unit no 3 : linear integrated circuit · 2. feedback & its types feedback used invariably in...

Unit No 3 : Linear Integrated Circuit1.Open Loop Configuration of Op-amp

2. Feedback & its types

3.Feedback Amplifier.

4.Concept of Negative Feedback

Advantages of Negative Feedback

5.Opamp Application-Linear & Non linear

6.Virtual Ground Concept

7.Ideal Inverting Amplifier7.Ideal Inverting Amplifier

8. Ideal Non-Inverting Amplifier

9.Voltage Follower

10.Summing – Inverting & Non-inverting Amplifier

11.Integrator

12.Differentiator

13.Comparator

14.Schmitt Trigger.

15.IC 555

16. Block Diagram of op-amp

International Institute of Information Technology, Prof.Dipak Raut

1 . Open Loop Configuration1 . Open Loop Configuration


1. Open Loop Configuration Simplest way to use an opamp is in open loop mode

Open loop gain AOL is very large, the Vo is either at its +Vsat or –Vsat forevery small values of Vd.

Due to high gain ,small voltage present at the input also gets amplified &opamp goes in saturation

Linear range is from a to b (Operating range)

Application in open loop behavior - voltage comparator & zero crossingdetector


2. Feedback & Its Types2. Feedback & Its Types


2. Feedback & Its Types

Feedback used invariably in amplifier to improve its performance & tomake it more ideal

Feedback : The part of output is sampled & fed back to input of amplifier.At input we have two signals i.e input signal & sampled output

Types

If input signal & sampled output is in-phase then feedback is called If input signal & sampled output is in-phase then feedback is called‘Positive Feedback’

If input signal & sampled output is out of phase then feedback is called‘Negative Feedback’


3. Feedback Amplifier


3. Feedback Amplifier

A triangular black box is a symbol of amplifier.

Opamp without feedback:

Let Vin & Vout be input & output of amplifier

Then voltage gain A of amplifier will be

A = V out / V in -----Gain without feedback

As gain of amplifier is greater than 1,Vout will be greater than Vin. As gain of amplifier is greater than 1,Vout will be greater than Vin.

Opamp with feedback:

The Vout is given back to input through Feedback network, it providesthe output Vf is called feedback output which is applied to input ofamplifier Vin.

Vin – input of amplifier

Vf – feedback signal

Vs – Original signal input

Such a amplifier is called ‘Feedback Amplifier’


Contd…

Gain of such a feedback amplifier is

Af = Vout / Vs

Af – Gain with feedback


4. Negative Feedback


4. Concept of Negative Feedback

Assume a sinusoidal input signal to amplifier which is assumed to beinverting amplifier meaning that out voltage is 180 degree out of phasewith respect to input voltage.

Let β is the feedback factor of feedback network

Vf = β.Vout or β = Vf / Vout

Where β lies between 0 & 1Where β lies between 0 & 1

Case 1 : If β = 0 , means no feedback signal applied , Vf = 0.

Case 2 : If β = 1 , means feedback signal applied , Vf = Vout.


Derive the expression of gain of amplifier with feedback in term of gain without feedback

Assuming feedback network is purely resistive,then Vf & Vout will be in phase. But amplifier is inverting type so Vout & Vs will be out of phase by 180 degree.

So Vf & Vs will be out of phase by 180 degree Input voltage Vin = Vs - Vf (Vs & Vf are 180 degree out of phase)

Vs = Vin + Vf.Gain with feedback

Af = Vout / VsAf = Vout / Vin + VfBut Vf = β.VoutBut Vf = β.VoutAf = Vout / Vin + β.VoutDivide N & DN by VinAf = (Vout/Vin) / (Vin/Vin + β.Vout/Vin)But Vout/Vin = A

Af =A/1+ βA. ---------General Feedback EquationAf < A

Af - Closed Loop Gain. For Inverting ampifier Vin = Vs – Vf (Vs & Vf out of phase)

For Non-Inverting ampifier Vin = Vs + Vf (Vs & Vf are in phase)


