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Universal Collage Of Engineering And Technology Subject : Analog Electronics NAME: EN.NO: Jay Bhavsar Yagnik Dudharejiya JAY Pandya Darshan Patel 130460109006 130460109013 130460109034 130460109043 Guidance by : Prof. Kapil Dave 1 Symbol of op-amp Analyze of op-amp circuit Packages of op-amp Pin configuration of op-amp Applications of op-amp Frequency response of op-amp Design of op-amp Power supplies of op-amp 2 Circuit symbol of an op-amp •Widely used •Often requires 2 power supplies + V •Responds to difference between two signals 3 Characteristics of an ideal op-amp •Rin = infinity •Rout = 0 •Avo = infinity (Avo is the open-loop gain, sometimes A or Av of the op-amp) •Bandwidth = infinity (amplifies all frequencies equally) 4 I+ V+ + + IV- - Vout = A(V+ - V-) - •Usually used with feedback •Open-loop configuration not used much 5 Vout = A(V+ - V-) Vout/A = V+ - VLet A infinity then, V+ -V- 0 6 V+ = VI+ = I- = 0 Seems strange, but the input terminals to an op-amp act as a short and open at the same time 7 •Write node equations at + and - terminals (I+ = I- = 0) •Set V+ = V- •Solve for Vout 8 Types of Packages: Small scale integration(SSI)<10 components Medium Scale integration of op-amp(MSI)<100 components Large scale integration (LSI)>100 components Very large scale integration (VLSI)>1000 components 9 I2 I1 = (Vi - V- )/R1 I2 = (V- - Vo)/R2 I1 set I1 = I2, (Vi - V-)/R1 = (V- Vo)/R2 but V- = V+ = 0 Gain of circuit determined by external components Vi / R1 = -Vo / R2 Solve for Vo Vo / Vi = -R2 / R1 10 TYPES: The flat pack Metal can or Transistor Pack The duel-in-line packages(DIP) 11 Very popular circuit 12 Current in R1, R2, and R3 add to current in Rf (V1 - V-)/R1 + (V2 - V-)/R2 + (V3 - V-)/R3 = (V- - Vo)/Rf Set V- = V+ = 0, V1/R1 + V2/R2 + V3/R3 = - Vo/Rf solve for Vo, Vo = -Rf(V1/R1 + V2/R2 + V3/R3) This circuit is called a weighted summer 13 Offset Null 1 2 Non-inverting Input 3 Vcc- 4 Inverting Input A741 8 7 6 5 NC Vcc+ Output Offset Null 14 As a integrator As a differentiator 15 I2 I1 = (Vi - V-)/R1 I2 = C d(V- - Vo)/dt I1 set I1 = I2, (Vi - V-)/R1 = C d(V- Vo)/dt Output is the integral of input signal. CR1 is the time constant but V- = V+ = 0 Vi/R1 = -C d(Vo)/dt Solve for Vo Vo = -(1/CR1)( Vi dt) 16 17 NON-INVERTING CONFIGURATION (0 - V-)/R1 = (V- - Vo)/R2 I I Vi But, Vi = V+ = V-, ( - Vi)/R1 = (Vi - Vo)/R2 Solve for Vo, Vo = Vi(1+R2/R1) 18 Vi = V+ = V- = Vo Vo = Vi Isolates input from output 19 Write node equations using: OR Write node equations using: V+ = V- model, let A I+ = I- = 0 Solve for Vout infinity Solve for Vout Usually easier, can solve most problems this way. Works for every op-amp circuit. 20 Rin = Vin / I, from definition VI V+ V Rin = Vin / 0 Rin = infinity 21 Rin = Vin / I, from definition I I = (Vin - Vout)/R V- I = [Vin - A (V+ - V-)] / R V+ Vout = A(V+ - V-) But V+ = 0 I = [Vin - A( -Vin)] / R Rin = VinR / [Vin (1+A)] As A approaches infinity, Rin = 0 22 Inverting configuration Non-inverting configuration Vi Rin = 0 at this point Vo /Vi = - R2/R1 Vo/Vi = 1+R2/R1 Rin = R1 Rin = infinity 23 DIFFERENCE AMPLIFIER Use superposition, set V1 = 0, solve for Vo (non-inverting amp) set V2 = 0, solve for Vo (inverting amp) Fig. A difference amplifier. 