBC = 0.6 V, VEC = 1.3 V

S. Qui tc chung gii loi bi ton ny nh sau:

i) Vi BJT NPN, ta xt phn cc thun/ngc ca JE v JC theo:

JE: VBE = VB VE = VBC + VCE = VCE VCB = VBCVEC JC: VBC = VB VC = VBE + VEC = VBE VCE

> 0 : phn cc thun < 0 : phn cc ngc

ii) Vi BJT PNP, ta xt phn cc thun/ngc ca JE v JC theo:

JE: VEB = VE VB = VEC+VCB = VECVBC =VCB VCE JC: VCB = VC VB = VCE + VEB = VEB VEC

> 0 : phn cc thun < 0 : phn cc ngc

p dng BJT NPN vo cc cu a), b), v c), ta c kt qu sau: Trng hp VBE [V] JE VBC [V] JC Min hot ng

a) 0.8 > 0 thun 0.8 0.4 = 0.4 > 0 thun Bo ha b) 2.1 1.4 > 0 thun 1.4 < 0 ngc Tch cc thun c) 1.2 < 0 ngc 0.6 < 0 ngc Tt

p dng BJT PNP vo cc cu d), e), v f), ta c kt qu sau: Trng hp VEB [V] JE VCB [V] JC Min hot ng

d) 0.9 0.4 > 0 thun 0.9 > 0 thun Bo ha e) 0.6 > 0 thun 0.6 0.4 > 0 thun Bo ha f) 1.3 0.6 > 0 thun 0.6 < 0 ngc Tch cc thun

2. Hy tm IE, VCE v VBC ca BJT trong mch hnh 1. S. Vi phn cc trong mch ta thy JE v JC u b phn cc ngc => BJT min tt => IE = IC = IB = 0 Nh vy (nu chn in th ti cc B l in th t): VCE = VC VE = 10V 15V = 5V Cn VBC = 10V

Hnh 1 Hnh 2 3. Hy tm IE, VEC v VCB ca BJT trong mch hnh 1. S. Vi phn cc trong mch ta thy JE c phn cc thun v JC c phn cc ngc => BJT min tch cc thun => IE = (20V0.7V)/39K = 0.495 mA

BT v BJT2

Nu chn in th t ti cc B th ta c VE = 0.7V v VC = 20V + 20K x IC 20V + 20K x IE = 20V + 20K x0.495mA = 10.1V

Suy ra: VEC = VE VC = 0.7V (10.1V) = 10.8V

VCB = VCVB = VC = 10.1V 4.Hy tm hiu sut pht e, h s vn chuyn min nn B v li dng E chung ca BJT NPN vi cc tham s sau: NDE = 1x1018cm3, NAB = 1x1016cm3, Dn = Dp, WB = 100nm, v Lp= Ln = 1m. S.

a) Hiu sut pht e Theo LT ta c

0

0

1 1 eEP E BeEN b B E

pI D WI n D L

vi pe0 = ni2/NDE ; nb0 = ni2/NAB DE = h s khuch tn ca ht dn thiu s ti E = Dp DB = h s khuch tn ca ht dn thiu s ti B = Dn LE = chiu di khuch tn ca ht dn thiu s ti E= Lp Thay cc biu thc trn vo e , ta c dng biu din khc ca e nh sau:

1 pAB BeDE n p

DN WN D L

T tnh c e = 1 (1016/1018)(1)(100/1000) = 1 10-3 = 0.999 Nh vy e = 0.999

b) H s vn chuyn min nn B Theo LT ta c

2112

Bn

B

WBL

vi WBn= b rng min nn phn trung ha WB LB = chiu di khuch tn ca ht dn thiu s ti B= Ln

Suy ra: B = 1 (1/2)(100/1000)2 = 0.995 Nh vy B = 0.995

c) li dng E chung Ta c: = Be = 0.995 x 0.999 = 0.994 => = /(1) = 0.994/(10.994) 166 Nh vy = 166

Hnh 3 Hnh 4 Hnh 5

BT v BJT3

5. Cho mch li LED hnh 3, vi BJT c =100. LED ny c in p dn VLED(on) = 1.5V v cn li vi dng ILED = 20mA. in p iu khin VI = 0V lm LED tt, VI = 5V lm LED sng. Hy tm:

a) Gi tr ca RC cho BJT min bo ha khi LED sng nu RB = 1K. b) Di gi tr ca RC cho BJT min bo ha khi LED sng nu RB = 1K. c) Di gi tr ca RC cho BJT tch cc thun khi LED sng nu RB = 1K. d) Di gi tr ca RB cho BJT min bo ha khi LED sng nu RC c tr cu b). e) BJT c l bao nhiu cho BJT min bo ha khi LED sng nu RB = 1K v RC =170.

S. a) Gi tr ca RC cho BJT min bo ha khi LED sng nu RB = 1K.

Khi bo ha ta c: VCC = 5V = RCIC + VLED + VCEsat vi IC = ILED =20mA

Suy ra RC = (VCCVLED VCEsat)/ILED = ( 5V1.5V0.2V)/20mA = 3.3V/20mA = 165. Nh vy RC =165 b) Di gi tr ca RC cho BJT min bo ha khi LED sng nu RB = 1K.

