Q1 A A certain application requires a oneshot with a pulse w

Q.1\\

A.\\ A certain application requires a one-shot with a pulse width of approximately 100 ms.

Using a 74121 below, If the exterior resistance RT is arbitrarily selected to be 40 k? , find the other component value?

B.\\

1. In the absence of a trigger pulse, what will be the state of a OS output?

2. What determines the tp value for a OS? Answer =

3. When a nonretriggerable OS is pulsed while it is in its quasi-stable state, the output

is not affected.

                                       a). True                                                               b). False

Q.2\\.

1. A 20-kHz clock signal is applied to a J-K flip-flop with J = K = 1.What is the frequency of the FF output waveform?                Answer =

2. How many FFs are required for a counter that will count 0 to 25510?    Answer =

3. What is the MOD number of the counter in question 2?       Answer =

4. What is the frequency of the output of the eighth FF when the input clock frequency is 512 kHz?

                                       Answer =

5. If this counter starts at 00000000, what will be its state after 520 pulses? Answer =

Q3\\

A. Using the diodes along with RB as shown in Figure below, calculate the values

of RA and RB necessary to get a 1 kHz, 25 percent duty cycle waveform out of

a 555. Assume C is a 0.1 F capacitor.

B.\\ Calculate the frequency and the duty cycle of the 555 astable multivibrator output for C = 0.001µ F, RA = 2.2 k? , and RB = 100 k? .

Solution

Q1)

A)

IC 74121 or 555 timer ,when used as one shot or monostable multivibrator,

pulse width tw = 0.33*C*R

Given:R=40KOhm & tw =100ms

C=tw/(0.33R) = (100*10^-3)/(0.33*40*10³) =7.575 uF [answer]

B)

1)

As Q is normally LOW,

when trigger is not applied to it,Q is in a state of 0 or LOW.

2)

exteranlly connected Resistance R_T & Capacitance C determines the value of tp.

3)

When output is in Quasi-stable state,Retriggering the trigger will not affect the output in anyway.

So the statement is True.

Q2)

2.1)

frequency of clock fc=20KHz,

When J=K=1 ,J-K Flip-Flop is in toggle mode ,so output will have time period twice that of clock as output requires two clocks to have same output.

=>so FF of output waveform will be f =fc/2 =20KHz/2 = 10KHz [answer]

2.2)

Need to count 0 -25510 ,so no. of bits in 25510 be n

n>=log₂(25510) =14.63

=>n=15

So no. of flip-flops require N=n=15 [answer]

2.3)

MOD number is M=25510 [ANSWER]

2.4)

For 8_thflipflop , frequency f=fc/2⁸ =512KHz/256 =2KHz [answer]

2.5) If this counter starts at 00000000, after 520 pulses,counter will increase by 520

so counter value will be 520 decimal or 208 Hex or \"000 0010 0000 1000\"

so counter value will be \"000 0010 0000 1000\" [answer]

Q3)

For circuit with diode,

Given:f=1KHz , duty cycle D=25% =0.25, C=0.1uF        ; Find R_A & R_B

t_L=0.94R_AC

t_H=0.94R_BC

duty cycle D= t_H/(t_H+t_L) = R_B/(R_A+R_B) =0.25

=>R_A =3R_B

frequency f=1/(t_H+t_L) =1/0.94*(R_A+R_B)C =1/0.94*4R_B*C =1/3.76R_AC

f=1/3.76R_AC

R_A=1/3.76fC =1/3.76*10³*0.1*10^-6 = 2.66*10³ Ohm [answer]

R_B=3R_B=7.97*10³ Ohm    [answer]

B)

C = 0.001µ F, RA = 2.2 k , and RB = 100 k

frequency f=1/(t_H+t_L) =1/{[0.693*(R_A+R_B)]+[0.693*R_B]}C =1/0.693(R_A+2R_B)C

f=1/[0.693(2200+2*100000)*0.001*10^-6) =7136.5 Hz [answer]

Duty Cycle D= t_H/(t_H+t_L) = (R_A+R_B)/(R_A+2R_B) =[(2200+100000) / (2200+2*100000)] =0.5054 =50.54 % [answer]