Digital circuits miscellaneous
- In a 4-bit weighted resistor D/A converter, the resistor value corresponding to LSB is 32-k ohm. The resistor value corresponding to MSB will be _________kΩ
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2n–1 R = 32,
where n = 4
∴ 23R = 32
⇒ R = 4k -ohmCorrect Option: D
2n–1 R = 32,
where n = 4
∴ 23R = 32
⇒ R = 4k -ohm
- An analog voltage is in the range of 0 to 8 V is divided in eight equal intervals for conversion to 3-bit digital output. The maximum quantization error is _________V
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Quantizing error = V/2N
= 8/23 = 1V.Correct Option: A
Quantizing error = V/2N
= 8/23 = 1V.
- For the Schmitt trigger oscillator shown in the given figure saturation output voltage are + 10 V and – 5 V. The frequency of oscillation is ___________Hz
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Switching voltage are
V+ = 10 = 5V, V+ = - 5 = - 2.5 V 2 2
The charging and discharging of capacitor is shown in the figure
∴ 5 = 10 + (– 2.5 – 10) e-t1/RC
⇒ e-t1/RC = 2/5 ⇒ t1 = RC ln 2.5
∴ – 2.5 = – 5 + (5 – (– 5)) e-t1/RC
∴ e-t1/RC = 1/4
⇒ t2 = RC ln 4,
T = t1 + t2 = RC ln 2.5 + RC ln 4
= RC ln 10 R
= 50 kΩ, C = 0.01 μF
⇒ T = 50k × 0.01μ × ln 10 = 1.15 ms
ƒ = 1/T = 869 HzCorrect Option: A
Switching voltage are
V+ = 10 = 5V, V+ = - 5 = - 2.5 V 2 2
The charging and discharging of capacitor is shown in the figure
∴ 5 = 10 + (– 2.5 – 10) e-t1/RC
⇒ e-t1/RC = 2/5 ⇒ t1 = RC ln 2.5
∴ – 2.5 = – 5 + (5 – (– 5)) e-t1/RC
∴ e-t1/RC = 1/4
⇒ t2 = RC ln 4,
T = t1 + t2 = RC ln 2.5 + RC ln 4
= RC ln 10 R
= 50 kΩ, C = 0.01 μF
⇒ T = 50k × 0.01μ × ln 10 = 1.15 ms
ƒ = 1/T = 869 Hz
- The switching circuit given in the figure can be expressed in binary logic notation as
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NA
Correct Option: A
NA
- Figure given below shows four D t ype FF's connected as a shift register using an X OR gate. The initial state and 3 subsequent states for 3 clock pulses are also given.
State QA QB QC QD after fourth clock pulse is
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Naming the FFs as 1 to 4 from Left to Right after the 3rd clock pulse FF3 is 0 and FF4 is 1 so that X OR output is 1 which is fed to QA.
So QA to QB → 0 QB to QC → 0. QC → QD → 0
After 4th clock pulse 1000Correct Option: D
Naming the FFs as 1 to 4 from Left to Right after the 3rd clock pulse FF3 is 0 and FF4 is 1 so that X OR output is 1 which is fed to QA.
So QA to QB → 0 QB to QC → 0. QC → QD → 0
After 4th clock pulse 1000