Cch 1: Xt 0 VCE VCEsat = 0.2V Ta c: VCE = VCC RCILED VLED hay 0 VCC RCILED VLED VCEsat = 0.2V Hay (VCC VLED VCEsat)/ILED RC (VCC VLED)/ILED

(5V 1.5 0.2)/20mA RC (5V 1.5V)/20mA 165 RC 175

Cch 2: JC c phn cc thun, ngha l VBC > 0 hay VB > VC Ta c: VB = VBEsat = 0.8V v VC = VCCRCICVLED = VCCRCILEDVLED Hay 0.8V > VCCRCILEDVLED = 5V RCx20mA1.5V RC > (5V1.5V0.8V)/20mA = 135 RC > 135 Ngoi ra cn iu kin VCE 0 => VCCRCILEDVLED 0 => RC (VCC VLED)/ILED= 175 Tm li: 135 < RC 175 Ch :

Cch 2 cho tr chnh xc hn v VCEsat = 0.2V l tr tiu biu m trong thc t c th thay i nhiu.

Thc t thng phn cc cho JC dn thun (VBC = 0.7V).

c) Di gi tr ca RC cho BJT tch cc thun khi LED sng nu RB = 1K. Khi tch cc thun th JC c phn cc ngc, ngha l VBC < 0 hay VB < VC Ta c: VB = VBEsat = 0.8V v VC = VCCRCICVLED = VCCRCILEDVLED Hay 0.8V < VCCRCILEDVLED = 5V RCx20mA1.5V RC < (5V1.5V0.8V)/20mA = 135 RC < 135 Ch : Thc t phn cc VCB > 0.4V

d) Di gi tr ca RB cho BJT min bo ha khi LED sng nu RC c tr b) Khi bo ha th IBsat > ICsat Vi IBsat = (VI VBEsat)/RB v ICsat = ILED Suy ra RB < (VI VBEsat)/ ILED = 100(5V0.8)/20mA = 21K Nh vy: 0 < RB < 21K

e) BJT c l bao nhiu cho BJT min bo ha khi LED sng nu RB = 1K v RC =170. Khi bo ha th IBsat > ICsat Vi IBsat = (VI VBEsat)/RB v ICsat = ILED

Suy ra > ICsat / IBsat = RB ILED/(VI VBEsat)= 1K x 20mA/(5V0.8V) = 4.76 Nh vy: > 4.76 (cc BJT thng dng u c t vi chc n vi trm => lun tha)

BT v BJT4

6. Thit k mch np pin theo hnh 4 c th chnh c dng np pin t 10mA n 100mA. C ngha l ta phi tm cc gi tr VCC, R1, R2 (bin tr) lm cho Q1 (c = 100) hot ng nh ngun dng hng chnh c IC = 10mA n 100mA. S. Gi s BJT min tch cc thun v dng m hnh tn hiu ln vi VBE=0.7V. Khi IC = IB vi IB=(VCCVBE)/(R1+R2) v thay vo yu cu vi IC: 10mA IC 100mA , ta c:

1 2

10 100CC BEV VmA mAR R

Bin tr R2 c th c chnh t 0 n R2, v IC max khi R2=0. Nh vy ta c th chn R1 bng:

1

100 CC BEV VmAR

hay 1 10 CC BER V V [] Ta c th chn VCC=12V (ch cn ln hn in p danh nh ca pin np c), suy ra R1=11.3K Thc t th ta phi dng in tr vi gi tr chun, do nn chn R1=12K (ngha l IC max hi nh hn 100mA). Gi tr ca R2 c tm nh sau:

2 110CC BEV VR R

mA

Thay cc tr s vo ta c R2 = 101.7K Nu dng in tr chun th chn bin tr 100K cho R2, khi IC min hi ln hn 10mA! Ch : Thc t th cc pin NiCd nh loi 9V c lm t 8 pin nh 1.2V, ngha l in p danh nh ca n l 9.6V. Khi pin c np y th mi pin nh c in p l 1.3V, dn n in p np y l 10.4V. Nh vy ch cn chn VCC > 10.4V + VCEsat th t yu cu! 7. Hy tm im tnh ca BJT trong mch hnh 5. Cho trc cc gi tr linh kin: R1 = 100K; R2 = 50 K; RC = 5 K; RE = 3 K; VCC = 15 V; VBE(on) = 0.7 V, v = 100. S. tin tnh ton ta s bin i mch cc nn thnh mch tng ng Thvenin nh sau:

Vi VBB=VCCR2/(R1+R2) v RB=R1R2/(R1+R2). Gi s BJT min tch cc thun, nu tnh ra khng ng th ta phi i li gi thit BJT min bo ha v tnh li! Ti mch nn-pht ta c phng trnh sau:

VBB = IBRB + VBE + IERE = [RB + ( + 1)RE ]IB + VBE Suy ra dng IB c tnh theo cng thc sau:

1BB BEB B EV VI

R R

T suy ra: IC = IB Ti mch cc thu, ta c: VCC = ICRC + VCE + IERE = ICRC + VCE + ( + 1)ICRE/ Suy ra:

1CE CC C C EV V I R R

BT v BJT5

Nu >> 1 th ta c CE CC C C EV V I R R Thay cc gi tr vo ta c: VBB = 5V; RB = 33.3 K; IB = 12.8 A; IC = 1.28 mA; VCE = 4.78 V > VCEsat = 0.2V (ng tch cc thun) Nh vy BJT c im tnh:

