Home / KTU Solved Questions / KTU B.Tech CS207-Electronic Devices & Circuits Question Paper with Answers

KTU B.Tech CS207-Electronic Devices & Circuits Question Paper with Answers

MODEL QUESTION PAPER  with Answers Prepared  by ktubtechquestions.com

THIRD SEMESTER B.TECH  DEGREE EXAMINATION  JANUARY 2017

CS207 ELECTRONIC DEVICES & CIRCUITS

Time: 3 Hrs                                                                                            Marks: 100

PART A

( Answer All Questions Each carries 3 Marks )

 

  1. Draw the circuit diagram of RC differentiator

ANS: A circuit in which output voltage is directly proportional to the derivative of the input  is known as a differentiating circuit. A differentiating circuit is a simple series RC circuit where the output is taken across the resistor R.The circuit is suitably designed so that the output is proportional to the derivative of the input.Thus if a d.c. or constant input is applied to such a circuit, the output will be zero.

 edc1

  1. Explain the working of positively biased Clipper

ANS: In a positive clipper, the positive half cycles of the input voltage will be removed. The circuit arrangements for a positive clipper are illustrated in the figure given below.

1

As shown in the figure, the diode is kept in series with the load. During the positive half cycle of the input waveform, the diode ‘D’ is reverse biased, which maintains the output voltage at 0 Volts. Thus causes the positive half cycle  to be clipped off. Dur­ing the negative half cycle of the input, the diode is forward biased and so the nega­tive half cycle appears across the output.

In Figure (b), the diode is kept in parallel with the load. This is the diagram of a positive shunt clipper circuit. During the positive half cycle, the diode ‘D’ is forward biased and the diode acts as a closed switch. This causes the diode to conduct heavily. This causes the voltage drop across the diode or across the load resistance Rto be zero. Thus output voltage during the positive half cycles is zero, as shown in the output waveform. During the negative half cycles of the input signal voltage, the diode D is reverse biased and behaves as an open switch. Consequently the entire input voltage appears across the diode or across the load resistance RL if R is much smaller than RL. Actually the circuit behaves as a voltage divider with an output voltage of [RL / R+ RL] Vmax = -Vmaxwhen R>> R

 

  1. Differentiate between Enhancement and Depletion types MOSFET

ANS: 

  1. Enhancement type mode MOSFET will be off for gate to source 0V as there exists no channel to conduct. Depletion type MOSFET conducts at 0V has positive cut off gate voltage so less preferred. Depletion MOS also conducts at 0V therefore has less useful application.
  2. Since the logic operations of depletion MOSFET is the opposite to the enhancement MOSFET, the depletion MOSFET produces positive logic circuits, such as, buffer, AND, and OR.
  3. The depletion MOSFET is free from sub-threshold leakage current and gate oxide leakage current.
  4. As a enhancement MOSFET shrinking in size, there is no way to stop the sub threshold leakage current diffused across from source to drain because the drain and source terminals are closer physically. This is not the problem with depletion type MOSFET because a pinched channel stops the diffusion current completely.

 

  1. Explain the transfer Characteristics of a JFET.

The graphical characteristics plot of the saturation drain current against the gate to source voltage is known as the transfer characteristics of JFET. It can be obtained from static characteristics very easily. The transfer characteristics of an n- channel is shown below

1

 

PART B

( Answer any TWO, Each carries 9 Marks )

 

  1. Explain the working of voltage Doubler circuit.

ANS:                  As its name suggests, a Voltage Doubler is a voltage multiplier circuit which has a voltage multiplication factor of two. The circuit consists of only two diodes, two capacitors and an oscillating AC input voltage (a PWM waveform could also be used). This simple diode-capacitor pump circuit gives a DC output voltage equal to the peak-to-peak value of the sinusoidal input. In other words, double the peak voltage value because the diodes and the capacitors work together to effectively double the voltage.

DC Voltage Doubler Circuit

4

So how does it work. The circuit shows a half wave voltage doubler. During the negative half cycle of the sinusoidal input waveform, diode D1 is forward biased and conducts charging up the pump capacitor, C1 to the peak value of the input voltage, (Vp). Because there is no path for capacitor C1 to discharge into, it remains fully charged and acts as a storage device in series with the voltage supply. At the same time, diode D2 conducts via D1 charging up capacitor, C2.

During the positive half cycle, diode D1 is reverse biased blocking the discharging of C1 while diode D2 is forward biased charging up capacitor C2. But because there is a voltage across capacitor C1 already equal to the peak input voltage, capacitor C2 charges to twice the peak voltage value of the input signal.