5. Advantages & Applications of Negative Feedback


5. Advantages & Applications of Negative Feedback Advantages :

1. It reduces the gain & makes it controllable

2. It reduces the possibility of distortion

3. It increases the bandwidth

4. It increases the input resistance of opamp

5. It decreases the out of resistnace of opamp5. It decreases the out of resistnace of opamp

6. It reduces the effect of temperature, power supply on gain ofthe circuit.

Application :

1. Feedback amplifier

2. Inverting & Non-Inverting amplifiers

3. V-I & I-V Converters

4. Instrumentation Amplifier


6. Virtual Ground Concept


6. Virtual Ground Concept We can make two assumption which are realistic & simplify the

analysis of op-amp circuit:

1. Zero Input Current :

The current drawn by either of terminals is zero.

2. Virtual Ground :

Vd between inverting & non-inverting terminal is essentially zero.

Even if Vo is in few mv, due to large A then Vd is almost zero.Even if Vo is in few mv, due to large AOL then Vd is almost zero.

Vo = Vd. AOL,

Vd = Vo / AOL , ( Vd=0, Because Aol is infinite)

Vd = 0 ;

V1 – V2 = 0

V1 = V2 ;

Thus voltage at both terminal are same.


Contd….

Fig shows the concept of virtual ground concept. Thick lineindicate the virtual short circuit between input terminals

If the non-inverting terminal is grounded, so by concept ofvirtual short, the inverting terminal is also at ground potential,though there is no physical connection between invertingterminal & ground. This is principle of virtual ground .terminal & ground. This is principle of virtual ground .


7. Inverting Amplifier


7. Ideal Inverting Amplifier Used amplifier is inverting type so Vo & Vin are 180 degree out of

phase.

By concept of virtual ground concept ,two terminal are always atsame potential.

At input side,

node B is grounded, node a is also at ground potential ,

VB = 0 ; so VA = 0.

I= (Vin – VA) / R1

I= Vin / R1 -----------eq.1

Opamp input current is always zero hence entire current I passesthrough Rf .

At output side, considering the direction of current,

I= (VA - Vo) / Rf = -(Vo / Rf) ------------eq.2

Equate eq.1 & eq.2

Vin / R1 = -(Vo / Rf) , Therefore Vo / Vin = -(Rf / R1)

Av = Vo/Vin=-(Rf/R1) -sign indicates negativefeedback International Institute of Information Technology,

Prof.Dipak Raut

Observations :

1. The output is inverted w.r.t input which is indicated by minus sign

2. The voltage gain is independent of open loop gain of opamp whichis assumed to be large

3. The gain can be obtained by selecting values of Rf & R1

4. If Rf > R1, the gain is greater than 1

If Rf < R1, the gain is less than 1

if Rf = R1, the gain is unity.


8. Non-Inverting Amplifier


8. Ideal Non-Inverting Amplifier

Vin is applied at non-inverting amplifier which amplifies withoutproducing any phase shift between Vin & Vo

At input side, Node B is at potential Vin, hence potential at A issame as at B which is Vin

by virtual ground concept,

VA = VB = VinVA = VB = Vin

At output side,

I = (Vo – VA)/Rf = (Vo – Vin)/Rf -------eq .1 (As VA = Vin)

At inverting input terminal,

I = (VA – 0)/R1 = Vin/R1. -------eq.2

As input current zero, entire current passes through R1

Equate eq.1 & eq.2

(Vo – Vin)/Rf = Vin/R1 Av = Vo/Vin = (1 + Rf/R1)


Observations :

1. In this amplifier input may not be applied directly to opamp terminalbut it may applied through some circuit.

2. Let Vin is input voltage applied to non-inverting amplifier throughsome resistive network such that voltage available at non-invertingterminal VB is different than Vin.

Then non-inverting amplifier always amplifies voltage available toits non-inverting terminal by factor (1 + Rf/R1).

Hence Vo = (1+Rf/R1).VB


9. Voltage Follower


9.Voltage Follower The circuit in which output voltage follows the input voltage

The node B is at potential Vin, Due to virtual ground concept

VB = VA = Vin

Node A is directly connected to output without resistivenetwork.

Vo = VA , So Vo = Vin

For this circuit , voltage gain is unity. For this circuit , voltage gain is unity.