VCEQ = 4.78 V ICQ = 1.28 mA IBQ = 12.8 A

Hnh 6 Hnh 7 8. Mch hnh 6 c 2 BJT ging nhau hon ton v c phn cc min tch cc thun. Hy tm:

a) Dng IX khi V1=V2 b) Hiu s V1V2 cho IC1 = 10IC2

S. a) Dng IX khi V1=V2

Ta c: IX = IC1 + IC2 vi IC1 = IC2 = ISexp(V1/VT) Hay IX = 2ISexp(V1/VT )=ISX exp(V1/VT ) Khi ging nh tng ng 1 BJT c dng bo ha ngc l ISX v din tch mt ct ngang gp i BJT ban u!

b) Hiu s V1V2 cho IC1 = 10IC2 Lp t s 2 dng IC ta c

11 2

2

1

2

TT

T

VV VV

VC SV

C VS

I I e eI

I e

suy ra: 11 2

2

ln CTC

IV V VI

V1V2 = VT ln10 = 25mV x ln10 58mV 300oK 9. Xt mch hnh 7, trong V1 biu din tn hiu c to ra t microphone, IS = 3 x 1016A, = 100, VA= v Q1 hot ng min tch cc thun.

a) Nu V1= 0, hy xc nh cc tham s tn hiu nh ca BJT? b) Nu V1= 1mV, hy tm nhng thay i trong dng thu v dng nn?

S. a) Tnh cc tham s tn hiu nh. Cho V1 = 0 ta tm im tnh ca BJT

Ta c IC = IS exp(VBE/VT) = 3 x 1016A exp(800mV/25mV) = 23.69 mA Suy ra h dn gm = IC/VT = 23.69mA/25mV =0.9476 mho V in tr vo r = /gm = 100/0.9476mho 103.53

b) Nu V1= 1mV, hy tm nhng thay i trong dng thu v dng nn? Vi mch tng ng tn hiu nh ta thy V1 = v, do ta c s thay i dng thu l:

IC = gmV1 = 0.9476 mho x 1mV = 0.9476 mA V s thay i dng nn l:

IB = IC / = 0.9476 mA /100 = 9.476 A (c th tnh IB = v / r = 1mV/(100/0.9476) = 9.476 A)

BT v BJT6

Mch tng ng tn hiu nh ca hnh 7

10. Hy xc nh in p ra Vo v dng IC trong mch sau. V tnh tip gi tr cc tham s tn hiu nh ca BJT ny. Gi s BJT c = 100 v VA = .

S. d tnh ton ta s v li mch vi tng ng Thvenin cho mch phn cc ti cc nn:

Nh vy ta c: Vo = 1.07V; IC = 1.93mA; gm = 0.072 mho ; v r =1.39K

BT v BJT (2011) 1

Chng 5 Bi tp v BJT (AY1112-S1)

(Gi s cc BJT u lm t Si v in p JE dn l 0.7V v bo ha:VBEsat=0.8 v VCEsat=0.2V)

1. Vi cc trng hp sau BJT hot ng min no? a) NPN: VCB = 0.7V, VCE = 0.2V b) NPN: VBE = 0.7V, VCE = 0.3V c) NPN: VCB = 1.4V, VCE = 2.1V

d) PNP: VEB = 0.6V, VCE = 4V e) PNP: VCB = 0.6V, VCE = 5.4V f) PNP: VCB = 0.9V, VCE=0.4V

2. Hy xc nh xem cc BJT trong hnh 5.1 c phn cc min hot ng no v tm IB, IC, VCE (bit BJT c = 100)?

Hnh 5.1 Hnh 5.2 Hnh 5.3 Hnh 5.4

3. Mch trong hnh 5.2 vi BJT c = 150 v VCC = 12V, RC = 1.5K. Hy xc nh gi tr ca RB BJT lm vic min: a) tch cc thun; b) bo ha.

4. Mch trong hnh 5.3 vi BJT c = 150, VCC = 12V, RC = 1.5K v RE = 500. Hy xc nh gi tr ca RB BJT lm vic min: a) tch cc thun; b) bo ha.

5. Mch trong hnh 5.4 vi BJT c = 150 v VCC = 12V, RC = 1.5K. Hy xc nh gi tr ca RB BJT lm vic min: a) tch cc thun; b) bo ha.

6. Hnh 5.5 cho thy cc in th o c ti cc chn B v E ca cc BJT. Hy tm ca mi BJT trong mch, bit cc BJT ang ch tch cc thun.

Hnh 5.5

BT v BJT (2011) 2

7. Xt BJT loi NPN ch tch cc thun. a) Hy xc nh IE, v cho BJT c IB = 5 A v IC =0.62 mA b) Hy xc nh IB, IC, v cho BJT c IE = 1.2 mA v =0.9915

8. Hy xc nh min lm vic ca BJT loi NPN c = 100 cho cc trng hp sau: a) IB = 50 A v IC = 3 mA. b) IB = 50 A v VCE = 5 V. c) VBE = 2 V v VCE = 1 V.

9. Mch hnh 5.6 thng dng xc nh v IS ca BJT. Cc diode trong mch c phn cc thun v c st p trn mi diode l 0.7V. Nu ngi ta o c IB = 6.28 A v IE = 9.29 mA, hy xc nh v IS.

