BEE
UNIT – 5Basic Electronics Q1) In a CB IE= 2mA, IC=1.5mA. Calculate IB?S1) IE =IB+IC2= IB+1.5IB=0.5mA Q2) In a CB current amplification factor is 0.9. If emitter current is 1.2mA. Determine the value of base current?S2) α = 0.9IE =1.2mAα = IC/ IE IC = α IE =0.9 x 1.2 = 1.08mAIE =IB+IC1.2= IB+1.08IB= 0.12mA Q3) In a CB connection IC=1.0mA and IB= 0.02mA. Find the value of current amplification factor?S3) IE =IB+IC =1+0.02 = 1.02mA
α = IC/ IE α = 1.0/1.02 = 0.98 Q4) In a CB connection the emitter current is 0.98mA. If the emitter circuit is open the collector current becomes 40
A. Find total collector current. α =0.92S4) ICBO=40
AIC = α IE+ICBO = (0.92 x 0.98x10-3) + 40x10-6IC =0.94mAQ5) In a common base connection, α = 0.95. The voltage drop across 3 kΩ resistance which is connected in the collector is 2.5 V. Find the base current.S5) IC = 2.5/3000 = 0.83mAα = IC/ IE IE = IC/α =0.83/0.95=0.87mAIE =IB+IC0.87 =IB+0.83IB=0.04mA Q6) Find the value of β if (i) α = 0.9 (ii) α = 0.98 (iii) α = 0.99.S6) 
= α/1- α = 0.9/1-0.9 = 9
= α/1- α = 0.98/1-0.98 = 49
= α/1- α = 0.99/1-0.99 = 99 Q7) The collector leakage current in a transistor is 200 μA in CE arrangement. If now the transistor is connected in CB arrangement, what will be the leakage current? Given that β = 120.S7) ICEO=200 μA
= 120α =
/1+
= 120/121=0.99ICEO=ICBO/1- αICBO= 1.6 μA Q8) For a certain transistor, IB = 18 μA; IC = 2 mA and β = 60. Calculate ICBO.S8) IC = 
IB+ICEOICEO= IC - 
IB= 2x10-3-(60x18x10-6) = 0.92mAα =
/1+
= 60/61=0.98ICBO= (1- α) ICEO = (1-0.98)x 0.92=15.08 μA Q9) Compare CB CE and CC configuration?S9)
Q10) Simplify f(X,Y,Z)=∏M(0,1,2,4)f(X,Y,Z)=∏M(0,1,2,4)using K-map. A10)
Therefore, the simplified Boolean function isf = (X + Y).(Y + Z).(Z + X) Q11) Simplify: F(P,Q,R,S)=∑(0,2,5,7,8,10,13,15) A11)
F = P’Q’R’S’ + PQ’R’S’ + P’Q’RS’ +PQ’RS’ + QSF = P’Q’S’ + PQ’S’ + QSF = Q’S’ +QS Q12) Simplify: F(A,B,C)=π(0,3,6,7) A12)
F = A’BC +ABC +A’B’C’ +ABC’F = BC + C’ ( A’B’ + AB )
A. Find total collector current. α =0.92S4) ICBO=40AIC = α IE+ICBO = (0.92 x 0.98x10-3) + 40x10-6IC =0.94mAQ5) In a common base connection, α = 0.95. The voltage drop across 3 kΩ resistance which is connected in the collector is 2.5 V. Find the base current.S5) IC = 2.5/3000 = 0.83mAα = IC/ IE IE = IC/α =0.83/0.95=0.87mAIE =IB+IC0.87 =IB+0.83IB=0.04mA Q6) Find the value of β if (i) α = 0.9 (ii) α = 0.98 (iii) α = 0.99.S6) = α/1- α = 0.9/1-0.9 = 9 = α/1- α = 0.98/1-0.98 = 49 = α/1- α = 0.99/1-0.99 = 99 Q7) The collector leakage current in a transistor is 200 μA in CE arrangement. If now the transistor is connected in CB arrangement, what will be the leakage current? Given that β = 120.S7) ICEO=200 μA = 120α = /1+= 120/121=0.99ICEO=ICBO/1- αICBO= 1.6 μA Q8) For a certain transistor, IB = 18 μA; IC = 2 mA and β = 60. Calculate ICBO.S8) IC = IB+ICEOICEO= IC - IB= 2x10-3-(60x18x10-6) = 0.92mAα = /1+= 60/61=0.98ICBO= (1- α) ICEO = (1-0.98)x 0.92=15.08 μA Q9) Compare CB CE and CC configuration?S9)
|
|
|
|
0 matching results found