FeedbackAmps-306

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ECE306
Lab 2, Feedback amplifiers
04.12.2018
LAB 2 FEEDBACK AMPLIFIERS
PRELAB ASSIGNMENT
1.
Read Sedra and Smith 5th ed.[1] ,sections 8.1,8.3,8.4 and 8.6.1 before coming to the lab.
Read the laboratory instructions and compare to S&S. 6th ed. sections 10.1,10.3, 10.4 and 10.6.
2.
Calculate the theoretical value of the voltage gain V2 /V1 for the circuit shown in Figure 2 of the
notes. Note that there is no load resistor.
3.
Calculate the theoretical value of the open loop voltage gain Vo'/Vs for the circuit shown in
Figure 5 of the notes.
4.
Calculate the theoretical value of the open loop feed-forward gain A for the circuit shown in
Figure 5 of the notes.
5.
Calculate the theoretical open loop output resistance Ro without feedback for the circuit shown
in Figure 5 of the notes.
6.
Calculate the theoretical loop gain Aβ for the circuit shown in Figure 5 of the notes.
7.
Calculate the theoretical value of the closed loop voltage gain Vo/Vs for the shunt-shunt
amplifier in Figure 4 of the notes.
8.
Calculate the theoretical closed loop output resistance with feedback Rof for the shunt-shunt
amplifier in Figure 4 of the notes.
9.
Calculate the theoretical open loop feed-forward gain A for the series-shunt amplifier in Fig. 9
of the notes.
10.
Calculate the theoretical open loop output resistance without feedback Ro for the series-shunt
amplifier in Fig. 9 of the notes.
11.
Calculate the theoretical loop gain Aβ for the series-shunt amplifier in Fig. 8 of the notes.
12.
Calculate the theoretical value of the closed loop voltage gain Vo/Vs using the circuit in Fig. 8 of
the notes.
13.
Calculate the theoretical closed loop output resistance with feedback Rof using the circuit in
Fig. 8 of the notes.
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ECE306
Lab 2, Feedback amplifiers
04.12.2018
LABORATORY INSTRUCTIONS
In this lab you will investigate the effects that negative feedback has on amplifiers. You will start with
a basic building block amplifier with know parameters. You will apply feedback to improve the
parameters. You will study two different types of amplifiers in this lab using similar evaluation
techniques. Part 1, part 2 and part 3 have a theory section and lab procedures.
Introduction: Feedback Amplifiers
Basic Feedback Amplifiers,
1. A block and β block
Amplifier Topologies
1. voltage-mixing voltage-sampling
2. current-mixing current-sampling
3. voltage-mixing current-sampling
4. current-mixing voltage-sampling
Part 1:
Basic amplifier module
Procedure 1: Build the basic amplifier. Measure the voltage gain and output impedance.
Part 2:
Shunt-shunt amplifier
Procedure 2: Build the A block of a shunt-shunt amplifier and measure the gain.
Procedure 3: Measure the open loop output impedance of the shunt-shunt amplifier.
Procedure 4: Measure the closed loop voltage gain of the shunt-shunt amplifier.
Procedure 5: Measure the closed loop output impedance of the shunt-shunt amplifier.
Part 3:
Series-shunt amplifier
Procedure 6: Build the A block of a series-shunt amplifier and measure the gain.
Procedure 7: Measure the output impedance of the series-shunt amplifier.
Procedure 8: Measure the closed loop voltage gain of the series-shunt amplifier.
Procedure 9: Measure the closed loop output impedance of the series-shunt amplifier.
List of Figures
Figure 1: The basic block diagram of a negative feedback amplifier.
Figure 2: The circuit schematic for the ‘Basic’ amplifier.
Figure 3: ‘Basic’ amp model.
Figure 4: The shunt–shunt topology model using the ‘basic’ amp.
Figure 5: The A circuit model, note the feedback resistor, β, is disconnected.
Figure 6: The circuit schematic for the shunt-shunt A block.
Figure 7: The circuit schematic for the shunt-shunt closed loop amp.
Figure 8: The series–shunt topology model using the ‘basic’ amp.
Figure 9: The A circuit model for the series–shunt amplifier.
Figure 10: The circuit schematic for the series-shunt A block.
Figure 11: The circuit schematic for the series-shunt closed loop amp.
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Lab 2, Feedback amplifiers
04.12.2018
Introduction
Negative feedback is used to improve the parameters of a basic amplifier building block. The simple
op-amp would be almost useless without it. Negative feedback has a number of desirable effects.
Among these are: controlled, well defined gain, extended bandwidth, improved input impedance, and
improved output impedance. Feedback amplifiers are generalized into the form shown in Fig. 1. The
A block usually has a significant amount of gain. The β block provides the negative feedback required
to achieve those desirable effects listed earlier. The amplifier topology presented in Fig. 1 allows
similar techniques to be used to evaluate the parameters and the performance of different types of
amplifiers. The different types are listed in table 1.
Negative feedback
Figure 1 shows the basic block diagram for negative feedback. The x signals can be voltage or
current. The signal types define the amplifier type as listed in Table 1. The gain of the feed-forward
network is A, and the gain of the feedback network is β.
xs
xi
Source
A
xo
Load
xf
β
Figure 1: The basic block diagram of a negative feedback amplifier.
The text makes use of the idea of separating the amplifier function A from the feedback function β so
that the general amplifier feedback equation can be used for any combination of x in voltage or
current. This closed loop gain, Af, equation is:
Af 
xo
A