The output voltage is equal to input voltage.

Vo α Vin

The output follows the input so this circuit is called ‘voltagefollower’.

It is also called Source Follower ,Unity Gain Amplifier, BufferAmplifier or Isolation Amplifier.

Advantages of voltage follower :

1. Very large input resistance , order of MΩ

Low output impedance almost zero. It can be used to connect high2. Low output impedance almost zero. It can be used to connect highimpedance source to a low impedance load to avoid loading effect.

3. It has larger bandwidth.

4. The output follows input exactly without phase shift.


10. Summing Amplifier


10. Summing Amplifier1. Inverting Summer In this circuit all input voltages v1, v2 & v3

are applied to inverting terminal through R1, R2 & R3.

As point B is grounded, due to virtual ground concept, node A is alsoat ground potential

VA = VB = 0

At input side,

I1 = V1 – VA /R1 = V1 / R1.

I2 = V2 / R2 , I3 = V3 / R3

Applying KCL at node A,

I = I1 + I2 + I3.

As opamp input current is zero, entire current I passes through Rf

Note : As input impedance of opamp is very large, so it is possibleto apply more than one input to inverting input termnal.This iscalled ‘Summer ‘ or ‘Adder’ circuit.


At output side,

I = (VA – Vo) / Rf = -(Vo / Rf)

I1 + I2 + I3 = -(Vo / Rf)

Put the values of I1 , I2 & I3

(V1/R1) + (V2/R2) + (V3/R3) = -(Vo/Rf)

Vo = -[(Rf/R1).V1 +(Rf/R2).V2 +(Rf/R3).V3 ]

Observation :

1.If R1 = R2 = R3 = R then V0 = -Rf/R1[V1 + V2 + V3]

2.If R1 = R2 = R3 = Rf = R then Vo = -[V1 + V2 + V3]2.If R1 = R2 = R3 = Rf = R then Vo = -[V1 + V2 + V3]

3.Negative sign indicates there is phase difference of 180 degreebetween input & output hence circuit called inverting amplifier.


2. Non-inverting Summer

The circuit gives non-inverted sum of input signals is callednon-inverting summing amplifier.

VA = VB , due to virtual ground concept

At input side,

I1 = V1 – VB / R1 & I2 = V2 – VB / R2

But as input current of opamp is zero, I1 + I2 = 0

(V1 – VB) / R1 + (V2 – VB) / R2 = 0

After simplificationAfter simplification

V1 / R1 + V2 / R2 = VB (1/R1 + 1/R2)

VB = (R2V1+R1V2)/(R1+R2) ---------eq.a

At node A, I = VA / R but VA = VB

I = VB / R -------------eq.1

At output side,

I = Vo – VA / Rf = Vo – VB / Rf . ------------eq.2

Equate eq.1 & eq.2


VB / R = Vo – VB / Rf

After simplification,

Vo / Rf = VB [1/R + 1 / Rf ]

After simplification,

Vo = VB [(R + Rf)/R]. --------eq 3

Put value of VB in eq 3

Vo = [(R2V1+R1V2).(R+Rf)] / [R(R1 + R2)]

Rearranging the equation

Vo = R2(R+Rf)/ R(R1+R2).V1 + R1(R+Rf) / R(R1+R2).V2Vo = R2(R+Rf)/ R(R1+R2).V1 + R1(R+Rf) / R(R1+R2).V2

This eq. shows that output is weighted sum of inputs.

If R1 = R2 = R = Rf we get Vo = V1 + V2


11. Integrator


11.Integrator Due to virtual ground concept

VB = 0, so VB = VA = 0As per 1st assumption input current is zero, so entire current pass

through R1,also through Cf.At input side ,I = Vin –VA/R1 = Vin / R1 ----eq.1At output side ,I = Cf. d / dt(VA – Vo)


I = Cf. d / dt(VA – Vo)= - Cf . dVo / dt -------------eq.2

Equate eq.1 & eq.2Vin / R1 = - Cf . dVo / dtIntegrating both sideʃ (Vin / R1 ).dt = - Cf ʃ (dVo / dt ).dt = - Cf.VoVo = -(1 / R1.Cf) ʃ Vin .dt + Vo (0) ---------------Integrator eq.Vo(0) – Constant of Integration , indicating initial output voltage.The equation show that output is -1/R1Cf times integral of input & R1Cf is

time constant of integrator.

