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8/12/2019 Devices Experiment 1- Inverting and Non-Inverting
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University of Technology, Jamaica
Faculty of Engineering and Computing
Laboratory Report
Title of Experiment: Non-Inverting , Inverting and Summing Amplifier
Experiment # : 1
Instructor : Mr. Thorpe
Course Name: Electronic Devices and Circuits 2
Programme : BENG Electrical
Submitted by: Yanique Gibbs
ID No. : 1206283
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Aim: To investigate the operations of an Operational Amplifier through used in configurations
of inverting , non-inverting and summing .
Theory:
An op-amp is a high-gain direct-coupled amplifier that is often powered by both a positive and a
negative supply voltage. This allows the output voltage to swing both above and below ground.
Figure a. Op-amp
Note the two inputs label + and - ; the positive input is called the non-inverting and the
negative input the inverting .In many applications , one of the amplifier inputs is grounded , so vo
is out of phase with the input if the signal is connected to the non-inverting terminal, and vo is
out of phase with the input if the signal is connected to the inverting input.
When a small input voltage is applied , this will cause the amplifier to be driven all the way to itsextreme positive and negative voltage limits . However resistors are connected to and around the
amplifier in such a way that the signal undergoes vastly smaller amplification. The resistors
cause gain reduction through signal feedback.
Circuits using op-amps and resistors can be configured to perform many useful operations such
as inverting , non-inverting and summing .
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Figure b. Inverting Amplifier
In this very useful application of an op-amp ,the non-inverting input is grounded , Vin, is
connected through Rin to the inverting input and feedback resistor Rfis connected between the
output and V2.
Kirchoffs node equation at V1 yields
V1/ R = 0
Kirchoffs node equation at V2 yields
VinV2 / Rin + VinV2 / Rf = 0
V1 = V2 yields
V1 = V2 = 0
Therefore the closed-loop gain as
Vo/Vin = - Rf / Rin
The negative sign in the equation indicates an inversion of the output signal with respect to the
input as it is 180
o
out of phase. This is due to the feedback being negative in value.
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Figure c. Non-Inverting op-amp
Figure c shows another useful application of an op-amp called the non-inverting configuration.
The signal Vinis connected directly to the non-inverting input and the resistor R2 is connectedfrom the inverting input to ground. Under the ideal assumption of infinite input impedance , no
current flows into the inverting input so Iin=If.
Thus , V1 / R2 = VoV1/ Rf
V0 = A(VinV1)
V1= VinVo/ A
Letting A = infinite , the term Vo/ A foes to zero
Thus , V1= Vin
Vin / R2 = Vo Vin / Rf
Solving for Vo / Vin / Rf
Vo / Vin = 1 + Rf / R2
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Figure d. Summing op-amp
It is possible to sum several input voltages in one op-amp circuit called a summing amplifier .
Figure d shows and inverting op-amp circuit that can be used to sum and scale three input
signals. Note that the signals V1, V2and V3 are applied through separate resistors R1, R2and
R3to the summing junction of the amplifier and that there is a signal feedback resistor Rf.
Following the same procedure we used to derive the output of the inverting amplifier having a
single input , we obtain for the three-input op-amp.
If= I1+ I2+ I3= - [ V1/Rin+ V2/Rin+ V3/Rin]
Inverting equation Vo= Rf/Rin* Vin
Then Vo = - Rf [ V1/Rin+ V2/Rin+ V3/Rin]
Equipment:
Solderless Breadboard
Cables & Wires
741 C Operational Amplifier
Several - Watt Resistors
Decade Capacitor Box (0-10 4-silicon diodes F )
9V Battery
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Diagrams:
Refer to lab manual (Figures 9.5, 9.6 and 9.7).
Method:
As stated in lab manual (Methods 9.5.1, 9.5.2 and 9.5.3).
Results:
Table 1 . Showing results for the Non-Inverting Amplifier
Ri = 9.73 k Vi = 0.4 V Vi (rms) = .141 V
Rf = 32.6 k Vo = 1.75V Vo(rms) = 1.23 V Input and Output voltage in
phase
Period (T) = 1ms Frequency = 1KHz
When Rf is short-circuit Vo equals to Vin.
0.4V = 0.4 V
Diagram 1. Showing Output and Input waveforms in-phase.
When function generator supplies 10 kHz @ 4V peak-to-peak
Ri = 9.73 k Vi = 4 V
Rf = 99.5 k Vo = 15.5V Slope = 96.96 x 10-
V/s
Gain = 11.23 % Error in gain = 2.04 %
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Table 2. Showing the results when the gain is 2 (simulated):
R1/ k Rf/ k Vi/v Vi/v(Rms) Vo/v Vo/v(Rms)
32.5 32.6 .4 .141 0.8 .2828
Table 3. Showing the results when the gain is 11:
R1/ k Rf/ k Vi/v Vi/v(Rms) Vo/v Vo/v(Rms)
9.73 99.5 4 1.41 15.5 10.96
Frequency= 10kH
Slew rate 96.96 x10-6
V/s: (rise/run)
Rise =4.15
Diagram 2. Showing a 4V peak to peak Square wave Input Waveform of Circuit in Figure 9.5with at 10kHz
Diagram 3.Showing the output and the effect of the slew rate.
