CHAPTER 5

Operational Amplifier

Operational AmplifiersBuffer AmplifierSumming AmplifierDifferential AmplifierPositive & negative Feedback

OPERATIONAL AMPLIFIER

Block Diagram of Op-amp

- A signal voltage at the inverting(–) input tends to produce an output voltage Vout that is in opposite polarity to Vi (Voltage at inverting input).

- A signal voltage at the non inverting(+) input tends to produce a Vout that is the same polarity as Vni (Voltage at non inverting input).

Ideal Op-Amp Model

Assume,AVOL is very large~approximate 20 000 Generally, Vo =AVOL(Vni-Vi) Vo =AVdiff Vdiff =Vo/AVOL =15/200000 = ~0.07mV(quite small) Since Rin is considered infinity (~1MOhm), Ii=Vdiff/Rin = ~0V/1MOhm = 0 A

Inverting amplifier

-the inverting amplifier, will amplify the input,Vs and invert(sonsang) the value which is going to be negative value.

Inverting amplifierIR1=VR1/R1

= (VS-Vi)/R1

= (VS-~0)/R1

= VS/R1

= 0.5/1kOhm = 0.5mAIRF= (Vi -VOUT)/ RF

= (0-VOUT)/ RF

= -VOUT/ RF

So, VOUT= -IRFRF

= -IR1RF(IRF=IR1) = - (0.5mA) (20kOhm) = -10V (Since it is inverting, the output will be –ve)Closed loop voltage gain,AVCL= VOUT /Vs = -10V/0.5V = -20AVCL= VOUT /Vs = -VS (RF/R1) /Vs AVCL= -RF/R1=-20

Non inverting amplifier

V-term= VsIR1=VS/R1

So, VS=IR1R1IRF= (VOUT-VS)/RFBut IR1= IRFVS/R1= (VOUT-VS)/RFVOUT-VS= (RF/R1)VSVOUT = (RF/R1)VS+VS = VS[(RF/R1)+1] VOUT/ Vs = (RF/R1)+1 = 1+ RF/R1 = AV So, AV= 1+ RF/R1Thus, for R1= 1kOhm and RF=20kOhm, Vs= 0.5V as before, the non inverting amplifier provide voltage gain of AV= 1+ 20k/1k = 21 VOUT= 21Vs = 21(0.5) = 10.5V

-the value,Vs will be amplify but not going to be invert.

Op amp summing amplifier

I1=V1/R1I2=V2/R2IT=(0-VOUT)/RF but IT= I1+ I2-VOUT/ RF =V1/R1+V2/R2-VOUT= [V1/R1+V2/R2] RF

Let R1=10kOhm, R2=20kOhm, RF=40kOhm, V1=1.2V, V2= -1.9VSo, -VOUT= [V1/R1+V2/R2] RF -VOUT= [1.2/10k+(-1.9)/20k] 40k =1V VOUT= -1V

Differential amplifier

-amplify the difference between two signal

-the difference between this two signal is considered an error

-this error is going to be Vout. The gain is 1 since it only comparing. VOUT=(V2-V1) X RF/R1

-Example:

V1=2.2V ,V2=1.5V ,RD=RF=86kohm, VOUT=?

R1=R2=10kohm

VOUT=(1.5-2.2)X(86k/10k)

= (0.7) X 8.6

= -6V

Integrator

-for standard amplifier,which use two resistor has output nearly instantaneous.-but output of integrator is not instantaneous, it is function of time, which is controlled by the resistor capacitor network, R1 and C1.

-When input signal change, the capacitor changes to a new value and the time it takes to charge to a new value is T, the integral time.

VOUT=-(1/R1C1) X ∫VIN dt

-Cut off frequency, fc=1/(2 RFC1)

- Stop act as integrator when f < fc and act as inverting amplifier.

Example. Sketch and label the values of the output signal Vout for Vin that is 1kHz square wave with a peak voltage of + 1 V (2Vp-p)Solution.

The input signal is specified as square wave varying between +1V and-1V at 1 kHz rate,T= 1/f = 1/1kHz = 1ms.which means that the input will be +1V for half time, or 0.5ms and at -1V at 0.5ms.

For t=0 to 0.5ms, Vi=1V

Vo1= -(1/R1C1) X ∫VIN dt + Vo(0)

=-1/(10kΩ x 0.01µF) x + 0

= -10000 x

= -10000 x 0.5ms

= -5V

For t =0.5ms to 1.0msVo2= -(1/R1C1) X ∫VIN dt + Vo1

=-1/(10kΩ x 0.01µF) x + (-5V)

= [-10000 x ] - 5V = [-10000 x (-0.5ms)] -5V = 0 V

[t] 0.5ms 0ms

t][ 1.0ms 0.5ms

ms 0.5

ms 01dt

ms 1.0

ms 0.5(-1)dt

Op Amp differentiator

-The output of differentiator is proportional to the rate of change of

input; VOUT= - (RFC1) x dVIN/dt

-Cut off frequency, fc=1/2 R1C1

-If f > fc , it stops acting as differentiator and act as inverting amplifier.

Example: Calculate VOUT in figure above where RF=2.2k Ohm and C= 0.001uF and where VIN is ramp input that goes from +5V to -5V in time given from figure shown above.

Solution.For t= 0 to 50usVout= - (RFC1) x dVIN/dt = -(2.2k)(0.001u) x 10/(50u) = -0.44V

For t=50us to 100usVout= -(2.2k)(0.001u) x [-10/(50u)] = 0.44V

For t=100us to 150usVout=-(2.2k)(0.001u) x 10/(50u) = -0.44V

For t= 150us to 200usVout= -(2.2k)(0.001u) x [-10/(50u)] = 0.44V

Voltage follower

-Output is connected directly to its inverting input, thus producing the output that is equal to the non inverting input voltage in both amplitude and polarity. Output = Input, so Gain, A=1

Offset value of op amp-the offset value is a unique value for each op amp.-the original value of new op amp normally not zero as what is suppose to be.-This offset value will interrupt the operation of the circuit if not modify appropriately.-the way to modify is add certain value of resistor across the op amp to achieve value of zero.

R1=y , R2= 2y , R3 =4y , R4=6y , R5=8y , R6=10y , R7=12y , R8=14y, R9=16y , R10=18y , R11= 20y, R12=22y, R13=24y, R14=26yV1=x , V2 =2x , V3=4x ,V4=8x, V8=10xNote: Please simplified your answer.a)Name types of all five op amp above. b)Give values/expression for current flows through R1,R2,R3,R4,R5 a)Give expression/value for V5, V6, V7 and V9, Vout

R1=12kΩ , R2=33kΩ , R3 =80kΩ , R4=220kΩ, R5=30kΩ , R6=10 kΩ , R7=12kΩ , R8=63kΩ, R9=20kΩ , R10=240kΩ , R11= 200kΩ,R12=10 kΩ,R13=12 kΩ, R14=10kΩV1=3V,V2= 5V, V3=4V, V4=5V,V10=3V

A) Name all op amp above

B) Find current go through R1,R2,R3 and R4

C) Find Vout

Consider the integrator circuit in Figure given. Theinput signal, Vi to the circuit is a 10kHz square wave with 2Vp-p. Determine each of the following:a) Sketch and label the values of the output signal,Vo.b) At what freq. will the circuit stop acting as an

integrator.c) Sketch the output signal,Vo when it is not acting as a

integrator and what it is acting now.

Example: Calculate VOUT in figure above where RF=1.7k Ohm,R1=10kOhm and C= 0.008uF and where VIN is ramp input shown in figure below.

At what freq. it will stop acting as differentiator?What its gain now?