A signal voltage at the inverting(–) input tends to produce an output voltage Vout that is in opposite polarity to Vi (Voltage at inverting input).
A signal voltage at the non inverting(+)
input tends to produce a Vout that is the
same polarity as Vni (Voltage at
non inverting input).
Ideal Op-Amp Model
Assume,A VOL is very large~approximate 20 000
Generally, Vo =A VOL (V ni -V i )
Vo =A V diff
V diff =Vo/A VOL
= ~0.07mV(quite small)
Since Rin is considered infinity (~1MOhm), I i =V diff /R in
= 0 A
Inverting amplifier -the inverting amplifier, will amplify the input,Vs and invert(sonsang) the value which is going to be negative value.
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
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.
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
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
for standard amplifier,which use two resistor has output nearly
-but output of integrator is not instantaneous, it is function of time, which is
controlled by the resistor capacitor network, R1 and C1.
-When input signal change, the capacitor changes to a new value and the time it takes to charge to a new value is T, the integral time.
V OUT =-(1/R1C1) X ∫V IN dt
Cut off frequency, fc=1/(2 R F C1)
- Stop act as integrator when f < fc and act as inverting amplifier.
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.
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
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.
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
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
Offset value of op amp
-the offset value is a unique value for each
-the original value of new op amp normally
not zero as what is suppose to be.
-This offset value will interrupt the operation of
Consider the integrator circuit in Figure given. The
input signal, Vi to the circuit is a 10kHz square
wave with 2Vp-p. Determine each of the following:
Sketch and label the values of the output signal,Vo.
At what freq. will the circuit stop acting as an integrator.
Sketch the output signal,Vo when it is not acting as a integrator and what it is acting now.
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?