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Table 4. Showing results for the Inverting Amplifier
Diagram 4. Showing the input and output waveforms produced from the inverting op-amp inorder to obtain a voltage gain of -1
Diagram 5. Showing the input and output waveforms produced from the inverting amplifierhaving a voltage gain of -2.2 and is out of phase by 180
o.
Gain of -10 Gain of -2.2 Gain of -1
Ri .986 k .986 k .986 k
Rf 9.72 k 2.2 k .986 k
Vi 0.4 V 0.4 V 0.4 V
Vi (rms) .141 V .141 V .141 V
Vo 3.94 V 845 mV 382 mV
Vo (rms) 1.39 V 0.299 V 0.135 V
Uncertainty in Gain 1.4 % 1.36 % 0 %
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Table 5. Showing results for the Inverting Amplifier with
Multiple Inputs
Diagram 4
Calculation and Uncertainties:
Sample Calculation for a non-inverting amplifier
( ) ( )
Inverting Amplifier with
Multiple InputsR1 1.2 k
R2 17.80 k
Rf 3.9 k
V1 0.4 V
V2 9.08 V
Vo 1.6V
V0 (rms) 0.5656
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Sample Calculation for a inverting amplifier
Vo = - (Rf / R) * Vin
Vo = - (9.72 k/ .986 k) * 0.4
Vo = 3.94 V
Sample Calculation for a inverting amplifier with multiple inputs.
VO1= -(Rf/R1)*Vin -(3.9k/1.2k)*.4v 1.3v
VO2= -(Rf/R2)*Vin -(3.9k/17.80k)*.4v .0876v
Vo= VO1 + VO2
Vo = 1.30876V
Theoretical Gain Vo = 1.6 V% Error in Vo = (1.61.30876)*100%/ 1.6 = 18.20%
Analysis of Results:
1. The derived theoretical input-output relationship for a non-inverting amplifier.Ii = If
Vi / Ri = (Vo- V-) / Rf
But Vo = (V+
- V-) A
Hence Vi / Ri = (Vi+
- Vi-) AV
-/ Rf
Vo = A (V+- Vi )
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Vi = Vi+- Vo /A
Vi= Vi+
Since Vi-= Vi
+, then we can replace Vi
-with Vi
+. Hence Vi
-/R can be written as Vi
+/ R.
Hence Vi+/ Ri = (Vo - Vi
+) / Rf
Vi+
*Rf = (VoVi
+) / R1
Vi+
*Rf = Vo * R1Vi
+* R1
Vi+
*Rf + Vi
+* R1 = Vo * R1
Vi(Rf + R1 ) = Vo * R1
Therefore Rf + R1 = Vo * R1 / Vi
Therefore Vo / Vi = Rf + Ri / Ri
Vo / Vi = 1 + Rf/ Ri
Gains for circuit depicted in Figure 9.5 of Lab sheet:
Rf= 32.6 k
Ri= 9.73 k
Theoretical Gain AV= (1 + (32.6 / 9.73)) = 4.35
Practical Gain AV= 1.85/0.4 = 4.35
% Error in Gain = (4.354.35)*100%/4.61 = 0%
Gains for circuit depicted in Figure 9.5 of Lab sheet with Rfshort circuited:
Rf= 0 k (Short Circuit)
Ri= 9.73 k
Theoretical Gain AV= (1 + (0 / 9.73)) = 1
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Practical Gain AV= 0.4/0.4 = 1
% Error in Vo= (0.40.4)*100%/0.4 = 0 %
To get a gain of 2 for circuit depicted in Figure 9.5 of Lab sheet:Theoretical Gain AV= (1 + 1) = 2
Rf/Ri= 1 so Rf= Ri
Rf= 32.6 k
Ri= 32.5 k
Practical Gain AV= 0.801/0.4 = 2.0025
% Error in Gain= (2.00252.00)*100%/ 2.00= 0.125%
To get a gain of 11 for circuit depicted in Figure 9.5 of Lab sheet:
Theoretical Gain AV= (1 + 10) = 11
Rf/Ri= 10 so Rf= 10*Ri
Rf= 32.6k
Ri= 3.260k
Practical Gain AV= 4.4 /0.4 = 11
% Error in Gain = (1111)*100%/11 = 0%
To get a gain of 11 @10kHz 4V peak-peak for circuit depicted in Figure 9.5 of Lab sheet:
Theoretical Gain AV= (1 + 10) = 11
Rf/Ri= 10 so Rf= 10*Ri
Rf= 99.5 k
Ri= 9.73 k
Practical Gain AV= 44.9 /4 = 11.22
% Error in Gain = (11.2211)*100%/11 = 1.96%
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2. The derived theoretical input-output relationship for a inverting amplifier.If + Iin= 0