Negative sign indicates that there is phase shift of 180 degreebetween input & output.

Sometimes Rcomp = R1 is connected at non-inverting terminal toprovide the bias compensation.


Input & Output Waveforms

i) Step input signal

Let the step input signal is applied with A magnitude

Mathematically it can be expressed

Vin(t) = A for t>=0

= 0 for t<0

From integrator eq. R1Cf = 1 & Vo(0) = 0

Vo(t) = - ʃ Vin .dt = - ʃ A .dt = -A ʃ dt = - A[t] (limit o to t) = -At

Vo(t) = -At ----output is straight line with slope of –A where A ismagnitude of step input.


Drawback of Ideal Integrator

1. The opamp gain is very high at zero frequency & in absence ofinput voltage. Vios get amplified & appears at output is called‘error voltage’

2. The Ib bias current also produces a capacitor charging current& add its effect in output error voltage.

3. Due to Vios & Ib & high opamp gain, out may achieve itssaturation level.

4. In presence of input voltage also, Vios & Ib produce an error due4. In presence of input voltage also, Vios & Ib produce an error dueto which true integration is not possible.

5. The output waveform may be distorted due to error voltage.

6. The bandwidth of ideal integrator is small hence can be usedfor very small range.

To overcome all these limitation , a practical integrator circuit isused


Practical Integrator

Rf – Connect Rf in parallel with Cf to reduce the low frequency gain.

Rcomp – Connect Rcomp at non-inverting amplifier to overcome biascurrent error.


Application of Practical Integrator

1. In Analog Computers

2. In solving the differential equation

3. In ADC

4. In various signal wave shaping circuits

5. In ramp generators.


12.Differentiator

Node B is at ground potential , due to virtual ground concept

VB = VA = 0

As input current is zero so entire amount of current I1 is flowing through Rf

At input side,

I1 = C1.d/dt(Vin – VA) = C1.dVin/dt ------eq.1I1 = C1.d/dt(Vin – VA) = C1.dVin/dt ------eq.1

At output side,

I1 = - Vo/Rf ----------------------------------eq.2

Equate eq. 1 & 2

C1.dVin/dt = - Vo/Rf

Vo = - C1.Rf.dVin/dt

Eq. shows that output is RfC1 times the differentiation of input, RfC1 is call time constant of the differentiator.


Input & Output Waveforms

i) Step input signal

Let the step input signal is applied with A magnitude

Mathematically it can be expressed

Vin(t) = A for t>=0

= 0 for t<0

From differentiator eq.

Vo(t) = - dVin / dt = - d(A )/dt = 0


Disadvantages

1. The gain of differentiator increases as frequency increases.thus at high frequency differentiator becomes unstable &

break into oscillations .There is possibility that opamp may go into saturation.

2. The input impedance Xc1 = (1/2ΠfC1) decreases as frequency increases .This makes circuit very much sensitive to noise .

3. Hence differentiator circuit suffers from limitation on its stablilty& noise problems , at high frequency.


Practical Differentiator

Cf II Rf & R1 in series with C1 ----The noise & stability at highfrequency can be corrected

Rcomp – Connect Rcomp at non-inverting amplifier biascompensation.


Application of Practical Differentiator

1. In Wave shaping Circuit – to detect the high frequencycomponents in the input signal.

2. In FM Demodulators.


13 Comparator Circuit which compares two signals

1. a signal voltage applied at one input of opamp

2. a known reference voltage at other input

Produce either high or low voltage depending on which input ishigher.