In other words, V(positive peak) + V(negative peak) as on the negative half-cycle, D1 charges C1 to Vp and on the positive half-cycle D2 adds the AC peak voltage to Vp onC1 and transfers it all to C2. The voltage across capacitor, C2 discharges through the load ready for the next half cycle.

Then the voltage across capacitor, C2 can be calculated as: Vout = 2Vp, (minus of course the voltage drops across the diodes used) where Vp is the peak value of the input voltage. Note that this double output voltage is not instantaneous but increases slowly on each input cycle, eventually settling to 2Vp.

As capacitor C2 only charges up during one half cycle of the input waveform, the resulting output voltage discharged into the load has a ripple frequency equal to the supply frequency, hence the name half wave voltage doubler. The disadvantage of this is that it can be difficult to smooth out this large ripple frequency in much the same way as for a half wave rectifier circuit. Also, capacitor C2 must have a DC voltage rating at least twice the value of the peak input voltage.

The advantage of “Voltage Multiplier Circuits” is that it allows higher voltages to be created from a low voltage power source without a need for an expensive high voltage transformer as the voltage doubler circuit makes it possible to use a transformer with a lower step up ratio than would be need if an ordinary full wave supply were used. However, while voltage multipliers can boost the voltage, they can only supply low currents to a high-resistance (+100kΩ) load because the generated output voltage quickly drops-off as load current increases.

By reversing the direction of the diodes and capacitors in the circuit we can also reverse the direction of the output voltage creating a negative voltage output. Also, if we connected the output of one multiplying circuit onto the input of another (cascading), we can continue to increase the DC output voltage in integer steps to produce voltage triplers, or voltage quadruplers circuits, etc, as shown.

 

  1. Explain the working of a simple sweep circuit using transistor as a switch with relevant sketch. 

The miller sweep circuit or miller integrator generator is a precise and linear ramp voltage using active devices providing required feedback, the effective time constant and supply voltage is enhanced.
Miller sweep circuit is the basic schematic of a widely used sawtooth generator.
The amplifier acts to increase the aiming potential; thus, linearity is improved and the output amplitude is increased. As the integral of a step function is a ramp, it is evident that this circuit would provide a sawtooth output.
The feedback can be negetive voltage or current type. The circuit of miller sweep generator is drawn below.

5
It uses a bjt,the switch shown in the fig. Is an ideal switch normally it is closed.
The transistor is in off condition. The timing capacitor is charged to a voltage Vcc via the resistor Rc and the switch .

6

The switch is opened at time t=0, then the base emitter voltage of the transistor increases to Vbe ON. This increases in the base emitter voltage through the capacitor c is coupled to the output.

  1. List the important features of 7805 voltage regulator IC. Draw a typical circuit to get 5V regulated voltage

IC 7805 is a DC regulated IC of 5V. This IC is very flexible and is widely employed in all types of circuit like a voltage regulator. It is a three terminal device and mainly called input , output and ground.

The pin explanation of the 7805 is described in the following table:

PIN NO. PIN DESCRIPTION
1 INPUT In this pin of the IC positive unregulated voltage is given in regulation.
2 GROUND In this pin where the ground is given. This pin is neutral for equally the input and output.
3 OUTPUT The output of the regulated 5V volt is taken out at this pin of the IC regulator.

In the circuit diagram C2 as well as C3 are filter capacitor while bypass capacitors are the C1 and C4.The electrolytic polarized capacitors are employed for this purpose. For the purpose of filter capacitors normally 10mfd value of the capacitor used. And in these projects we also used 100mfd value of the capacitor. While in all kinds of circuit the value of bypass capacitor is 0.1 mfd. And in generally un-polarized mainly disc capacitors employed for this purpose.

Currently we have the circuit for the 5V DC positive regulation and we are also familiar with the component values used in the circuit. In the table below we have mentioned the value in detail of all the components used in the circuit of 5V DC positive regulator.

 

 

PART C

( Answer All Questions Each Carries 3 Marks )

  1. List the advantage and disadvantage of Fixed biasing circuit

Merits:

  • It is simple to shift the operating point anywhere in the active region by merely changing the base resistor (RB).
  • Very few number of components are required.

Demerits:

  • The collector current does not remain constant with variation in temperature or power supply voltage. Therefore the operating point is unstable.
  • Changes in Vbe will change IB and thus cause RE to change. This in turn will alter the gain of the stage.
  • When the transistor is replaced with another one, considerable change in the value of β can be expected. Due to this change the operating point will shift.