Hnh 5.6 Hnh 5.7 10. Cho mch hnh 5.7 vi BJT c = 80 v IS = 5x1014A. Hy tm cc gi tr in tr RB v RC sao cho BJT

lm vic ch tch cc vi IC = 8mA v VCE = 5V (bit VCC = 12V). 11. Cho trc cu to ca 3 BJT hnh 5.8. Gi s cc BJT ny c ch to cng vt liu Si v c tham s ch

to ging nhau tr cc tham s c thay i nh trong hnh 5.8. Hy so snh gi tr hiu sut pht e, h s vn chuyn min nn B, v ca BJT 1 vi 2 BJT cn li?

Hnh 5.8

12. BJT c in tr ra ro = 225 K IC = 0.8 mA. Hy tm:

a) in p Early VA. b) p dng kt qu ca a) tm in tr ra ro IC = 0.08 mA v IC = 8 mA..

13. Xt BJT ch tch cc thun c IC = 1 mA khi VCE = 1 V, v VBE c gi khng i. Hy xc nh IC khi VCE = 10 V nu: a) VA = 75 V b) VA = 150 V

BT v BJT (2011) 3

14. Hy tm hiu sut pht e, h s vn chuyn min nn B v ca BJT NPN vi cc tham s sau: NDE = 1x1018cm3, NAB = 1x1016cm3, Dn = Dp, WB = 100nm, v Lp= Ln = 1m.

Hnh 5.9 Hnh 5.10 Hnh 5.11 Hnh 5.12 15. Cc in p phn cc trong mch hnh 5.9 l V+=3.3V v V = 3.3V. in th o c ti collector l

VC=2.27 V. Hy xc nh IB, IC, IE, v . 16. Cc in p phn cc trong mch hnh 5.10 l V+= 5V v V = 5V. Gi s BJT ny c =85. Hy xc nh IB,

IC, IE v VEC. 17. Vi mch hnh 5.11, hy tm RE v RC BJT c IEQ=0.125 mA, VECQ=2.2V. Bit BJT c =110. 18. Vi mch hnh 5.12, BJT c = 0.992. Hy tm RE IE = 1mA. T tm tip IB, IC v VBC.

Hnh 5.13 Hnh 5.14 Hnh 5.15

19. Cho mch hnh 5.13 vi BJT c =100, ngi ta mun LED tt khi VI = 0V v LED sng vi ILED= 15mA

v VLED = 1.5V khi VI=5V. Hy tnh cc gi tr linh kin RB1 v R1 cho cc trng hp sau: a) LED sng vi BJT ch bo ha. b) LED sng vi BJT ch tch cc thun.

20. Cho mch hnh 5.14 vi BJT c =150, hy tm IC, IE v VC cho cc trng hp sau: a) VB = 0.2V; b) VB = 0.9V; c) VB = 1.5V; v d) VB = 2.2V.

21. Cho mch hnh 5.15 vi BJT c =80, hy xc nh a) VI c VCEQ = 6V. b) Di tr s ca VI c: 3V VCEQ 9V.

BT v BJT (2011) 4

22. Cho mch hnh 5.16 vi BJT c IS = 6 x 1016A v >> 1, hy tm VX trong trng hp: a) khng dng xp x VBE ; b) dng xp x VBE=0.7V.

Hnh 5.16 Hnh 5.17 23. Cho mch hnh 5.17 vi BJT c IS = 7 x 1016A v =100, nu R1 = 10K hy tm VB cho IC = 1mA

(bit collector c ni vi ti ln ngun dng vn duy tr BJT ch tch cc thun) trong trng hp: a) khng dng xp x VBE ; b) dng xp x VBE=0.7V.

24. Hy xc nh im hot ng v m hnh tn hiu nh ca Q1 cho cc trng hp trong hnh 5.18. Gi s BJT c IS = 8 x 1016A, =100 v VA =.

Hnh 5.18

Hnh 5.19 Hnh 5.20

BT v BJT (2011) 5

25. Cho mch hnh 5.19 vi BJT c IS = 5 x 1017A , hy tm VX trong trng hp: a) VA = ; b) VA =5V. 26. Cho mch hnh 5.20 vi BJT c IS = 1 x 1017A v VA=5V, hy tm s thay i trong dng IC nu VCC thay

i t 2.5V n 3V.

Hnh 5.21 Hnh 5.22 27. Cho mch hnh 5.21 vi RE = 600, RC = 5.6K, R1 = 250K, R2 = 75K, = 120 v VA =.

a) Tnh im tnh Q ca BJT (ICQ v VCEQ). b) Tnh cc tham s H ca m hnh tn hiu nh. c) Tnh h dn gm v li p AC AV = v0/vS d) in tr nhn cc nn ca BJT Rib.

