bab 41_op amp

CHAPTER 4OPERATIONAL AMPLIFIER(PENGUAT KENDALIAN)Ref:080114HKNOperational Amplifier*

Operational Amplifier

Op-AmpIntroduction of Operation Amplifier (Op-Amp)Analysis of ideal Op-Amp applicationsComparison of ideal and non-ideal Op-AmpNon-ideal Op-Amp consideration

Ref:080114HKNOperational Amplifier*


PengenalanRef:080114HKNOperational Amplifier*Menguatlan voltan DC dan AC.

Mempunyai gandaan yang tinggi.

Cip silikon terdiri daripada komponen seperti perintang, pemuat dan transistor.


Ref:080114HKNOperational Amplifier*DIP Dual In Line


Ref:080114HKNOperational Amplifier*An ideal opamp has the following properties:

i. infinite open-loop gain ii. infinite bandwidth iii. zero input current iv. zero output impedance v. zero offset voltage


Ref:080114HKNOperational Amplifier*Kegunaan Umum- Mempunyai bidang ruang dari 0Hz 1MhzContoh: 709, 101, 741, 301

Kegunaan DC dan prestasi rendah- Mempunyai galangan masukan yang sangat tinggi- Voltan offset masukan rendahContoh: H0052, 108

Kegunaan AC dan prestasi tinggi- Mempunyai bidang ruang yang lebar- Contoh: H0063


Ref:080114HKNOperational Amplifier*Untuk kuasa dan voltan tinggiMenggunakan satu talian bekalan kuasa dan boleh memacu beban secara terus.Contoh: H0004, H0021, M124

Boleh diprogram- Penguat kendalian khas yang boleh diprogram- Contoh: 4250


Ref:080114HKNOperational Amplifier*Sebagai litar senilai komputer analog yang boleh melaksanakan fungsi Pencampur, pendarab, pembahagi, pengkamil dan pengkerbedaan.

Penjana gelombangDiguna pada Wein Bridge Oscillator bagi menjana gelombang sain.Mengguna litar-litar pengkamil dan pengkerbedaan untuk menghasilkan gelombang berbentuk segiempat dan segitiga.

Penapis aktif RC Menapis isyarat iaitu membenarkan isyarat pada frekeunsi tertentu saja melaluinya.

Penganda kepada isyarat-isyarat- Menguatkan isyarat audio/video/radio frekeunsi- Menguatkan isyarat digit yang dihantar melalui jarak jauh.


Ref:080114HKNOperational Amplifier*Penguat kendalian akan rosak sekiranya kadaran maksima yang berikut dilepasi (exceeded)

Kuasa pelepasan maksima 310mW

Julat suhu kendalian 0o C hingga 70o C

Bekalan Voltan maksima 18 volt

Voltan masukan pembezaan maksima iaitu antara terminal 30V.

Voltan masukan mod sepunya maksima 12V, pada bekalan kuasa 15V.

Julat suhu simpanan 55o C hingga 125o C


Ref:080114HKNOperational Amplifier*Saiz yang kecil

Keboleharapan yang tinggi

Harga yang murah

Boleh bekerja pada julat suhu yang rendah

Voltan dan arus offset yang rendah.

Mengganda dan menguatkan isyarat

Sebagai litar senilai untuk komputer analog contoh : pencampur , pembahagi, pengkamil, pengkerbeda.

Penjana gelombang isyarat


Operational Amplifier (Op-Amp)Very high differential gainHigh input impedanceLow output impedanceProvide voltage changes (amplitude and polarity)Used in oscillator, filter and instrumentationAccumulate a very high gain by multiple stages



IC ProductRef:080114HKNOperational Amplifier*DIP-741Dual op-amp 1458 device


Single-Ended InputRef:080114HKNOperational Amplifier* + terminal : Source terminal : Ground 0o phase change + terminal : Ground terminal : Source 180o phase change


Double-Ended InputRef:080114HKNOperational Amplifier* Differential input 0o phase shift change between Vo and Vd

Qu: What Vo should be if, (A) (B)Ans: (A or B) ?


DistortionRef:080114HKNOperational Amplifier*The output voltage never excess the DC voltage supply of the Op-Amp


Ref:080114HKNOperational Amplifier*Common-Mode Operation Same voltage source is applied at both terminals Ideally, two input are equally amplified Output voltage is ideally zero due to differential voltage is zero Practically, a small output signal can still be measured

Note for differential circuits:Opposite inputs : highly amplifiedCommon inputs : slightly amplified Common-Mode Rejection


Common-Mode Rejection Ratio (CMRR)Ref:080114HKNOperational Amplifier*Differential voltage input : Common voltage input : Output voltage :Ad : Differential gainAc : Common mode gainCommon-mode rejection ratio:Note:When Ad >> Ac or CMRR Vo = AdVd


CMRR ExampleRef:080114HKNOperational Amplifier*What is the CMRR?Solution :NB: This method is Not work! Why?(1)(2)


Op-Amp PropertiesRef:080114HKNOperational Amplifier*Infinite Open Loop gain The gain without feedbackEqual to differential gainZero common-mode gainPratically, Gd = 20,000 to 200,000(2) Infinite Input impedanceInput current ii ~0AT- in high-grade op-amp m-A input current in low-grade op-amp(3) Zero Output Impedanceact as perfect internal voltage sourceNo internal resistanceOutput impedance in series with loadReducing output voltage to the loadPractically, Rout ~ 20-100


Frequency-Gain RelationRef:080114HKNOperational Amplifier*Ideally, signals are amplified from DC to the highest AC frequencyPractically, bandwidth is limited741 family op-amp have an limit bandwidth of few KHz.Unity Gain frequency f1: the gain at unityCutoff frequency fc: the gain drop by 3dB from dc gain AdGB Product : f1 =Ad fc20log(0.707)=3dB