 

9.What is the use of DC load line

The DC load line is the load line of the DC equivalent circuit, defined by reducing the reactive components to zero (replacing capacitors by open circuits and inductors by short circuits). It is used to determine the correct DCoperating point, often called the Q point. A load line is used in graphical analysis of nonlinear electronic circuits, representing the constraint other parts of the circuit place on a non-linear device, like a diode or transistor. It is usually drawn on a graph of the current vs the voltage in the nonlinear device, called the device’s characteristic curve. A load line, usually a straight line, represents the response of the linear part of the circuit, connected to the nonlinear device in question. The points where the characteristic curve and the load line intersect are the possible operating point(s) (Q points) of the circuit; at these points the current and voltage parameters of both parts of the circuit match

 

10  Discuss the Barkhausen criterion for feedback oscillators.

 

  • A small change In DC power supply or noise component in oscillator circuit can start oscillation and to maintain oscillation in circuit must satisfy Barkhausen’s criterion.
  • Barkhausen’s criterion states that,
  • The loop gain is equal to unity in absolute magnitude, that is, | βA|=1
  1. | βA| > 1: In this condition, feedback is greater than the input voltage Thus addition of input wave and feedback wave will result in larger amplitude wave and as oscillation goes on the amplitude will increase and this can be harmful for device.
  2. | βA| < 1: In this condition, feedback is less than the input voltage Thus addition of input wave and feedback wave will result in smaller amplitude wave and as oscillation goes on the amplitude will gradually decrease and oscillations will die out.
  3. | βA|=1, In this condition, feedback equal to the input voltage Thus addition of input wave and feedback wave will result wave having amplitude of input and as oscillation goes on the amplitude will remain constant and hence a sustained oscillation is achieved.
  • The phase shift around the loop is zero or an integer multiple of 2π∠βA=2πnnє0,1,2,…2π∠βA=2πnnє0,1,2,… Complex value of βA is given by

βA=1+j0

In above expression imaginary part is zero because we assume phase shift zero or 360º now if phase shift isn’t zero then |βA|≠1|βA|≠1, which is not suitable condition for oscillation. For phase shift equal to 180º | βA|=1 but input and feedback signal will be out of phase and they will cancel each other hence phase shift must be an integer multiple of 2π

  1. Draw the equivalent circuit of crystal oscillator

The equivalent electrical circuit for the quartz crystal shows a series RLC circuit, which represents the mechanical vibrations of the crystal, in parallel with a capacitance, Cp which represents the electrical connections to the crystal. Quartz crystal oscillators tend to operate towards their “series resonance”.

5

 

 

 

 

PART D

( Answer any TWO, Each carries 9 Marks )

  1. Draw the circuit of a RC coupled amplifier and explain the working.

5

a coupling capacitor Cis used to connect the output of first stage to the base i.e. input of the second stage and this continues when more stages are connected.

Since here the coupling from one stage to next is achieved by a coupling capacitor followed by a connection to a shunt resistor, therefore, such amplifiers are known as resistance-capacitance coupled amplifier or simply RC coupled amplifier.

The resistances R1, Rand Rform the biasing and stabilisation network.

The emitter bypass capacitor offers a low resistance path to the signal. Without this capacitor the voltage gain of each stage would be lost.

The coupling capacitor CC  transmits a.c. signal but blocks d.c. This prevents d.c. interference between various stages and the shifting of operating point.

Working of RC Coupled Amplifier

When a.c. signal is applied to the base of the first transistor, it is amplified and appears across its collector load RC.

Now the amplified signal developed across Ris given to the base of the next transistor through a coupling capacitor CC .

The second stage again amplifies this signal and the more amplified signal appears across the second stage collector resistance.

In this way the cascaded stages amplify the signal and the overall gain is considerably increased.

However, the total gain is less than the product of the gains of individual stages. It is because, when a second stage follows the first stage, the effective load resistance of first stage is reduced due to the shunting effect of the input resistance of second stage. This reduces the gain of the stage which is loaded by the next stage.

To explain it better, let us take an example of  3-stage amplifier. The gain of first and second stage will be reduced due to loading effect of the next stage. But the gain of the third stage which has no loading effect due to subsequent stage, remains unchanged.

The overall gain is equal to the product of the gains of three stages.