28. Cho mch hnh 5.22 vi BJT c = 100 v VA =. a) Tnh im tnh Q ca BJT (ICQ v VCEQ). b) Tnh cc tham s H ca m hnh tn hiu nh. c) Tnh h dn gm v li p AC AV = v0/vS

Hnh 5.23

29. Cho mch hnh 5.23 vi BJT c = 100 v VA =. Hy tm m hnh tn hiu nh hnh v hnh T cho mi trng hp.

30. Cho mch hnh 5.24 vi BJT c = 100 v VA =. Hy tm m hnh tn hiu nh hnh v hnh T cho mi trng hp.

Hnh 5.24

BT v MOSFET1

Chng 7 Bi tp v MOSFET

Ch : Trong cc BT sau, ta gi s cc N-EMOS c nCox=200A/V2 v VTN=0.4V; cc P-EMOS c pCox=100A/V2 v VTP= -0.4V 1. Xc nh min hot ng ca M1 trong hnh sau:

2. Xc nh min hot ng ca M1 trong hnh sau:

3. Mch hnh 3 dng 2 MOSFET ging nhau mc ni tip. Gi s 2 MOSFET ny hot ng

min in tr, chng t rng chng tng ng nh 1 MOSFET Meq vi Weq v Leq l bao nhiu?

Hnh 3 Hnh 4

BT v MOSFET2

4. Mch hnh 4 dng MOSFET lm R thay i c. Phi chn W/L l bao nhiu mch ny cho Vout= 95%Vin (gi s ln ca Vin < 0.1V, RL=100 v VG=1.8V).

Hnh 5 Hnh 6 Hnh 7 Hnh 8 Hnh 9

5. Vi mch hnh 5, gi s =0, tnh W/L cho M1 min bo ha? T xem c g xy ra nu M1 c dy lp oxide gp i gi tr c?

6. Vi mch hnh 6, gi s =0, hy tm gi tr ti thiu ca VDD cho M1 khng i vo min triode?

7. Vi mch hnh 7, gi s =0, hy tm quan h gia cc tham s mch M1 hot ng cnh min bo ha (VDS=VDSsat).

8. Vi mch hnh 8, gi s =0, tm W/L c dng phn cc ID=I1. 9. Vi mch hnh 9, gi s =0, tm dng phn cc ID.

10. Vi mch sau, hy v c tuyn IX theo VX khi VX thay i t 0 n VDD=1.8V, gi s =0. Trn c tuyn xc nh tr VX m ti MOSFET thay i min hot ng.

Hnh 10

11. Vi mch sau hy v mch tng ng tn hiu nh nu cc MOSFET c =0.01V-1 v W/L=20/0.18.

Hnh 11

BT v MOSFET3

12. Hy v mch tng ng tn hiu nh ca cc mch sau: (gi s 0)

13. Tm min hot ng ca cc P-EMOS trong cc mch sau:

14. Mch hnh 14 dng cc N-EMOS c VTN=1V, nCoxW/L=2A/V2, v = 0. Hy tm cc gi

tr ca VS1, VD2 v IS.

Hnh 14 Hnh 15 Hnh 16 15. Mch hnh 15 dng 2 NMOS ging nhau v cc tham s Iref=1mA, R=7K, v VTN=1V.

a) Nu RL=2K, khi dng I bng bao nhiu? b) Gi tr ca RL l bao nhiu dng I gi khng i.

DCBDBTOTAY1415-S1trang 1/6

HBK TpHCMKhoa TBMT GVPT: H Trung M

Bi tp n thi mn Dng c bn dn AY1415-S1 (Ngoi cc bi tp cc chng BJT, JFET v MOSFET, SV lm thm cc BT sau)

Ch : Thi cui k khng c chng JFET 1. Vi cc trng hp sau BJT hot ng min no?

a) NPN: VCB = 0.7V, VCE = 0.2V b) NPN: VBE = 0.7V, VCE = 0.3V c) NPN: VCB = 1.4V, VCE = 2.1V

d) PNP: VEB = 0.6V, VCE = 4V e) PNP: VCB = 0.6V, VCE = 5.4V f) PNP: VCB = 0.9V, VCE=0.4V

2. Vi hnh 1 cho trc VCC = 12V v BJT c = 100, hy tm RB v RC cho BJT min tch cc thun c im hot ng DC (im tnh Q) l ICQ = 3mA v VCEQ = 6V. Gi s gi RB vi gi tr va tm c, hy tm gi tr RC mch vn min tch cc thun.

Hnh 1 Hnh 2 Hnh 3 Hnh 4

3. Vi hnh 2, hy tm RB, RC v RE cho BJT min tch cc thun c im hot ng DC (im tnh Q) l ICQ = 4 mA v VCEQ = 4V. Bit VCC = 12V, VE = 4V v BJT c = 100. 4. Vi hnh 3, hy tm RB v RC cho BJT min tch cc thun c im hot ng DC (im tnh Q) l ICQ = 2 mA v VCEQ = 4V. Bit VCC = 10V v BJT c = 100. 5. Vi hnh 4, hy tm cc gi tr ca RC v RE cho BJT ch tch cc thun (bit =100) vi IC = 2mA v VC = 4.3 V. 6. Hy tm im lm vic DC (ICQ v VCEQ) ca BJT trong cc mch sau v cho bit n ang ch no. Gi s cc BJT c = 50 v bit BJT nu ch tch cc thun c |VBE| = 0.7 V..

Hnh 5

7. Xt BJT loi NPN ch tch cc thun. a) Hy xc nh IE, v cho BJT c IB = 5 A v IC =0.62 mA b) Hy xc nh IB, IC, v cho BJT c IE = 1.2 mA v =0.9915

8. Hy xc nh min lm vic ca BJT loi NPN c = 100 cho cc trng hp sau: a) IB = 50 A v IC = 3 mA. b) IB = 50 A v VCE = 5 V. c) VBE = 2 V v VCE = 1 V.