24 DIFFERENCE AMPLIFIER Vo1 = -(R2/R1)V1 Vo2 = (1 + R2/R1) [R4/(R3+R4)] V2 Add the two results Vo = -(R2/R1)V1 + (1 + R2/R1) [R4/(R3+R4)] V2 25 Vo = -(R2/R1)V1 + (1 + R2/R1) [R4/(R3+R4)]V2 For Vo = V2 - V1 Set R2 = R1 = R, and set R3 = R4 = R For Vo = 3V2 - 2V1 Set R1 = R, R2 = 2R, then 3[R4/(R3+R4)] = 3 Set R3 = 0 26 When measuring Rin at one input, ground all other inputs. Rin at V1 = R1, same as inverting amp Rin at V2 = R3 + R4 27 Add buffer amplifiers to the inputs Rin = infinity at both V1 and V2 28 where w0 = 1/RC (a) Magnitude response of (single time constant) STC networks of the low-pass type. 29 OPEN-LOOP FREQUENCY RESPONSE OF OP-AMP Open-loop gain at low frequencies Break frequency(bandwidth), occurs where Ao drops 3dB below maximum Unity gain frequency, occurs where Ao = 1 (A = 0dB) 30 •Open-loop op-amp •Inverting and non-inverting amplifiers •Low-pass filter •High-pass filter 31 FREQUENCY RESPONSE OF OPEN-LOOP OP-AMP Open-loop op-amp: ft = Ao fb where Ao is gain of op-amp 32 Inverting or noninverting amplifier: ft = |A| fb, where A = gain of circuit - 20 dB/dec A = - R2 / R1, inverting A = 1 + R2/R1, non-inverting |A| fb ft 33 Z2 Z1 Vi A = - Z2 / Z1 1 R2 R2 1 R1 A sC 1 R2 R1 1 sCR2 sC •At large frequencies A becomes zero. •Passes only low frequencies. 34 Low-pass filter: C acts as a short at high frequencies, gain drops to zero at high frequencies, ft |A| fb. fb = 1/2pR2C Due to external C Due to Vi - 20 fb 35 Z2 Z1 C VVi A = - Z2 / Z1 A R2 R1 1 sC •At large frequencies A becomes - R2 / R1. •Passes only high frequencies. 36 High-pass filter: C acts as an open at low frequencies, gain is zero at low frequencies, C Vi Due to external capacitor fL = 1/2pR1CDue to op-amp bandwidth 37 C Design the circuit to obtain: High-frequency Rin = 1KW High-frequency gain = 40dB lower 3 dB frequency = 100Hz •Rin = R1 + 1/sC. At high frequencies, s becomes large, Rin R1. Let R1 = 1KW •A = - R2 / (R1 + 1/sC). At high frequencies, s becomes large, A R2 / R1 . A = 40dB = 100, 100 = R2 / 1KW, R2 = 100KW. •fL = 1/2pR1C C = 1/2p R1 fL, 1/2p(1KW)100 = 1.59mF C= 38 20 DB/DECADE (DUE TO CAPACITOR) -20 DB/DECADE (DUE TO OP-AMP) FL = 100HZ 39 OUTPUT OF HIGH-PASS FILTER IN EXAMPLE 40 C2 C1 BANDPASS FILTER • Both C2 and C1 act as shorts at high frequencies. • C2 limits high-frequency gain • C1 limits low-frequency gain • The gain at midrange frequencies = - R2 / R1 •fL = 1/2pR1 C1 •fH = 1/2pR2 C2 -20 dB/decade (due to C2) bandwidth fL fH 41 Saturation: Input must be small enough so the output remains less than the supply voltage. Slew rate: Maximum slope of output voltage. Response time of op-amps are described by a slew rate rather than a delay. 42 PSRR:(Power supply rejection ratio) CMRR:(Common mode rejection ratio) 43 Definition: The change in an op-amp input offset voltage (Vios) caused by variation in the supply voltage and it is called as “power supply rejection ratio”. It is also called a “SVRR”, AND “PSS”. Equation: PSRR:Vios/v 44 Definition: It is the ratio of common mode gain and differential gain. Equation : Vcm=(V1+V2)/2 Vo=AdVd+AcmVcm Where Ad=Differential gain and Vcm= Common mode gain CMRR=(Acm/Ad) Final equation : Vo=[V1-V2+Vcm/CMRR]Ad 45 www.google.com 1. Op-Amp and Linear integrated Circuit technologyRamakant A Gayakwad, PHI Publication 2. Digital Fundamentals by Morris and Mano, PHI Publication 3. Micro Electronics Circuits by SEDAR/SMITH.Oxford Pub F.COUGHLIN, FREDERICK F. DRISCOLL 4. Operational Amplifier and Linear integrated Circuits By K.LAL kishore. 5. Fundamentals of Logic Design by Charles H. Roth Thomson 46 Thank you 47