l01 operational amplifier

Lecture 1 Op-AmpIntroduction of Operation Amplifier (Op-Amp)Analysis of ideal Op-Amp applicationsComparison of ideal and non-ideal Op-AmpNon-ideal Op-Amp consideration

Operational Amplifier

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 ProductDIP-741Dual op-amp 1458 device


Single-Ended Input + terminal : Source terminal : Ground 0o phase change + terminal : Ground terminal : Source 180o phase change


Double-Ended Input Differential input 0o phase shift change between Vo and Vd

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


DistortionThe output voltage never excess the DC voltage supply of the Op-Amp


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)Differential voltage input : Common voltage input : Output voltage :Gd : Differential gainGc : Common mode gainCommon-mode rejection ratio:Note:When Gd >> Gc or CMRR Vo = GdVd


CMRR ExampleWhat is the CMRR?Solution :NB: This method is Not work! Why?(1)(2)


Op-Amp PropertiesInfinite 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 RelationIdeally, 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 GdGB Product : f1 = Gd fc20log(0.707)=3dB


GB ProductExample: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV

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


Ideal Vs Practical Op-Amp

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 gain


Noninverting AmplifierKirchhoff node equation at V+ yields,

Kirchhoff node equation at V yields,

Setting V+ = V yields

or


Noninverting amplifier Noninverting input with voltage divider Voltage follower Less than unity gain


Inverting AmplifierKirchhoff node equation at V+ yields,



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.


Multiple InputsKirchhoff node equation at V+ yields,




Inverting IntegratorNow 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


Op-Amp IntegratorExample:

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


Op-Amp Differentiator


Non-ideal case (Inverting Amplifier)3 categories are considering

Close-Loop Voltage Gain Input impedance Output impedanceEquivalent Circuit


Close-Loop GainApplied KCL at V terminal,

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


Close-Loop GainWhen 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


Input Impedance


Input ImpedanceFinally, we find the input impedance as,Since,, Rin become, Again withNote: The op-amp can provide an impedance isolated from input to output


Output ImpedanceOnly 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,


l01 operational amplifier

Documents

voltage differential

common voltage input

input differential input

dual opamp

highgrade opamp

amplified output voltage

ideal opamp properties

loadreducing output