9. H1x1

10. v V11. 12. 13.

14. Ccchn

15. Gi

Hy tm hiu017cm3, NA

Cho mch

VLED = 2 V kCho mch Cho mch

a) khnCho mch

Cho mch c t CC v Cng v khi pha) Tnh ib) Tnh cc) Tnh hd) in trs c 2 BJTa) Mun cb) Nu IR =c) Mun c

BJT phd) Trng

gp 2 l

u sut pht AB = 1x1015c

Hnh 6 hnh 6 vkhi VI=5V. hnh 7 v hnh 8.(a)

ng dng xp hnh 8.(b)

hnh 9 vCE ng vai

hn tch DCm tnh Q c

c tham s h dn gm v vo nhn T c cng c dng in= 2 mA th

c dng ra Ii tha iu

g hp 2 BJTn din tch

D

e, h s vncm3, Dn = D

i BJT c =Hy tnh ci BJT c =

) vi BJT cp x VBE ; ) vi BJT c

Hnh 9 i VCC = 5V,tr ghp v

C s h mcca BJT (ICQh ca m hn li p A cc nn c

10c tnh. n chun IR =in tr ti

IOUT c sai su kin g? T khng c c

min pht c

DCBDBTO

n chuyn mDp, WB = 12

H=100, ngc gi tr lin=100, hy x IS = 5 x 10 b) d

IS = 5 x 10

, RE = 600 bypass tn ch chng. Q v VCEQ).nh tn hiu n

AC AV = Vouta BJT.

VCC = V

= 1mA th RRL phi th

s so vi IRcng c tnca Q1.

OTAY1415

min nn B v20nm, v Lp

Hnh 7 i ta mun Lnh kin RB1 vxc nh VI016A v >dng xp x017A , hy t

, RC = 5.6Khiu AC, ng

nh. t/Vin

VEE = 6V v

R = ? a iu kinkhng vt

nh th IOUT =

5-S1trang

v ca BJTp= Ln = 2m

(a

(b

LED tt khi v R1 cho c VCEQ

>> 1 ( cVBE =0.7V.tm VX tron

K, R1 = 25gha l khi v

n g mcht qu 4% (n

= ? Nu IR =

2/6

T NPN vi m.

a)

)

VI = 0V v o LED sng= 6V.

IE IC), h

ng trng h

Hn50K, R2 = v m hnh

VBEQ

h trn vn lngha l (IR= 1mA v di

cc tham s

Hnh 8 LED sng

g vi BJT y tm VX tr

p: a) VA =

nh 10 75K, = tn hiu nh

Q = 0.7V, VA

ngun dnIOUT)/IR 4in tch mi

sau: NDE =

vi ILED = 2 ch borong trng ; b) VA =

120 v VA = ta s nt t

VA= , = 1

ng? 4%) th cn pht ca Q

=

20mA o ha.

g hp: 5V.

=. tt

00.

a Q2


16. N-JFET c IDSS = 10mA v VTH = 4V. Hy cho bit min hot ng ca JFET ny nu ngi ta o c cc in th ti D, G v S so vi t trong cc trng hp sau:

a) VD = 5V, VG = 3.5V, v VS = 4V. b) VD = 4V, VG = 1V, v VS = 5V. c) VD = 5.5V, VG = 1V, v VS = 5V.

17. P-JFET c IDSS = 10mA v VTH = 4V. Hy cho bit min hot ng ca JFET ny nu ngi ta o c cc in th ti D, G v S so vi t trong cc trng hp sau:

a) VD = 4V, VG = 3V, v VS = 2V. b) VD = 4V, VG = 2V, v VS = 1V. c) VD = 5V, VG = 1V, v VS = 2V.

18. N-JFET c VTH = 4 V, IDSS = 10 mA v VA = . a) Vi VGS = 2 V, hy tm VDS ti thiu dng c hot ng min bo ha. Tnh ID vi VGS = 2 V

v VDS = 3 V. b) Cho VDS = 3 V, hy tm s thay i trong ID tng ng vi s thay i VGS t 2 n 1.6V. c) Vi VDS nh, tnh gi tr ca rds VGS = 0 V v VGS = 3 V. d) Nu VA = 100 V, hy tm in tr ra ro ca JFET khi hot ng trong min bo ha vi dng in ID l

1 mA, 2.5 mA, v 10 mA.

Hnh 11 Hnh 12 Hnh 13 Hnh 14 19. JFET trong mch hnh 11 c VTH = 3 V, IDSS = 9 mA, v VA = V. Hy tm tt c cc gi tr in tr VG = 5V, ID = 4 mA, v VD = 11V. Gi s chn dng 0.05 mA chy qua mch chia p v VDD = 15V. 20. Mch hnh 12 c VTH = 5 V v IDSS = 10 mA.

a) Nu RD = 1 K, hy tm gi tr VDD cho JFET vn ch bo ha? b) Nu RD = 1 K, hy tm gi tr VDD cho JFET vn ch triode? c) Nu VDD =12V, hy tm gi tr RD cho JFET vn ch bo ha? d) Nu VDD=12V, hy tm gi tr RD cho JFET vn ch triode?