 

  1. Discuss the effect of negative feedback on the performance of an amplifier.

 

Fig. 13.3 *shows the principles of negative voltage feedback in an amplifier. Typical values have been assumed to make the treatment more illustrative. The output of the amplifier is 10 V. The fraction mv of this output i.e. 100 mV is fedback to the input where it is applied in series with the input signal of 101 mV. As the feedback is negative, therefore, only 1 mV appears at the input terminals of the amplifier. Referring to Fig. 13.3, we have, Gain of amplifier without feedback,

2

The following points are worth noting :

  • When negative voltage feedback is applied, the gain of the amplifier is **reduced. Thus, the gain of above amplifier without feedback is 10,000 whereas with negative feedback, it is only 100.
  • When negative voltage feedback is employed, the voltage actually applied to the amplifier is extremely small. In this case, the signal voltage is 101 mV and the negative feedback is 100 mV so that voltage applied at the input of the amplifier is only 1 mV.
  • In a negative voltage feedback circuit, the feedback fraction mv is always between 0 and 1.
  • The gain with feedback is sometimes called closed-loop gain while the gain without feedback is called open-loop gain. These terms come from the fact that amplifier and feedback circuits form a “loop”. When the loop is “opened” by disconnecting the feedback circuit from the input, the amplifier’s gain is Av, the “open-loop” gain. When the loop is “closed” by connecting the feedback circuit, the gain decreases to Avf , the “closed-loop” gain.

* Note that amplifier and feedback circuits are connected in series-parallel. The inputs of amplifier and feedback circuits are in series but the outputs are in parallel. In practice, this circuit is widely used.

** Since with negative voltage feedback the voltage gain is decreased and current gain remains unaffected, the power gain Ap (= Av × Ai ) will decrease. However, the drawback of reduced power gain is offset by the advantage of increased bandwidth

 

 

  1. Explain the working of a Wien oscillator. Derive an expression for the frequency of oscillations.

 

The Wien bridge oscillator is used to find unknown values of components. In most of the cases this oscillator is used in the audios. The oscillators are designed simply, size is compressed and it has stable in frequency output. Hence the maximum output frequency of the Wien bridge oscillator is 1MHz and this frequency is from the phase shift oscillator. The total phase shift of the oscillator is from the 360° or 0°.

It is a two stage amplifier with RC bridge circuit and the circuit has the lead lag networks. The lags at the phase shift are increasing the frequency and the leads are decreasing the frequency. In additional by adding the Wien Bridge oscillator at a particular frequency it becomes sensitive. At this frequency the Wien Bridge is balance the phase shift of 0°. The following diagram shows the circuit diagram of the Wienbridge oscillator. The diagram shows R1 is series with the C1, R3, R4 and R2 are parallel with the C2 to from the four arms.

From the above diagram we can see the two transistors are used for the phase shift of 360°and also for the positive feedback. The negative feedback is connected to the circuit of the output with a range of frequencies. This has been taken through the R4 resistor to from the temperature sensitive lamp and the resistor is directly proportional to the increasing current. If the output of the amplitude is increased then the more current is offered more negative feedback.

6

Wien Bridge Oscillator Operation

The circuit is in the oscillation mode and the base current of the first transistor is changed randomly because it is due to the difference in voltage of DC supply. The base current is applied to the collector terminal of the first transistor and the phase shift is about the 180°. The output of the first transistor is given to the base terminal of the second transistor Q2 with the help of the capacitor C4. Further, this process is amplified and from the second transistor of collector terminal the phase reversed signal is collected.

The output signal is connected to the phase with the help of the first transistor to the base terminal. The input point of the bridge circuit is from the point A to point C the feedback of this circuit is the output signal at the second transistor. The feedback signal is given to the resistor R4 which gives the negative feedback. In this same way the feedback signal is given to the base bias resistor R4 and it produces the positive feedback signal.

By using the two capacitors C1 and C2 in this oscillator, it can behave continuous frequency variation. These capacitors are the air gang capacitors and we can also change the values of the frequency range of the oscillator.

 

PART E

( Answer any FOUR, Each carries 10 Marks )

 

  1. Explain the characteristics of op-amps
Infinite Voltage Gain An ideal op amp will have infinite voltage gain. Op amps are devices that many times are used to function as amplifiers. A voltage is input into the op amp and as output, it produces the voltage amplified. An ideal op amp will produce mega-gain, practically, it will be able to produce infinite gain. It will amplify the signal infinite times over so that we can have as much gain as we’d ever need.

