- Xanh da trai, xanh la, den, nhO vang.
B. Hay pMn tfch m~ch dlinh Iw diu, vai tra cua b9 h?c trong nguon thu cap 9 Phan Uch m9-ch 6n ap bu dung transistor.
9. Phan Uch m9-ch 6n ap bu dung transistor.
A'" , _ I. HE DEM VA MA
Chucing 3
KYTHUAT56
1. Khiii ni~m ve h~ dem
H¢ dem lil t6 hqp cae qui tac gQi va hieu dien cae con so co gia tf! xac dinh.
Cha so lit ohung ky hi¢u dung d~ bi~u dien m(>t con so.
C6 h¢ thong dem theo vi tef va M thong dem kh6ng theo vj trf. (H<;: dem
muai dung eha so A R~p va h¢ dem theo ehu so La rna). 2. Chuy~n d6i so giG'a cae h~ dem
2.1. Chuyfn d6i so tu h¢ th:)p ph3ll sang h¢ nh~ phan
+ Chuyen ctOi so nguyen: Lay so can d6i chia licn tiep eha 2. Khi chia chi lay thuang la so nguyen. C.hia eha lai khi nh~n duqc ket qua phcp chia lil kh6ng thi. so ohi phao nh;}n dugc lil cae so du cua cae phep chia va dQC tu duai len.
Vi dl,l: Din chuyen deli so J 9 h¢ th~\p phao sang h~ fih! phan ta thvc hi¢n
nhu sau: 19:2 =9 du 1 9:2 =4 du 1 4:2 =2 duO 2:2 =1 du 0 1:2 =0 du 1
Ket qua :19 iJ h? th~p pMn bling 10011 h? nhi phAn Tuang II! ta co (9)]0 = (1001),
(21)", = (10101),
+ Chuy~n deli s6 sau dau phAy: Liiy s6 cAn chuyen doi nh3n voi 2 duqc tich s6. Lay s6 truoc ctau phiy cua tich lam chi! s6 cua h? 2. (Chi! s6 sau difu phay cua tich dAu titn duqc viet ngay sau dau phAy). L~i lay phAn tfch sau dau phiiy nhAn v6i 2 va l~i lam nhu tren ta duqc chi] so nhi philn tiep theo.
Vf dl,l: (0,75)10 d6i sang h¢ nhi philo.
0,75 x 2 = II'~ f'
0,5X2=~
NhuVilY (0,75)", = ([1,11), 0,35 x 2 = 0,7
Va nhu vf.j.y (0,35)10 = 0,01 6 h~ nhi phan
0,7x2=1,4
I
'" 1 j'
I
va (0,7) 6 he th~p phan = (0,11) he nhi phan,
2.2. Chuyen doi so tll h¢ nhj phan ve h¢ th*p phan
+ Chuyen doi so nguyen
Vi d~ I: (1010), = (7)", Ta c6: I 0 I ° \11-:Ol':=O L - 1 . 2 =2 0,2' , = ° L -_ _ _ _ 1.2 = 8 Vi d~ 2: (11011), = (?),,' Ta c6: I I ° I I 4 J 2 I II 1.2 + 1.2 + 0,2 + 1.2 + 1.2 16 + 8 + ° + 2 + I = (27)10' = (10)10
+ Chuytn d6i pharr sau Mu phiiy Vi d~ I: (0,1 00), = (?)IO' I\~ .J .2 _J 1.2 + 0,2 + 0.2 1 1 , + ° + ° = = (0,5)10 2 2
Vi d~ 2: (0, 1 1), = ('1)'0
t~, 1
1.2 + 1.2 = 2 1 1
+ 4 = 0,5 + 0,25 = (0,75)", 3, Mii BCD (Binary Codec Decimal)
- Dinh nghia: MIT BCD la rna dung bon bit de bi6u dien mOt chfr s6 th~p phan
Wang ling.
- C6 10 chi! s6 th~p phan tiT 0 den 9, VI vQ.y c6 rna BCD chuan luang ung
nhu sau:
o rna BCD Iii 0000 1 rna BCD Iii 0001 2 rna BCD Iii 0010 3 rna BCD Iii 0011
4 rna BCD lit OlOO
5 rna BCD Iii 0101
6 rna BCD lit 0110
7 rna BCD lit 0111 8 rna BCD lit 1000 9 rna BCD 1" 100 I
- M<)t so b h9 th~p phan dugc bieu dien bang rna BCD nhu sau: Vi d~ (75)", = (0111 O!OI)IlCD
(309)", = (0011 00001001 )OCD
- Tu rna BCD doi vc h~ th~p phan:
Vi du: (0110 OlOO)IlCD = (64)", Vi d~: (1000 OOIO)IlCD = (82)", Ii, cAc CONG LOGIC CO nAN