1Operational Amplifiers•Here we see two matcheddifferential amps cascadedto form a basic OP-AMP.•The differential pair canceltemperature drifts andcommon mode noise at theinput.•First built to perform mathoperations, analoguecomputers.2Differential Amplifiers•VOUT =AV (V1-V2)•+Vcc and -VEE bias the inputs and outputto 0 volts.•Common mode noise refers to same signpickup on V1 and V2.•The differential pair cancels common-mode signals in transistors Q1 and Q 2 eg, temperature drifts, noise,etc.+VccRCRCRE-VEEV1ΔVQ1 Q2VC2V2VC1noiseT-drift V1-V2123LM7414Ideal Op Amp•Very high or Infinite open loop gain. AV~100K•Very high or Infinite input impedance ~100MΩ.•Very low or zero output impedance. ~1Ω•Zero DC output voltage when inputs grounded. -+ZinZout5Negative Feedback• High gain amplifiers can becomeunstable in open loop configuration.• A portion of the output β can befeed back to the input to correct thissituation.• For Av>>1 the closed loop gainAvβ ∼ 1/βis only a function of the feedbackloop and β easily fixed. Av=VoutVin!!VoutVin "#Vin!=VoutVin( )1 "#VoutVin( )=Av#1 "#Av#!!!!Av#=Av1 +#Av!!!!!! Av#!!1#!!!!!!!when!! Av>> 1!!!!-+ZinZout−β6Inverting Amp1. V+=V- = 0 ( since no current flows thru Zin )2. Vin- I R1= 03. 0 - I R2=Vout4. AVβ = Vout/ Vin = -R2/R1 ( combining 2 and 3 )5. Zin = R1 (point-A at virtual ground). R1 should be fixed and not too small!-+ZinVinVout =-(R2/R1)VinR2R10VIVirtualgroundAI7Non-Inverting Amp1. VA=Vin ( No current flows thru Zin so ΔV+- = 0 )2. Vout- I R2= VA=Vin3. Vin - I R1 = 04. Vout = Vin (1+R2/R1)5. Input impedance = Zin + R1 ~ Zin-+ZinVinVout = Vin (1+R2/R1)R2R1IAVA ΔV+-8Voltage Follower-+ZinZoutVinVout1. ΔV+- =0 no current flows no voltage drop!2. Vin=VA=Vout3. Voltage Follower is used as an impedancematching device from a hi to low impedancedevise. It also acts as a current booster or buffer amp.VAΔV+-Zout lowZin hi9Difference Amp-+ZinV1Vout=(V2-V1) RF /R1R2R1AV2R3RF10Integrator/Differentiator-+V1RV2C Vout= !1/RC Vin"dtR-+V1CV2 Vout= !RC ddtVinintegratordifferentiator11Voltage SummerV1V2V3-+Vout = -(V1/R1+V2/R2+V3/R3)RFRFR1IR2R3•Inputs V1,V2,V3 are summed with gain factor Ri/RF I=1,2,3.12Frequency ResponseAVBand width ~ 1MHz-6db fhi flo We build an amplifier with gain of 1000,what will fhi be?Ans.- Based on the graph:fhi ~ (106 /gain) --> fhi~1 kHz13Frequency Response(2)With a voltage gain of 190, the amplifier will have that gain up to the frequency response Reached to about 6KHz below. Not very good for Hi-Fi reproduction! But if we were to reduce the voltage gain to just 20 then the frequency response would extend to 50KHz. If we needed an amplifier to have a gain of 190 then we would be far better off to have two cascaded amplifier; each with a voltage gain of just 14. The total gain would then be a little over 190 and the frequency response would be flat to over 70KHz.fmax ~ (106 /gain)14Instrumentation AmpgainbufferbufferAn unwanted common mode noise signal VCM appears on inputs 1and 2 with a weak transducer signal VS .The VCM is cancelled by the Difference Amp resulting in a noise-freeVS signal.Vout = Av(Vs+ - Vs-) + ACMRR(Vcm+ + Vcm-)Low-to-high impedance matchhigh input impedancelow output impedancelow input impedancehigh output impedancehigh output impedancelow input impedanceLow-to-high impedance match15INSRUMENATION AMPLIFIERInstrumentation amplifiers are actually made up of 2 parts: a buffered amplifier XOP1, XOP2 and a basic differential amplifier XOP3. The differential amplifier part is often essential when measuring sensors. Why? A sensor produces a signal between its terminals. However, for some applications, neither terminal may be connected to the same ground potential as your measuring circuit. The terminals may be biased at a high potential or riding on a noise voltage. The differential amplifier rescues the signal by directly measuring the difference between the sensor’s terminals.The buffered amplifier XOP1 and XOP2 not only provides gain, but provides impedance matching from a lowImpedance transducer to the high impedance Differentiral amplifier stage XOP3. SIGNAL GAINThe instrumentation amp offers two useful functions: amplify the difference between inputs and reject the signal that’s common to the inputs. The latter is called Common Mode Rejection (CMR). The signal gain is accomplished by XOP1 and XOP2 while XOP3 typically forms a differential gain of 1. You can calculate the overall gain by where R1=R3 and R5/R4 = R7/R6. http://www.ecircuitcenter.com/Circuits/instamp1/instamp1.htm16CMRR•The common mode rejection ratio (cmrr) reflects the ability of the op-amp to reject noise common to both inputs. • cmrr = Av /Acmrr = Ndb Consider an amplifier with cmrr=80db and Av=180 gain. If a differential signal of 2mVis applied along with an unwanted common mode signal of 10mV, what is theamplitude of each at the outputVdiff = 180 (2mV) = 360mVACMRR = Common mode gain =Av/80db = 180/10000 = 0.018Vcmrr = 0.018 (20 mV ) = 0.36 mVVout = Vdiff + Vcmrr = 360mV + 0.36mVVout = Av(V+ - V-) + ACMRR(V+ + V-)x Av+-17Analogue Computationdy2dt2+ k2y = 0-+RFR1RF d2ydt2R+-C !1RC"#$%&'!!yR+-C!1RC"#$%&'2y RC = 1RF/ R1= k2 ˙ ˙ y + k2y ! 0 +k2y +d2ydt2dydt18Analogue Computation !!y + a!y + by + c = 0!!!!!!!y = !(a!y + by + c) !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! "### $###Sum min g! Integratora = 1 / R3Cb = 1 / R2Cc = 1 / R1C !!!!!!!!!!!!!!!!!!!! "# $#Integrator1 = 1 / R'C !!!!!!!!!!!!!!!!! "# $#Inverter1 = R / R-+CR1R3 !!!y +a!y+byR2+c +!y-+! yC'-++ yR'RRVOUTloop!backsVCC19Amp Distortion20180o1)High gain for +feedback cancellation2) 180o+60o+60o+60o=360o- Music Tremolo-