21. JFET trong hnh 13 c IDSS = 1mA v VTH = 4V. Cho trc VDD = VSS = 5V v RD= 0, hy tm ID v VS vi: a) IQ = 0.5mA; b) IQ = 2mA 22. JFET trong hnh 14 c IDSS = 0.25 mA v VTH = 2V. Bit VDD = VSS = 6V, hy tm im tnh Q cho JFET vi: a) RD = 0 v RS = 100 K; b) .RD = 0 v RS = 10 K; c) RD = 22 K v RS = 100 K. 23. Vi mch hnh 14, JFET c cc tham s VTH= 3.5 V, IDSS = 18 mA, v VA = V. Cho trc VDD = VSS = 15V, IQ = 8mA v RD= 0.8K, hy tm VDS. 24. Vi mch hnh 14, JFET c cc tham s VTH = 3 V, IDSS = 10 mA, v VA = V. Cho trc VDD = 12V v VSS = 0, ta mun mch hot ng nh ngun dng 5 mA th phi chn RS = ?. Khi c yu cu g vi RD? 25. Vi mch hnh 11 c R1=150K, R2=50K, RD=RS=2K, IDSS =4mA, VTH = 2V v VDD=12V. Hy tm

a) im tnh (VGSQ, IDQ) v h dn gm ca n. b) Tn s fT nu JFET c Cgs = 20pF v Cgd =5pF.

26. Cho trc N-JFET c IDSS = 10 mA v VTH = 4V v gi s in p vo VIN c ln < 0.1V. Thit k mch chia p (cc D c ni vi in tr 1K ln VIN v cc S c ni xung t) chnh c bng in p VGS v in ra VOUT = VDS.

a) Hy tm biu thc VOUT/VIN. b) C th chnh c VOUT/VIN = 0.5 khng? Gii thch.


27. Hy cho bit min hot ng ca N-EMOS c VTN = 0.4V nu ngi ta o c cc in th ti D, G v S so vi t trong cc trng hp sau:

a) VD = 2 V, VG = 0.7 V, v VS = 0.5 V b) VD = 0.5 V, VG = 1 V, v VS = 0.5 V c) VD = 2 V, VG = 1.5 V, v VS = 0.5 V

28. Hy cho bit min hot ng ca P-EMOS c VTP = 0.4V nu ngi ta o c cc in th ti D, G v S so vi t trong cc trng hp sau:

a) VD = 0 V, VG = 2 V, v VS = 2 V b) VD = 0.3 V, VG = 0 V, v VS = 1 V c) VD = 3 V, VG = 0.6 V, v VS = 2 V

29. Tnh ID v VDS ca N-EMOS trong hnh 15. Bit MOS c VTN = 1V v n ox WC L = 0.5 mA/V2.

Hnh 15 Hnh 16 Hnh 17 30. Tnh ID v VDS ca N-EMOS trong hnh 16. Khi EMOS c

a) VTN = 4 V v n oxWCL

= 2 mA/V2; b) VTN = 2 V v n ox WC L = 4 mA/V2.

31. Tnh ID v VDS ca N-EMOS trong hnh 17. Bit EMOS c VTN = 1V v n ox WC L = 0.5 mA/V2.

32. Tnh ID v VDS ca N-EMOS trong hnh 18. Bit EMOS c VTN = 1V v n ox WC L = 0.5 mA/V2.

Hnh 18 Hnh 19 Hnh 20 33. Mch hnh 19 c VDD = 10V, R1 = 800K, R2 = 500K, RD = 4K, RS = 1K v dng N-EMOS c VTN=1V, n ox

WCL

=100 A/V2 , v VA = 200V. Hy tnh im tnh Q (ID, VDS) v VGS ca N-EMOS ny (khi tnh IDQ ta tnh gn ng vi =1/VA=0) v tm cc tham s gm v in tr ra ro. 34. Xt mch hnh 20 vi vi=0, N-EMOS c VTN = 2V, n ox WC L = 0.5mA/V

2, ta mun c ID = 0.4mA khi phi phn cc VGG bng bao nhiu? Hy tm gii hn in tr ti ca mch ny cho MOSFET vn min bo ho vi VGG va c tm ra?


TD mt s cu hi trc nghim Cu 35: Hnh 21 vi N-EMOS c n ox WC L =1000A/V

2, VTN=0.7V, VG=5V v VD=0.2V. Mch cho dng in a) I = 840 A b) I = 800 A c) I = 760 A d) I = 720 A e) 4 S trn u sai Cu 36: Vi mch hnh 22, gi s =0, VTN=0.5V, n oxC = 2mA/V2, M1 vn min bo ha khi W/L c tr s a) > 4.5 b) 1.3 c) 5.2 d) 4.8 e) 4 S trn u sai Cu 37: Hnh 23 l mch gng dng in vi M1 v M2 c cc tham s: 1 22n ox n oxM MC C ; VTN,M1=VTN,M2; M1=M2=0. Mun c Ix=3Ibias th (W/L)M2/(W/L)M1 bng a) 4 b) 6 c) 3/2 d) 2/3 e) 4 S trn u sai Cu 38: Mch hnh 24 c R1=800K, R2=500K, RD=4K, RS=1K v dng N-EMOS c VTN=1V,

n oxWCL

=2mA/V2 , v VA=150V. MOSFET ny c gm (khi tnh IDQ cho =0) v in tr ra ro l (gm; ro) a) (3.5mS;94.3K) b) (3.2mS;80K) c) (2.85mS;53K) d) (2.52mS;94.34K) e) 4 S trn u sai