x s 1  Aβ
(1)
Feedback topologies
The feedback network samples from the output and mixes it back into the input. If the output is a
voltage, then it must be sampled in shunt. If it is a current, then it must be sampled in series. If the
input is a voltage, then it must be mixed in series. If it is a current, then it must be mixed in shunt. A
feedback amplifier is denoted either by its input–output type (e.g. voltage-to-voltage) or its mixing–
sampling type (e.g. series–shunt). The two descriptions are equivalent. Figures 8.XX taken from [1].
helps understand the series / shunt configurations.
Typical amplifier type
Voltage Amplifier
Current Amplifier
Transconductance Amp
Gain
VO/VIN
IO/IIN
IO/VIN
Reference topology name
voltage-mixing voltage-sampling
current-mixing current-sampling
voltage-mixing current-sampling
3
Alternate topology name
series–shunt
shunt–series
series–series
ECE306
Lab 2, Feedback amplifiers
Transimpedance Amp
VO/IIN
current-mixing voltage-sampling
04.12.2018
shunt–shunt
Table 1: The four basic feedback amplifier topologies.
series- shunt
shunt-series
series-series
shunt-shunt
Figures 8.XX: The four basic feedback topologies
In this lab, we will investigate two different feedback amplifier topologies: a current-mixing voltagesampling (shunt–shunt) and a voltage-mixing voltage-sampling (series–shunt). See Table 1. In
voltage-sampling current-mixing, "voltage-sampling" refers to sampling the feedback voltage at the
output and "current-mixing" refers to summing the source and feedback currents at the input. In
(series–shunt), "series" refers to the mixing configuration at the input and "shunt" refers to the
sampling configuration at the output.
Basic amplifier
Figure 2 shows the ‘basic’ amplifier block we will use throughout this lab. This basic building block
amplifier is used instead of simply an op-amp. The reason for using the ‘basic’ amp is because a
simple op-amp has near ideal parameters. That means it is almost impossible to measure them or the
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ECE306
Lab 2, Feedback amplifiers
04.12.2018
parameters of a feedback amplifier constructed with it. In other words we will use a poor op-amp for
the following experiments. The parameters we are concerned about measuring are the input and
output impedance and the gain. This building block will be used in conjunction with other circuitry to
form a feedback amplifier. The other circuitry connected to this ‘basic’ amplifier will determine both
the A and the β of the feedback amplifier. As you will see the ‘basic’ amp becomes the A block when
you include the loading affects of the source, load and β impedances. First let’s analyze then build the
‘basic’ amp. We need to determine the parameters of interest, namely, the input and output
impedance and the gain. We can determine them from analyzing the circuit within the rectangle which
is the ‘basic’ amp. Then we will build it and measure them.
Rf
Vplus
100K
Vminus
4
5
Vminus
2
V1
GND
U1
Vo
V2
R1
1K
C1
0.1
C2
0.1
1K
7
1
3
Ro
6
LF356
+
1K
-
Rin
GND
Vplus
Basic Amplifier
Ceramic Bypass Caps
Caps in uF
1.0 uF Bypass Caps are better
Mount close to power pins
Figure 2: The circuit schematic for the ‘Basic’ amplifier.
Inside the basic amp is a standard inverting op-amp circuit with additional resistors, R1 and RO. We
will get to RO later. R1 does not influence the parameters that we are concerned with. It is there to
keep the DC offset of the basic amplifier low. Most op-amp circuits result in a lower DC offset when
each of the two input terminals is loaded by the same impedance. This is from ‘op-amps 301’. On to
the parameters of interest. You probably know that the closed loop gain of this circuit is usually given
as in Eq. 2.
VO
Rf