It produces output voltage either +Vsat or –Vsat

Comparator uses open loop configuration of opamp Comparator uses open loop configuration of opamp

Two Types –1. Non-Inverting Comparator

2. Inverting Comparator


Signal 1 : Vin at non-inverting terminal

Signal 2 : Vref at inverting terminal ; for basic comparator Vref = 0

Comparator compares Vin & Vref

Case I : If Vin > Vref then Vo = +Vcc = +Vsat

Case II : If Vin < Vref then Vo = -Vee = -Vsat

Reason of opamp goes into saturation

AOL of Opamp is very very high even for very small Vin , so opamp saturates


Apply Vref other than 0

Below waveform shows for

Case 1 : If Vin > Vref i.e Vref is positive

Case 2 : If Vin < Vref i.e Vref is negative

If we apply Vref other than zero to inverting input terminal ,tripping of comparator can be changed as per requirements.


Application of Comparator

1. Zero crossing detector

2. Level detector

3. Window detector

4. Duty cycle controller

5. Pulse generator


14.Schmitt Trigger Inverting mode produces opposite polarity output.

The output is feedback to non-inverting input which is of samepolarity as that of output. This ensures positive feedback.

Operation

Case I : When Vin > Vref then Vo = - Vsat

Case I : When Vin < Vref then Vo = + Vsat

The output voltage always at either at +Vsat or –Vsat but voltageat which it changes its state can be controlled by resistance R1 &at which it changes its state can be controlled by resistance R1 &R2.Thus Vref can be obtained as per requirment.

R1 & R2 forms potential divider network.

1. + Vref:VUT = Vo.(R2/R1+R2) = +Vsat.(R2/R1+R2)

2. - Vref : VLT= Vo.(R2/R1+R2) = -Vsat.(R2/R1+R2)

VUT : When Vo = +Vsat , +Vref is for positive saturation is calledupper threshold Voltage.

VLT : When Vo = -Vsat , -Vref is for negative saturation is calledlower threshold Voltage


15. IC555

For threshold > 2/3Vcc – FF –SET – Q –High –Transistor ON –goes in saturation ---C Discharge through Rb.C

For trigger < 1/3Vcc – FF –RESET – Q – Low –Transistor OFF --- C Charges through (RA+RB).C


16 . Block Diagram of Op-amp16 . Block Diagram of Op-amp


Commercial IC opamp consists of following blocks

1. Input Stage : 1.This stage requires High input impedance to avoid loading effect on sources. 2. It also requires low output impedance.

All such a requirements are achieved by Dual input unbalanced output differential amplifier.

3.This stage provides most of the voltage gain of amplifier.

2. Intermediate Stage : 1.The overall gain requirement is very high. Input stage alone can 1.The overall gain requirement is very high. Input stage alone can not provide such a high gain. To provide an additional voltage gain this stage consist of dual input balanced output differential amplifier. 2.This stage consists of chain of cascaded amplifiers called multistage amplifiers.

3. Level-Shifting Stage : 1.All stages are directly coupled to each other.As opampamplifies dc signal also. Hence stage by stage dc level increases well above ground potential.


2.Such a high dc voltage level may drive transistor intosaturation, this may cause distortion in output due to clipping

3.Hence before output stage it is necessary to bring high dcvoltage level to zero volts with respect to ground.4.Thus this stage brings dc level down to ground potential, whenno signal is applied to input terminal

4. Output Stage :1.Basic requirement of this stage is low output impedance , largeac output voltage swing & high current sourcing & sinkingac output voltage swing & high current sourcing & sinkingcapability.2. Such a requirement is achieved by push-pul complementaryamplifier.

3. This stage raises current supplying capability of opamp


17 . Idea Vs Practical 17 . Idea Vs Practical Characteristics of Op-amp


Sr.No Parameter Symbol Ideal Practical

1 Open Loop Gain AOL ∞ 2×105

2 Output Impedance Zout 0 75Ω

3 Input Impedance Zin ∞ 2MΩ

4 Input Offset Voltage Iios 0 20 nA

5Input Offset Current

Vios 0 2 mv5 Vios 0 2 mv

6 Bandwidth B.W ∞ 1 MHz

7 CMRR ρ ∞ 90 dB

8 SlewRate S ∞ 0.5 V/μsec

9 Input Bias Current Ib 0 80 nA

10Power Supply

Rejection RatioPSRR 0 30 μv/V


unit no 3 : linear integrated circuit · 2. feedback & its types feedback used invariably in...

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