Iin = - If
Vin / Rin= -Vo/ Rf
Vo= -Rf* Vin/ Rin
Vo/Vin= - Rf/ Rin
3. The derived theoretical input-output relationship for a inverting amplifier withmultiple inputs.
It= I1+ I2+ I3+. In
But It + If = 0
Hence It = - If
Also, I1= V1/R1, I2= V2/R2, In= Vn/Rn
And If = Vo / Rf
Therefore from It = - If , we can write these in terms of voltage and resistance
So, It = - If ; V1 / R1 + V2/ R2 + .. Vn/ Rn = Vo / Rf
If R1 R2 R3
Then Vo = - Rf (V1 / R1 + V2/ R2 + .. Vn/ Rn )
If R1 = R2 = R3 , then :
V1 / R1 + V2/ R2 + .. Vn/ Rn = - Vo / Rf
Rf / R1 (V1 + V2 + .. Vn) = Vo
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Discussion :
The theory and the practical application for a non-inverting amplifier coincide as the input and
the output signals were found to 90 degrees out of-phase ; as noted in Diagram 1. Also all output
signals from the non-inverting amplifier had the same frequency as the input signals. From table
1. an input of .4V was applied to the circuit and a output of 1.75 V . It can be clearly seen that
the signal was amplified. However when the feedback resistor was short out of the circuit , the
voltage applied was equal to the measured at Vo. It can be noted that the feedback takes part of
the amplified output , such that the gain is constrained much more by the feedback network and
less by the open loop gain. In this case however the amplifier acts as an buffer and only passes
the input signal to the output.The gains were measured by using the formulaVout= Vin(1+ (Rf/Ri) as highlighted in the theory. According to the equation , an increase in Vin
with brig rise to an increase in the output voltage ; as they are proportional . However they will
be a slight decrease in Vo because as the frequency increases , the gain decreases , thus resulting
in a change in Vo. The gain of the amplifier was found to be 4.35 with a percentage error of zero.
To determine the gain for the other non-inverting circuits the same equation was applied. The
uncertainties for the gain of 2 and 11 were found to be o and 1.96 %. The slew rate of the of the
circuit was found to be 96.96 x 10-6
V/s . This was done by taking the frequency and then tused
to calculate the time period. The difference between the peak and the trough were measured and
then divided by the time period calculated. The slew rate was however smaller that the accepted
slew rate of 0.5V/ s . This meant that the change between the input and output signal was very
slow. As the faster the slew rate the faster the change in output signal. The discrepancy in
reading may have been due to the 16V which was supplied to the circuit as the amplifier was
unable to handle the amplitude of the output.
From the theory is was noted that the inverting amplifier input and output signals will be out of
phase by 180 degrees and producing a negative gain which coincides with the experiment done.
The gain was calculated by using the equation Vo/ Vi= -Rf/Ri as derived in the theory above. To
obtain a gain of -10 , resistors .986 kand 9.72 k were chosen for R1 and Rf respectively.
These values were substituted into the equation to produce the gain of -10. When applied to the
circuit the uncertainty in reading were calculated to be 1.4 %. To obtain a gain of -2.2, the
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resistors Rf and Ri were used as 2.2 k and0.986 k respectively. The uncertainty in the gain
was found to be 1.36 %. To obtain a voltage gain of -1, the resistors Rfand Riwere used as .986
k and .986k respectively. The uncertainty was calculated to be zero . The output waveforms
for the inverting circuit were observed to be is out of phase with the input signals by 180o.
The summing amplifier is similar that of and inverting amplifier but just with multiple inputs. To
measure the gain ,the method of superposition had to be introduced . Which is done by
measuring output voltage and signal at a time . After obtaining the voltages V01and V02 , the
gain of each input signal was calculated by equation -Rf/Ri and then summed . The theoretical
output voltage was determined by the use of the equation Vo= (G1V1+ G2V2). The outputs
obtained were out of phase with the input, this is so because a summing amplifier is an inverting
amplifier, as was earlier stated.
Conclusion :
The experiment was proven to be true as the experiment values coincide with that of
experimental values; however slight discrepancies in values were noted. It was seen that the op-
amps with a combination resistors can be configured to perform many useful operations such as
inverting , non-inverting and summing .
References:
A.V.Bakshi. (2009). Electrical Networks.India: Technical Publications Pune.Peter Y. Yu, M. C. (2007). Fundamentals of ectification and smoothingr: Physics and
Applications.Germany: Springer.
Boylestad, Robert; Nashelsky, Louis (1998).Electronic Devices and CircuitTheory, Prentice Hall
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Edward Hughes, Ian McKenzie Smith, et al,Hughes Electrical and Electronic Technology10
thEdition,
http://www.electronics-tutorials.ws/diode/diode_1.html
http://www.electronics-tutorials.ws/diode/diode_1.htmlhttp://www.electronics-tutorials.ws/diode/diode_1.htmlhttp://www.electronics-tutorials.ws/diode/diode_1.html