The document discusses operational amplifiers and their applications. It describes the basic op-amp configuration, ideal op-amp model, and applications such as inverting amplifier, non-inverting amplifier, summing amplifier, differential amplifier, integrator, differentiator, and voltage follower. It also discusses offset adjustments and multiple op-amp circuits.
This presentation contains the basic information you need to know about operational amplifier.
I have tried to cover all the basic info. If anything is left out or you have any suggestions i will appreciate it.
this presentation is based on basic description of inverting and non-inverting amplifiers using op-amps and their medical use, hope it helps students :)
The performance obtainable from a single-stage amplifier is often insufficient for many applications, hence several stages may be combined forming a multistage amplifier. These stages are connected in cascade, i.e. output of the first stage is connected to form input of second stage, whose output becomes input of third stage, and so on.
thank u
Hansraj MEENA
This presentation contains the basic information you need to know about operational amplifier.
I have tried to cover all the basic info. If anything is left out or you have any suggestions i will appreciate it.
this presentation is based on basic description of inverting and non-inverting amplifiers using op-amps and their medical use, hope it helps students :)
The performance obtainable from a single-stage amplifier is often insufficient for many applications, hence several stages may be combined forming a multistage amplifier. These stages are connected in cascade, i.e. output of the first stage is connected to form input of second stage, whose output becomes input of third stage, and so on.
thank u
Hansraj MEENA
It’s a power electronics project. It is able to give output voltage(DC) more and less than input voltage as per requirement.
We can generate variable DC voltage which is less than input, but, the special things about this converter is, it has capability to produce variable DC voltage as high as twice the input voltage.
We have specially designed and manufactured inductor for this project.
Introduction to operational Amplifier. For A2 level physics (CIE). Discusses characteristics of op amp, inverting and non inverting amplifier, and voltage follower, and transfer characetristics, virtual earth , etc
It’s a power electronics project. It is able to give output voltage(DC) more and less than input voltage as per requirement.
We can generate variable DC voltage which is less than input, but, the special things about this converter is, it has capability to produce variable DC voltage as high as twice the input voltage.
We have specially designed and manufactured inductor for this project.
Introduction to operational Amplifier. For A2 level physics (CIE). Discusses characteristics of op amp, inverting and non inverting amplifier, and voltage follower, and transfer characetristics, virtual earth , etc
Fundamentals of oprational Amplifiers.pptxadityaraj7711
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Understand the “magic” of negative feedback and the characteristics of ideal op amps.
Understand the conditions for non-ideal op amp behavior so they can be avoided in circuit design.
Demonstrate circuit analysis techniques for ideal op amps.
Characterize inverting, non-inverting, summing and instrumentation amplifiers, voltage follower and first order filters.
Learn the factors involved in circuit design using op amps.
Find the gain characteristics of cascaded amplifiers.
Special Applications: The inverted ladder DAC and successive approximation ADC
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6. Inverting amplifier -the inverting amplifier, will amplify the input,Vs and invert(sonsang) the value which is going to be negative value.
7. Inverting amplifier I R1 =V R1 /R 1 = (V S -V i )/R 1 = ( V S -~0)/R 1 = V S /R 1 = 0.5/1kOhm = 0.5mA I R F = (V i -V OUT )/ R F = (0-V OUT )/ R F = -V OUT / R F So, V OUT = -I RF R F = -I R1 R F (I RF =I R1 ) = - (0.5mA) (20kOhm) = -10V (Since it is inverting, the output will be –ve) Closed loop voltage gain ,A VCL = V OUT /Vs = -10V/0.5V = -20 A VCL = V OUT /Vs = -V S (R F /R 1 ) /Vs A VCL = -R F /R 1 =-20
8. Non inverting amplifier V-term= Vs I R1 =V S /R 1 So, V S =I R1 R 1 I RF = (V OUT -V S )/R F But I R1 = I RF V S /R 1 = (V OUT -V S )/R F V OUT -V S = (R F /R 1 )V S V OUT = (R F /R 1 )V S +V S = V S [(R F /R 1 )+1] V OUT / Vs = (R F /R 1 )+1 = 1+ R F /R 1 = A V So, A V = 1+ R F /R 1 Thus, for R 1 = 1kOhm and R F =20kOhm, Vs= 0.5V as before, the non inverting amplifier provide voltage gain of A V = 1+ 20k/1k = 21 V OUT = 21Vs = 21(0.5) = 10.5V -the value,Vs will be amplify but not going to be invert.
9. Op amp summing amplifier I1=V1/R1 I2=V2/R2 IT=(0-V OUT )/RF but IT= I1+ I2 -V OUT / RF =V1/R1+V2/R2 -V OUT = [V1/R1+V2/R2] RF Let R1=10kOhm, R2=20kOhm, RF=40kOhm, V1=1.2V, V2= -1.9V So, -V OUT = [V1/R1+V2/R2] RF -V OUT = [1.2/10k+(-1.9)/20k] 40k =1V V OUT = -1V
10. 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. V OUT =(V 2 -V 1 ) X R F /R 1 -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
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12. 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.
13. 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 1.0ms Vo2= -(1/R1C1) X ∫VIN dt + Vo1 =-1/(10kΩ x 0.01µF) x + (-5V) = [-10000 x ] - 5V = [-10000 x (-0.5ms)] -5V = 0 V
14. 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.
15. Solution. For t= 0 to 50us Vout= - (RFC1) x dVIN/dt = -(2.2k)(0.001u) x 10/(50u) = -0.44V For t=50us to 100us Vout= -(2.2k)(0.001u) x [-10/(50u)] = 0.44V For t=100us to 150us Vout=-(2.2k)(0.001u) x 10/(50u) = -0.44V For t= 150us to 200us Vout= -(2.2k)(0.001u) x [-10/(50u)] = 0.44V
16. 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
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21. Example: Calculate VOUT in figure above where RF=1.7k Ohm,R1=10kOhm and C= 0.008uF and where V IN is ramp input shown in figure below. At what freq. it will stop acting as differentiator?What its gain now?