(2)
V1
Rin
This gain is given as a ratio of VO to V1 like the voltage amp topology. This equation is correct. Note
VO is inside the basic amp block. This amp can also be represented as a trans-impedance amp. We
can also call it a shunt–shunt or a current-mixing voltage-sampling circuit. Trans-impedance has a
VO/IIN gain equation. Well it does have this equivalent gain but we like to think about the inverting amp
gain as VO/V1. The trans-impedance gain equation for the basic amp is given in Eq. 3 and the input
and output impedances are given in equations 4 & 5.
VO
 Rf
Iin
(3),
V1
 Rin
Iin
(4),
V2
 RO
IX V10
See the appendix if you want to know how we get these equations. It’s easy.
5
(5)
ECE306
Lab 2, Feedback amplifiers
04.12.2018
So let’s calculate the gain as given in Eq. 1 using A and β for the inverting op-amp block to see if we
get Eq. 3 as a final result. A = -100,000,000 and β = -1/100,000. See the appendix to understand
where these values come from. It’s hard but worth the effort to understand it. Equation 1 gives:
Af 
A
100M
100M
100M



 99,900 VA  100K VA
100M
1  Aβ
1  100K
1  1000
1001
(6)
Equation 6 demonstrates that the gain equation and analysis technique (appendix) do agree.
We can draw a new model for our ‘basic’ amp using the parameters we just determined in our
analysis. Figure 3 shows the new ‘basic’ amp model. This amp model makes it easier to analyze our
feedback amplifier configurations.
Ro 1K
V1
Rin
+
V1
GND
1K
V2
-100V1
GND
Basic Amplifier
Figure 3: ‘Basic’ amp model.
From now on we will present circuits with the basic amp model
shown in Fig. 3 followed by the actual circuit built for you on the
proto boards from Fig. 2. We will not show the bypass capacitors
in the following circuits but they are used.
Note picture (1) to the right is the proto-board with the basic
amplifier circuit. V+ (1) and V- (7) are the power supply pins.
+Vin (6) is grounded in Fig. 2. –Vin (3) is the input V1 in Fig. 2.
Picture 1. Basic Amp Prototype
GND (2) is where the bypass capacitors are grounded. NC (4,5)
are not connected to anything on the board. Finally VO‘ (8) is V2 in Fig. 2 and is usually referred to as
VO’ in the text. There are 2 small wires sticking up from the board acting as test points for the power
supply voltages that appear at the VCC (7) and VDD (4) pins of the op-amp.
1.
The “basic amp” circuit shown in Fig. 2 is already built for you. See Picture 1. Measure the
voltage gain V2/V1 at 1 kHz and compare with the theoretical value. Measure the -3dB
frequency, with respect to 1KHz, and the unity gain frequency for the basic amplifier. Measure
the output resistance of the basic amplifier, as described below.
It is easy to measure the output impedance. Measure V2 (VO’) for a 1KHz sine wave input, pick an
input level that doesn’t clip the output signal, with no load. Make sure you are not clipping with the
scope. Now measure V2 (VO’) with a resistor loading the output of the ‘basic’ amp. Keep the load
resistor somewhere between 100Ω and 10KΩ. Assume the voltage source inside the ‘basic’ amp
model is ideal. i.e. 0Ω with a real output impedance. With the above measurements you have
everything you need to calculate Ro using Kirchoff’s laws. You are not going to be told how to
calculate this. Figure it out. You should also be able to guess RO from Figure 3. Measure it even
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ECE306
Lab 2, Feedback amplifiers
04.12.2018