Hnh 21 Hnh 22 Hnh 23 Hnh 24 Cu 39: Mch gng dng in hnh 25 vi cc BJT c cng VBEQ = 0.7V, VA= , = 100 v c im cu to ging nhau, ch khc v din tch min pht ca Q2 gp 2 ln din tch min pht ca Q1. Gi s gi tr RL vn lm cho Q2 ch tch cc thun, VCC =3V v RREF =2.3K, khi dng qua RL l a) 1.91mA b) 1.96mA c) 2.01mA d) 2.06mA e) 4 S trn u sai Cu 40: Vi hnh 26, ngi ta o c cc in th VB = 4.3V v VC = 1V. T suy ra BJT ny c l a) 49 b) 50 c) 66 d) 70 e) 4 S trn u sai Cu 41: Hnh 27 vi R1=10K, R2=5K, RC =RE=1K, =100, VCC=15V v VBE=0.7V. BJT c tr s hie l a) 601 b) 621 c) 646 d) 692 e) 4 S trn u sai Cu 42: Hnh 27 vi R1=10K,R2=5K,RE=1K,=100,VCC=15V v VBE=0.7V. BJT vn tch cc thun khi RC: a) < 2255 b) < 2352 c) < 2453 d) < 2557 e) 4 S trn u sai Cu 43: Mch hnh 27 vi R1=10K, R2=5K, RC =RE=1K, =100, VCC=15V v VBE=0.7V. Nu BJT c Cbe=20pF v Cbc=5pF th tn s ct fT c tr s l a) 32.71MHz b) 30.15MHz c) 28.54MHz d) 26.49MHz e) 4 S trn u sai Cu 44: BJT trong hnh 28 c phn cc ch a) Tt b) Tch cc ngc c) Tch cc thun d) Bo ha e) 4 S trn u sai Cu 45: Mch khuch i hnh 29 dng BJT c VBEQ=0.7V. Ngi ta chn tr s R1 v R2 sao cho VCE=3V v IC=1.5mA khi =150. Nu =200 th im tnh DC mi l (VCE; IC) a) (1V; 2.5mA) b) (2.5V; 2mA) c) (2V; 2mA) d) (2V; 2.5mA) e) 4 S trn u sai


Hnh 25 Hnh 26 Hnh 27 Hnh 28 Hnh 29

Cu 46: Vi hnh 30 c RL = 2K, JFET c phn cc min bo ha th VDD ti thiu bng a) 11 V b) 9 V c) 7 V d) 5 V e) 4 S trn u sai Cu 47: Vi hnh 31 c R1=140K, R2=60K, RD=2.7K, RS=2K, IDSS=8mA, VTH = 4V v VDD=20V; JFET ny lm vic vi im tnh (VGSQ, IDQ) c tr xp x l a) (-0.8V; 2.65mA) b) (-1.30V; 3.65mA) c) (-1.70V; 4.7mA) d) (-1.5V; 3.68mA) e) 4 S trn u sai Cu 48: Cng s liu vi cu trn th JFET trong mch ny c tn s ct fT l (bit Cgs=20pF v Cgd=2pF) a) 25.08MHz b) 22.17MHz c) 21.34MHz d) 19.54MHz e) 4 S trn u sai Cu 49: Xt hnh 32, JFET c cc tham s VTH = 3.5 V, IDSS = 18 mA, v VA = V. Khi VDS c gi tr l a) 7.43 V b) 8.6 V c) 1.17 V d) 1.17 V e) 4 S trn u sai Cu 50: JFET trong hnh 33 (c R= 1K, IDSS=20mA v VTH = 4V) c dng lm in tr c iu khin bng p, ta mun c t l chia p Vout/Vin = 1/5 th phi phn cc JFET vi VGS l a) 1.2 V b) 1.6 V c) 2.0 V d) 2.4 V e) 4 S trn u sai

IDSS = 2mA VTH = 3V

Hnh 30 Hnh 31 Hnh 32 Hnh 33

Cu 51: Mt diode n p c in p nh thng VBR= X1 Volt (X1 >0), khi nhit tng th VBR= X2 Volt (0 < X1 < X2). T ta suy ra diode ny n p da trn c ch nh thng:

a) do nhit b) ng hm c) thc l d) Zener e) c 4 S trn u sai

Cu 52: Cho mch n p hnh sau vi VS = 8V, R = 50 v Zener c VZ=5V, IZmin=10mA v IZmax = 50mA. cho mch vn cn n p th in tr ti RL phi thuc di gi tr:

a) 125 < RL < 250 b) 125 < RL < 500 c) 100 < RL < 250 d) 100 < RL < 500 e) c 4 S trn u sai

DCBD_AY1415-HD_on_thi_LT_cuoi_kyDCBD-Ch05-Bai_tap_giai_san_ve_BJT_2011DCBD-Ch05-Bai_tap_ve_BJT_2011DCBD-Ch07-Bai_tap_ve_MOSFETDCBD_AY1415-Bai_tap_on_thi_cuoi_ky

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