 

Gain Independent of Frequency In an ideal op amp, the gain that the op amp produces will be independent of frequency. This means that regardless of the frequency of the input signal going into the op amp, the gain that is produced will be constant and good across all frequencies.
Zero Input Voltage Offset In an ideal op amp, if no voltage is applied to the inverting and noninverting input pins, the op amp will output a voltage of 0, since there is no difference at all of the voltage applied to the 2 input pins.
 

Slew Rate

The Slew Rate of an op amp describes how fast the output voltage can change in response to an immediate change in voltage at the input. The higher the value (in V/µs) of slew rate, the faster the output can change and the more easily it can reproduce high frequency signals

.

  1. Discuss any 2 applications of op-amps

As Subtractor

A summing amplifier can be used to provide an output voltage that is equal to the difference of two  voltages.Such a circuit is called a Subtractor and is shown if fig.3.

7

Fig.3

As we can see, this circuit will provide an output voltage that is equal to the difference between V1 and V2.

The voltage Vis applied to a standard inverting amplifier that has unity gain.

Because of this, the output from the inverting amplifier will be equal to –V1.

This output is then applied to the summing amplifier, also having unity gain along with V2.

Thus output from second OP-Amp is given by:

8

The gain of the second stage in the Subtractor can be varied to provide an output that is proportional to the difference between the input voltages.

OP-Amp Integrator

An integrator is a circuit that performs integration of the input signal. The most important application of an integrator is to produce a ramp output voltage.Fig.4 shows the circuit of an OP-Amp integrator.
When a signal is applied to the input of this circuit, the output-signal waveform will be the integration of input-signal waveform.It consists of an OP-Amp, input resistor R and feedback capacitor C.

8

Circuit Analysis

Since point A in fig.4(i) is at virtual ground, the virtual ground equivalent circuit of operational integrator will be as shown in fig.4 (ii).Because of virtual ground and infinite impedance of the OP-Amp, all of the input current flows through the capacitor i.e.

8

8

This equation shows that the output is the integral of the input with an inversion and scale multiplier of 1/RC.

Output Voltage

If a fixed voltage is applied to the input of an integrator, the output voltage grows over a period of time, providing a ramp voltage.

The output ramp voltage is opposite in polarity to the input voltage and is multiplied by a factor 1//RC.

 

 

  1. Draw the functional block diagram of 555 timer.

8

The internal resistors act as a voltage divider network, providing (2/3)Vcc at the non-inverting terminal of the upper comparator and (1/3)Vcc at the inverting terminal of the lower comparator. In most applications, the control input is not used, so that the control voltage equals +(2/3) VCC. Upper comparator has a threshold input (pin 6) and a control input (pin 5). Output of the upper comparator is applied to set (S) input of the flip-flop. Whenever the threshold voltage exceeds the control voltage, the upper comparator will set the flip-flop and its output is high. A high output from the flip-flop when given to the base of the discharge transistor saturates it and thus discharges the transistor that is connected externally to the discharge pin 7. The complementary signal out of the flip-flop goes to pin 3, the output. The output available at pin 3 is low. These conditions will prevail until lower comparator triggers the flip-flop. Even if the voltage at the threshold input falls below (2/3) VCC, that is upper comparator cannot cause the flip-flop to change again. It means that the upper comparator can only force the flip-flop’s output high.

To change the output of flip-flop to low, the voltage at the trigger input must fall below + (1/3) Vcc. When this occurs, lower comparator triggers the flip-flop, forcing its output low. The low output from the flip-flop turns the discharge transistor off and forces the power amplifier to output a high. These conditions will continue independent of the voltage on the trigger input. Lower comparator can only cause the flip-flop to output low.

From the above discussion it is concluded that for the having low output from the timer 555, the voltage on the threshold input must exceed the control voltage or + (2/3) VCC. This also turns the discharge transistor on. To force the output from the timer high, the voltage on the trigger input must drop below +(1/3) VCC. This turns the discharge transistor off.

A voltage may be applied to the control input to change the levels at which the switching occurs. When not in use, a 0.01 nano Farad capacitor should be connected between pin 5 and ground to prevent noise coupled onto this pin from causing false triggering.

Connecting the reset (pin 4) to a logic low will place a high on the output of flip-flop. The discharge transistor will go on and the power amplifier will output a low. This condition will continue until reset is taken high. This allows synchronization or resetting of the circuit’s operation. When not in use, reset should be tied to +VCC.

 

  1. Write a short note on Binary weighted resistor

Binary Weighted Resistor DAC

8

Figure-1 mentions block diagram of binary weighted resistor DAC. It utilizes summing Op-Amp circuit. Weighted resistors are used to distinguish each bit from MSB to LSB. Transistors are used to switch between Vref and ground (bit high or low).