though you can guess. You will use this output impedance measurement technique later in the lab
and it will be good to know you are doing it right.
Shunt–shunt amplifier
For the first part of the experiment, we use the ‘basic’ amp to form a current-to-voltage (shunt–shunt)
feedback amplifier as shown in Figure 4.
The closed loop gain of this current-to-voltage feedback amplifier, From table 1, is voltage out
divided by current in. A transconductance equation as given in Eq 7. This equation comes from the
analysis technique used in Fig. 8.20 in [1]. See the appendix of this lab.
A
Vo
Is
(7)
where IS is the current delivered by the source voltage VS.
Rf b
100K
Rs 1K
Ro 1K
Vs
+
Rin
Vo
V1
1K
GND
-100V1
GND
Basic Amplifier
Figure 4: The shunt–shunt topology model using the ‘basic’ amp.
A circuit
To measure the open loop gain of the feed-forward, A, network, we must form the A circuit as shown
in Fig. 5. This consists of the basic amplifier loaded by the feedback and source impedances but with
the feedback function disconnected. (Fig. 8.20[1])
Rs 1K
Ro 1K
Vs
Rin
GND
Rf b
100K
+
V1
1K
Vo'
-100V1
-
Rf b
100K
Basic Amplifier
GND
Figure 5: The A circuit model, note the feedback resistor, β, is disconnected.
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Lab 2, Feedback amplifiers
Rf
04.12.2018
100K
2
Vs
Rf b
100K
GND
R1
Ro 1K
6
LF356
U1
Vo'
Vo
Rf b
7
1
3
+
1K
-
Rin
4
5
Vminus
Rs 1K
100K
1K
Vplus
GND
Basic Amplifier
Figure 6: The circuit schematic for the shunt-shunt A block.
We will use Fig. 5 with to calculate the theoretical open loop gain of the A circuit. We start with VS and
end up with VO’. This amplifier has the gain of VO’/IS. Where IS is the current delivered by the source
to the amplifier. If we define the voltage source as VS with a source impedance of RS then we can
generate an IS of VS/RS with an impedance of RS. With this definition the current into the amplifier is
IS, the current feeds RS, RFB and RIN in parallel to generate V1.
A
VO '
IS
VO '  VO
and
Rf b
R f b  RO
IS 
VS
RS
(9)
V1  IS
REQ
Rin
V1  IS
RSR f b
RSRin  RSR f b  RinR f b
so
A
VO '
RS
VS
and VO  100V1
REQ is RS, Rfb and Rin in parallel
REQ 
(8)
(10)
RSRinR f b
RSRin  RSR f b  RinR f b
so
(11)
Express 8 using 9, 10 and 11
A
VO '
V RSR f b
V RSR f b
I
RSR f b
RSR f b
RS  O
 100 1
 100 S
VS
VS R f b  RO
IS R f b  RO
IS RSRin  RSR f b  RinR f b R f b  RO
A  100
RSR f b
RSR f b
RSRin  RSR f b  RinR f b R f b  RO
(13)
for Rfb >> RO and Rin
A  100
RS
 50Rin  50,000 VA
2
Note From 8 IS 
(14)
VS
I
1
so S 
VS R S
RS
(15)
lets see what VO’/VS is
8
(12)
ECE306
Lab 2, Feedback amplifiers
04.12.2018
VO ' VO ' IS
I
RSR f b
RSR f b 1

 A S  100
VS
IS VS
VS
RSRin  RSR f b  RinR f b R f b  RO RS
For Rfb >> RO and Rin
(16)
VO '
RSR f b
RSR f b 1
RS

 100
 50 VV
VS (R S  Rin )R f b R f b RS
RS  Rin
(17)
Figure 8.20[1] shows the Norton equivalent of Figure 5 in which IS divides between RS, Rfb, and Rin
and thus reduces the gain to about Rin/Rfb of what it would be if Rin were infinite. Thus for small Rin a
reasonable, close enough, gain of the feed-forward network, A, is given in Eq.14
2.
Connect the circuit shown in Figure 6. Set the frequency of Vs to 1 kHz. Measure the open
loop voltage gain Vo'/Vs.
Open loop output resistance Ro (using the circuit in Fig. 6)
3.
Use the method described above (p5).
Calculating Aβ
The feedback network samples voltage in shunt and injects current into RS||Rin in shunt, Fig. 8.XX.
The gain of the feedback network is given by:
β
1
1

