in t c bnChng 4 . Mch phn cc Transistor lng cc ni

Mch Transistor Transistor hot ng ta phi cp in DC cho cc cc B,C,E ( phn cc) xc nh im tnh iu hnh Q ( IB, IC, VCE).

Hai mch transistor c bn: Khuch i Giao hon

ty theo dng mch ta c cch phn cc tng ng .

I.Phn cc mch khuch i rp CE1.Phn cc c nh p dng nh lut Kirchhoff v th ta c :Mch nn - pht: VCC = RBIB + VBE (1) IB = ( VCC VBE) / RB (2)Mch thu - pht: IC = IB (3) VCC = VCE + RCIC (4) VCE = VCC RCIC (5)

ng thng ti tnh:Phng trnh ng ti tnh: IC = ( VCC VCE ) / RC Ic (mA) ICM = DCLL ( -1/RC) VCC/RC ICQ Q IBQ

0 VCEQ VCC VCE (V)

Phan cc co nh co RETa co:

Khi T tang, IC tang VE =REIE tang VBEgiamIB giam IC giam lai, can bang lai. RE la ien tr on nh nhiet

2. Phn cc bng cu phn th v REMch in:

R1, R2 in tr phn cc. RC in tr cp in RE in tr n nh nhit .

L mch rt c thng dng.

Mch in tng ngTheo nh l Thevenin:

VBB = [R2 / (R1+ R2)] VCC (1) RB = R1R2 /( R1+R2) (2) Theo nh l Kirchhoff: VBB = RBIB + VBE + REIE (3)

(4)

Mch thu pht :

v :

im tnh iu hnh cho bi ( 4), (5), (8)

ng thng ti tnhPhng trnh ng Ic (mA)

ti tnh: ICM DCLL( - 1 / (RC + RE )

ICQ Q IBQ

vi:

0 VCEQ VCC VCE(V)

Vai tr ca in tr n nh nhit REKhi nhit T tng , ICBO tng , IC tng VE tng VBE = (VBB VE) gim IB gim IC gim li chng li s gia tng ni trn, gi Transistor khng h.

Cch mc RE c gi l mch hi tip m lm mch n nh nhit v ci tin cc i lng khc tt hn( di thng, tng tr,nhiu, bin dng).

3.Phn cc bng in tr cc thu-nnMch in thu-nn:

VCE = RBIB +VBE (1) IB = (VCE-VBE) / RB (2) v : (3) Mch thu pht: VCC= RC(IC +IB) +VCE (4) VCE =VC = VCC RC( IC+IB) (5) Ch : Trong (1) nu cha bit VC th phi tnh t VCC= RC ( IC + IB) + RBIB + VBE (1). IB = ( VCC VCE) / [ RB +( +1)RC] (2)

ng ti tnh DCLLPhng trnh DCLL:

Ic ( mA)

ICM DCLL ( - 1/ RC)

ICQ Q IBQICM =VCC/ RCVCEM = VCC

0 VCEQ VCC VCE(V)

Vai tr ca in tr hi tip RB c c s n nh nhit tt hn, cn kt hp c 2 in tr RB v RE ( xt on sau).Khi nhit T tng IC tng VCE gim VB gim IB gim IC gim chng li s gia tng trn, lm mch n nh nhit .y l loi mch thng s dng cc mch tin khuch i Micro( my vi m)

4.Phn cc bng in tr hi tip RB v REMch thu - nn :

VCC= RC(IC+IB)+ RBIB+ + VBE+ REIE (1) = RC( +1)+RBIB + VBE+ REIE

IC = IB (3)Mch thu pht: VCC= RC(IC+IB)+ VCE + REIE VCE = VCC ( RC + RE ) IC ( 4)

ng thng ti tnh DCLLPhng trnh DCLL:

Ic ( mA)

DCLL( - 1/ (RC + RE) ICM

ICQ Q ICM =VCC/ ( RC+RE)

0 VCEQ VCC VCE(V)

5.H s n nh nhit SKhi nhit thay i, cc thng s transistor thay i nh sau: ICBO tng gap i khi nhit tng ln 10oC.[ 8oC ( Si); 12oC(Ge)]. tng gp i khi nhit tng 50oC ( Si) ; 80oC ( Ge). VBE gim theo 2,2mV / oC [ -2,5mV / oC (Si); - 1,6mV / oC ( Ge) ]. Vy dng thu l hm s: IC = f ( ICBO, , VCE )

S thay i dng thu cho bi:

Cc h s n nh nhit:

H s n nh nhit trong mch phn cc bng cu phn th v RE.Ta c:

VBB = RBIB + VBE + REIE= RBIB + VBE + RE(IB+IC) = = VBE+ RBIB+REIC+REIB= VBE+ (RB+RE)IB+REIC (1) IC = + (2).Thay (2) vo (1):

Hay:

Sp xp li:

Hay:

Tnh c:

Do : nn:

SI cng nho mch cng n nh ( 1- 11), SI = 11 l tr s ti u.

SI cng nh cng

Tng t:

V trong cng thc vn cn cha c ICBO, VCE,

nn ta c th tnh theo cch sau:

Suy ra:

Do :

Hay:

Ch : Do cch tnh cc h s n nh phc tp ,nn ta ch xt h s SI ca mch trn . KHi gii quyet SI tt th cc s n nh khc tong i c gii quyt.