Output voltage from binary weighted resistor DAC is expressed as follows:
Vo = { RF/(2N-1*R)}* {2N-1*VN-1 + 2N-2*VN-2 + ……. + 21*V1 + 20*V2 }

Advantages:
• It is Simple in Construction.
• It provides fast conversion.
Disadvantages:
Disadvantages:
1) When number of binary input increases, it is not easy to maintain the resistance ratio.
2) Very wide ranges of different values of resistors are required.
For high accuracy of conversion, the values of resistances must be accurate.
3) Different current flows through resistors, so their wattage ratings are also different.
4) Accuracy and stability of conversion depends primarily on the absolute accuracy of the resistors and tracking of each other with temperature.
eg. For 10 digit converter
small resistance value = 10 kΩ and
large resistance value = 5.12 MΩ
It is very difficult and expensive to obtain stable precise resistances of such value.
5) Since ‘R’ is very large, op-amp bias currents gives a drop which offsets output.
6) Resistances of switches may be comparable with smallest resistor.

 

  1. Why Positive feedback is used in Schmitt trigger?

 

A Schmitt trigger circuit is also called a regenerative comparator circuit. The circuit is designed with a positive feedback and hence will have a regenerative action which will make the output switch levels. Also, the use of positive voltage feedback instead of a negative feedback, aids the feedback voltage to the input voltage, instead of opposing it. The use of a regenerative circuit is to remove the difficulties in a zero-crossing detector circuit due to low frequency signals and input noise voltages.

Shown below is the circuit diagram of a Schmitt trigger. It is basically an inverting comparator circuit with a positive feedback. The purpose of the Schmitt trigger is to convert any regular or irregular shaped input waveform into a square wave output voltage or pulse. Thus, it can also be called a squaring circuit.

8

As shown in the circuit diagram, a voltage divider with resistors Rdiv1 and Rdiv2 is set in the positive feedback of the 741 IC op-amp. The same values of Rdiv1 and Rdiv2 are used to get the resistance value Rpar = Rdiv1||Rdiv2 which is connected in series with the input voltage. Rpar is used to minimize the offset problems. The voltage across R1 is fedback to the non-inverting input. The input voltage Vi triggers or changes the state of output Vout every time it exceeds its voltage levels above a certain threshold value called Upper Threshold Voltage (Vupt) and Lower Threshold Voltage (Vlpt).

Let us assume that the inverting input voltage has a slight positive value. This will cause a negative value in the output. This negative voltage is fedback to the non-inverting terminal (+) of the op-amp through the voltage divider. Thus, the value of the negative voltage that is fedback to the positive terminal becomes higher. The value of the negative voltage becomes again higher until the circuit is driven into negative saturation (-Vsat). Now, let us assume that the  inverting input voltage has a slight negative value. This will cause a positive value in the output. This positive voltage is fedback to the non-inverting terminal (+) of the op-amp through the voltage divider. Thus, the value of the positive voltage that is fedback to the positive terminal becomes higher. The value of the positive voltage becomes again higher until the circuit is driven into positive saturation (+Vsat). This is why the circuit is also named a regenerative comparator circuit.

 

20.Explain any 2 types of  Analog to Digital Converters.

Sigma-Delta ADC

8

  • Oversampled input signal goes in the integrator
  • Output of integration is compared to GND
  • Iterates to produce a serial bitstream
  • Output is serial bit stream with # of 1’s proportional to Vin

Advantages

  • High resolution
  • No precision external components needed

Disadvantages

  • Slow due to oversampling

Dual Slope converter

 

  • The sampled signal charges a capacitor for a fixed amount of time
  • By integrating over time, noise integrates out of the conversion.
  • Then the ADC discharges the capacitor at a fixed rate while a counter counts the ADC’s output bits. A longer discharge time results in a higher count.

Advantages

  • Input signal is averaged
  • Greater noise immunity than other ADC types
  • High accuracy

Disadvantages

  • Slow
  • High precision external components required to achieve accuracy

 

 

 

©® ktubtechquestions.com

 

x

Check Also

KTU Previous Questions with Answers Engineering Mechanics

KTU Previous Questions with Answers Engineering Mechanics January 2016 10032     ...

KTU Previous Questions with answers of Computing and Problem Solving

KTU Previous Questions with answers of Computing and Problem Solving Reg. No:………….                                                        ...