Rf b
100,000
A
V
(17)
Note: The theoretical loop gain Aβ, is much lower than the usual gain for most op-amp based
feedback amplifiers. We have chosen Aβ low so that we will be able to measure changes caused by
using feedback and compare measured with theoretical values. For the basic amplifier if we
substituted an operational amplifier with a typical gain of 100,000, Aβ would be very large and it
would be difficult to measure output resistance with feedback Ro etc.
With an A of - 50,000 VA and β  
1 A
we get an Aβ = 0.5.
100kΩ V
An alternate way to calculate the loop gain, Aβ, from Fig. 4 is to short circuit the input and break the
feedback loop between VO and Rfb. Then the gain around the loop is:
Aβ 
RS || Rin
1KΩ || 1KΩ
500
1
100 
100 
100 
100  0.498  0.5
RS || Rin  R f b
1KΩ || 1KΩ  100KΩ
100,500
201
Closed loop voltage gain
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ECE306
4.
Lab 2, Feedback amplifiers
04.12.2018
Measure the closed loop voltage gain Vo/Vs using Fig. 7.
Rf b
100K
Rf
100K
Rin
4
5
Vminus
Rs 1K
1K
2
6
LF356
3
+
Vo
U1
7
1
R1
GND
Ro 1K
-
Vs
1K
GND
Vplus
Basic Amplifier
Figure 7: The circuit schematic for the shunt-shunt closed loop amp.
Closed loop output resistance Rof
The theoretical closed loop output resistance with feedback Rof from Eq. 8.37 [1]
Rof 
Ro
1  Aβ
(18)
5.
Using the circuit in Fig. 7, measure the closed loop output resistance Rof by measuring the
open circuit and loaded output voltages as you did before.
Series–shunt amplifier
For the second part of the experiment, we use the basic amplifier to form a voltage-to-voltage
(series–shunt) feedback amplifier as shown in Figure 8.
T1
Vg
1
1:1
3
Ro 1K
+
Vs
GND
2
-4
Rin
Vo
+
V1
1K
-100V1
-
R2
Basic Amplifier
10K
R1
1K
GND
Figure 8: The series–shunt topology model using the ‘basic’ amp.
Here we must use a transformer to isolate the signal generator reference from the ground of the
amplifier and the oscilloscope. The closed loop gain of this voltage-to-voltage feedback amplifier is
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Lab 2, Feedback amplifiers
Vo
Vs
Af 
04.12.2018
(19)
Note VS is not the output voltage of the signal generator. The methodology used to find A and β is
presented in Fig. 8.11 [1] shown in the appendix of this lab. The gain of the feedback network is:
β
R1
R1  R 2
(20)
The A circuit
To find the gain of the feed-forward network, A, we must form the A circuit as shown in Fig. 9. This
consists of the ‘basic’ amplifier loaded with the feedback impedances but with the feedback function
disconnected. The feed-forward gain is:
A
Vo '
Vs
(21)
6.
Connect the circuit shown in Fig. 10. Set the frequency of VS to 1 kHz and measure the
open loop voltage gain A. Remember you can’t ground either side of the VS signal or you will
change the circuit significantly when measuring it. Use the DVM or V1-V2 on the scope. Pin 1 on
the transformer is the red wire, 2 is black, 3 is blue and 4 is yellow.
T1
Vg
1
1:1
3
Ro 1K
+
Vs
GND
2
-4
Rin
Vo'
+
V1
1K
-100V1
-
R2
Basic Amplifier
909
R1||R2
10K
R1
1K
GND
Figure 9: The A circuit model for the series–shunt amplifier.
Note that V1<Vs because of the attenuation caused by Rin and R1||R2. To calculate the theoretical
gain in Fig. 8, Vs is divided by the attenuator formed by R1||R2 and the input resistance of the basic
amplifier Rin. We also need to know the attenuation factor for VO’.
Input attenuation is:
V1
Rin
1KΩ


 0.524
VS Rin  R1 || R2 1KΩ  0.909kΩ
And the output attenuation is:
11
(22
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Lab 2, Feedback amplifiers
04.12.2018
VO '
R1  R2
11KΩ


