Question

1. For the transistor circuit shown in figure 1, what is: a. Vce when VIN = OV? b. What minimum value of IB is required to saturate the transistor if Boc is 200? c. Calculate the maximum value of Rs when Vin = 5V. VCC 10V RC 1.0 K RB Q1 VIN=0V 15C02MH-TLE GND Figure

Answer 1

(a)Given that vcc=10v ,vin=0v then vce=vcc=10v

(b)Let vbe =0.7v then vcc-ic rc-vbe=0

Ic=(vcc-vbe)/rc=9.3mA

Ib=ic/b=46.5uA

(c)Rb=(vin-vbe)/rb=92.47k ohm

I Given that the haunista Cicunt Shown en tigue (a) given that cc = lou, Vinou 2 Rc -ks So there is no input voltage and no Cure ent passes across the Base tammal vin-ou s th I B=0A IC = BIB = B(o)=0A We know that We can KUL across the Collecta to emelles loop so then Vec-Ic Rc-UCE = 0 Vee - o CRC) - UCE 20 Ucc = UCE => VCE = Vcc=100 VCE - VCC =10V (b) let us Counter that UBE=0.7V (silicon) & given Bac=200 collecta to base miller apply Kul across the Ncc - Ic Rc-UBE=0 vee- Icke TUBE 10 = IC (IK) +0.7 >) IC= 10-0.7 t =9.3MA 92, I c = BIB IB - Ic - 93 m 200 - 465YLO A = 46.5uA IB = 465UA to Saluate the traunstor et the minimum Value of IB is reami ed - Bac in 200 is 46.5 MA Maasinun Value of RB When vin-su in- I BRB TUBE (Apply kur acloss input to base emila) Vina 46.5MY RB 40.7 = 5 RB=15-0.7)|46.544, 43/4654 = 22.47KS [RB = 92.47 3 10² = 92 47 KM ]