GB ProductRef:080114HKNOperational Amplifier*Example: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Ad = 20V/mV

Sol:Since f1 = 10 MHzBy using GB production equationf1 = Ad fcfc = f1 / Ad = 10 MHz / 20 V/mV = 10 106 / 20 103 = 500 Hz


Ideal Vs Practical Op-AmpRef:080114HKNOperational Amplifier*

IdealPracticalOpen Loop gain A105Bandwidth BW10-100HzInput Impedance Zin>1MOutput Impedance Zout0 10-100 Output Voltage VoutDepends only on Vd = (V+V)Differential mode signalDepends slightly on average input Vc = (V++V)/2 Common-Mode signalCMRR10-100dB


Ideal Op-Amp ApplicationsAnalysis Method :

Two ideal Op-Amp Properties:The voltage between V+ and V is zero V+ = VThe current into both V+ and V termainals is zero

For ideal Op-Amp circuit:Write the kirchhoff node equation at the noninverting terminal V+ Write the kirchhoff node eqaution at the inverting terminal V Set V+ = V and solve for the desired closed-loop gainRef:080114HKNOperational Amplifier*


Ref:080114HKNOperational Amplifier*1. Noninverting AmplifierKirchhoff node equation at V+ yields,

Kirchhoff node equation at V yields,

Setting V+ = V yields

or


Ref:080114HKNOperational Amplifier*Noninverting amplifier Noninverting input with voltage divider Voltage follower Less than unity gain


Ref:080114HKNOperational Amplifier*2. Inverting AmplifierHow it work :

Differential input voltge (Vdiff) is virtually zero, so the ve input terminal voltage (Vi) is at virtual ground potential.

Input current (Ii) is virtually zero.


Inverting AmplifierKirchhoff node equation at V+ yields,



Ref:080114HKNOperational Amplifier*Notice: The closed-loop gain Vo/Vin is dependent upon the ratio of two resistors, and is independent of the open-loop gain. This is caused by the use of feedback output voltage to subtract from the input voltage.


Output OffsetOutput offset voltage is the output of an operational amplifier when the two inputs are shorted together (and often tied to ground).

Its used to minimize the offset voltage error due to bias current and the offset voltage at the amplifier output and closed-loop gain times the offset voltage at the amplifier input will be equal.Ref:080114HKNOperational Amplifier*R3 = R1 // R2


Ref:080114HKNOperational Amplifier*VA = VB = 6V B

A


3. Multiple Inputs (Summing)Kirchhoff node equation at V+ yields,





Ref:080114HKNOperational Amplifier*R4 = R1 // R2 // R3


4. Inverting IntegratorRef:080114HKNOperational Amplifier*Now replace resistors Ra and Rf by complex components Za and Zf, respectively, therefore

SupposingThe feedback component is a capacitor C, i.e.,

The input component is a resistor R, Za = RTherefore, the closed-loop gain (Vo/Vin) become:

whereWhat happens if Za = 1/jC whereas, Zf = R? Inverting differentiator


4. Voltage Follower Ref:080114HKNOperational Amplifier* The unique noninverting amplifier.

The output connected directly to its inverting input.

The gain is always 1.Vout = Vin

Zin =


5. Op-Amp IntegratorThe integrator is mostly used in analog computers, analog-to-digital converters and wave-shaping circuits.

Integrator (and inverts) the input signal Vin(t) over a time interval t, t0 < t < t1, yielding an output voltage at time t = t1 Ref:080114HKNOperational Amplifier*


Op-Amp IntegratorRef:080114HKNOperational Amplifier*Example:

Determine the rate of changeof the output voltage.

Draw the output waveform.Solution:(a) Rate of change of the output voltage (b) In 100 s, the voltage decrease


6. Op-Amp DifferentiatorRef:080114HKNOperational Amplifier*2. The charge on the capacitor equals Capacitance x Voltage across the capacitor1. I flowing through the capacitor will be given as3. The rate of change of this charge is4. but dQ/dt is the capacitor current I5. from which we have an ideal voltage output for the op-amp differentiator is given as:


Op-Amp DifferentiatorRef:080114HKNOperational Amplifier*


Op-Amp DifferentiatorExample: Calculate Vout in figure below where Vin is ramp input that goes from -5 V to -5V in 50us.

Ref:080114HKNOperational Amplifier*Solution:Vout = -(Rf.C) x dVin/dt = -(2.2k)(0.001u) x 10/50u = -0.44V


Compare Integrator and DifferetiatorRef:080114HKNOperational Amplifier*


Non-ideal case (Inverting Amplifier)Ref:080114HKNOperational Amplifier*3 categories are considering

i) Close-Loop Voltage Gainii) Input impedanceiii) Output impedanceEquivalent Circuit


i) Close-Loop GainRef:080114HKNOperational Amplifier*Applied KCL at V terminal,

By using the open loop gain,The Close-Loop Gain, Av


Close-Loop GainRef:080114HKNOperational Amplifier*When the open loop gain is very large, the above equation become,Note : The close-loop gain now reduce to the same form as an ideal case


ii) Input ImpedanceRef:080114HKNOperational Amplifier*


Input ImpedanceRef:080114HKNOperational Amplifier*Finally, we find the input impedance as,Since,, Rin become, Again withNote: The op-amp can provide an impedance isolated from input to output


iii) Output ImpedanceRef:080114HKNOperational Amplifier*Only source-free output impedance would be considered, i.e. Vi is assumed to be 0Firstly, with figure (a),By using KCL, io = i1+ i2By substitute the equation from Fig. (a),R and A comparably large,


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