6.Phn cc transistor pnpThng c 2 dng phn cc thng dng:

Ch nn c khi tht quen vi mch transistor npn.Xem gio trnh TCBXem bi tp 2.9 v 2.10

Cc cch phn cc bng ngun n dng, gng dng s xt chng IC

II.Phn cc mch Transistor Giao hon1. iu kin phn cc giao hon Khi ngng (off): Ic = 0 VCE = VCC (1) Khi bo ho: VBE = 0,7V v ICbh) = VCC / RC ( 2) bo ho Ic(mA) Icbh Q2 =Vcc/Rc ngng Q1 0 VCC VCE(V)

ng biu din hFE theo dng IC

c bo ho su ( chc chn bo ho) phi c: IB > IBbh (3)

Thng chn:

hay:

2.Mch o ( Inverter , NOT)a. Dng 1: Tho IB > IBbh RB < RC Khi Vi = 0V= ViL , Transistor ngng. in th ng ra Vo = VoH = Vcc = logic 1 Khi Vi = Vcc = ViH , Transistor dn bo ho , Vo = 0V = VoL=logic 0Ta c bng chn l: c gi l cng NOT hay Transistor o

AF0110

b. Dng 2Mch in: Tho iu kin (4): RB < RCKhi Vi =0V VBE = 0V Transistor ngng VO = VCC= VOH =logic 1Khi Vi >0V Transistor dn bo ho VO = 0,2V = VOL=logic 0 Vy mch l cng NOT

ng dng cng NOTMch iu khin LED(a),iu khin R-le(Relay) (b):

3. Cng Logic h DTLa.Cng NAND Gm NOT+ AND hay AND + NOT

A F B

BAF001011101110

Phn gii cng NAND

2 diod dn

Q ngng

D1 dn Q ngng

D2 dn Q ngng

2 diod ngng Q dn bh

BAVo0V0VVcc=5V0V5V5V5V0V5V5V5V0,2V

b.Cng NOR h DTL Gm cng NOT+ OR

A B F

BAF001010100110

Phn gii cng OR

2 diod ngng, Q ngng D1 dn, Q dn D2 dn, Q dn

2 diod dn,Qdn

BAVo0V0VVcc =5V0V5V0,2V5V0V0,2V5V5V0,2V

4.Cng Logic h TTLa .Cng NOR Khi A=B=0Q1,Q2 ngng Vo = Vcc= VoH = logic 1Khi A=Vcc , B=0 Q1 dn , Q2 ngng Vo=O,2V = VoL = logic 0KHi A=0,B=Vcc Q! ngng, Q2 dn Vo= 0,2V = VoL=logic 0Khi A=B=VCC Q1, Q2 dn Vo=0,2V = VoL = logic 0

b. Cng NAND ng ra n ccMch n gin ( xem hnh sau)Hin nay t s dngCch hot ng:

A=B=0V : Q1 dnQ2 ngngVo =5V=1A=5V,B=0V: Q1 dnQ2 ngng Vo=5V A=0V,B=5V: Q1 dnQ2 ngng Vo=5VA=B =5V: Q1 ngngQ2 dnVo=0,2V

Cng NAND chun (h TTL)

Khi A=B=0 Q1dn Q2ngng Q3 ngng, Q4 dn Vo = 2,4 3,6 V = = VoH = logic 1 Khi c hoc A hoc B xung 0 Q1 dn, Q2, Q3 ngng, Q4 dn Vo = VoH = logc 1KHi A=B=Vcc ni B-E 1 ngng, nhng ni C-B1 dn, Q2 dn Q3 dn,Q4 ngng Vo=0,2V = VoL

Giai thch cch hot ng:KHi A=B=0V

Q! dn Q2 ngng Q3 ngng, Q4 dn VO = 2,4V 3,6V= = VOH = = logic 1Khi ch c 1 ng vo ln cao v 1 ng vo thp: tng t trn VO=VOH

Khi A=B = Vcc = ViH

ni B-E1 ngng, nhng ni B-C1 dn Q2 dnQ3 dn v Q4 ngng VO = 0,2V 0,4V= = VOL = = logic 0

Bang s thatCong NAND co ngo cho phep ( Enable)

C B A F

1 x x 1 cam

0 0 0 1 A

0 0 1 1 B F

0 1 0 1 C 0 1 1 0

c. IC h TTLNi 2 ng vo A v B ca cng NAND li vi nhau ta c cng NOT (IC h TTL)

d. c tnh chung (chun) ca h IC TTL ViH = 2V Vcc VOH = 2,4V IiH = 40uA IOH = -400uA

ViLmax = 0.8V VOLmax= 0,4V IiL=-1,6mA IOL = 16mAFAN OUT = 16mA/1,6 mA= 400uA/40uA = 10

FAN OUT

OUT INVOHmax= 5V ViHmax = 5V NM 0,4VVOHmin=2,4V ViHmin=2V NM ViLmax=0,8V VOLmax=,4V0 0,4V 0VCac ac tnh khac se trnh bay sau.

Vung khongcho phep

Dai bat nh

o mien nhieu(Noise immunty)-Le nhieu VOH VIH

VOL VIL VNH= VOHmin VIHmin=2,4 2 =0,4V

Khi VOH = 2,4V thuc vao cong tai sau th ,cong nay hieu la mc cao va hoat ong ung. Khi co xung nhieu am > 0,4V th ViH < 2V, nen ri vao vung bat nh va cong tai se hoat ong sai. VNL= ViLmax VOLmax= 0,8V- 0,4V = 0,4 VTng t, khi xung nhieu dng>0,4V th ViL> 0,8V nen ri vao vung bat nh va cong tai hoat ong sai.Xung nhieu am khong anh hng .