 0.916
VO RO  R1  R2 12KΩ
(23)
Now we can evaluate Eq. 21.
A
Vo ' 0.917VO 0.917( 100)V1
91.7



 48.05 VV
Vs V1 0.524
V1 0.524
1 0.524
(24)
Collecting terms from 21, 22 and 23 we get Eq. 25:
A
VO'
Rin
R1  R2

(-100)  0.524  0.917  ( 100)  48.05 VV
VS Rin  R1 || R2 Ro  R1  R2
(25)
Same answer that Eq. 24 gives how nice.
Rf
100K
3
2
3
Vs
GND
2
4
R3
Ro 1K
6
LF356
U1
Vo'
Vo
7
1
1:1
+
1
1K
-
Vg
Rin
4
5
Vminus
T1
1K
R2
Vplus
10K
Basic Amplifier
R1
R1||R2
909
1K
GND
Figure 10: The circuit schematic for the series-shunt A block.
Open loop output resistance Ro (using the circuit in Fig. 10)
7.
Measure the open loop output resistance RO as you did before by measuring VO’ and VOL’.
Once again measure VO’ with the 10KΩ and 1KΩ resistors connected to the output then add an
additional RL to get the new VO’.
Calculating Aβ
Again note the theoretical loop gain Aβ, is much lower than that for most feedback amplifiers. We
have chosen it low so that we will be able to measure changes caused by using feedback and
compare measured with theoretical values.
The feedback network samples voltage in shunt and injects voltage into Vs in series. The gain of the
feedback network is given by
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ECE306
Lab 2, Feedback amplifiers
04.12.2018
R1
1kΩ

 0.09
R1  R 2
1kΩ  10kΩ
Aβ = -48.05(-0.0909) = 4.37
β
(26)
An alternative way to find the loop gain Aβ is to short circuit the input Vs and break the feedback loop
between Vo and R2. Then the gain around the loop is
βA 
R1 || Rid
1kΩ || 1kΩ
(-100)  
(-100)  0.0437( 100)  4.37
R1 || Rid  R2
1kΩ || 1kΩ  10kΩ
(27)
Closed loop voltage gain (using the circuit in Fig. 11)
8.
At 1 kHz, measure the closed loop voltage gain Vo/Vs.
Rf
100K
3
2
3
Vs
GND
2
4
R3
Ro 1K
6
LF356
1K
Vo
U1
7
1
1:1
+
1
1K
-
Vg
Rin
4
5
Vminus
T1
R2
Vplus
10K
Basic Amplifier
R1
1K
GND
Figure 11: The circuit schematic for the series-shunt closed loop amp.
Closed loop output resistance Rof (using the circuit in Fig. 11)
The theoretical closed loop output resistance with feedback Rof from Sedra and Smith equation (8.19)
Ro
Rof 
(28)
1  Aβ
9.
Measure the closed loop output resistance ROf as you did before by measuring VO’ and VOL’.
Once again measure VO’ with the 10KΩ and 1KΩ resistors connected to the output then add an
additional RL to get the new VOL’.
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Lab 2, Feedback amplifiers
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POST LAB QUESTIONS
1.
For the circuit shown in Fig. 2 compare the measured voltage gain V2/V1 at 1 kHz with the
theoretical value. Also compare the measured output resistance of the basic amplifier with the
theoretical value. Plot the “open-loop” gain of the basic amp building block.
2.
Compare the measured value of the open loop voltage gain Vo'/Vs to the theoretical value of
the open loop voltage gain of the circuit shown in Fig. 5.
3.
Compare the measured value of the open loop output resistance Ro without feedback to the
theoretical value of same for the circuit shown in Fig. 5.
4.
Compare the measured value of the closed loop voltage gain Vo/Vs for the shunt-shunt
amplifier to the theoretical value of same for the circuit shown in Fig. 4.
5.
Compare the measured value of the closed loop output resistance with feedback Rof for the
shunt-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 4.
6.
Compare the measured value of the open loop feed-forward gain A for the series-shunt
amplifier to the theoretical value of same for the circuit shown in Fig. 10.
7.
Compare the measured value of the open loop output resistance without feedback Ro for the
series-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 10.
8.
Compare the measured value of the closed loop voltage gain Vo/Vs for the series-shunt
amplifier to the theoretical value of same for the circuit shown in Fig. 9.
9.
Compare the measured value of the closed loop output resistance with feedback Rof for the
series-shunt amplifier to the theoretical value of same for the circuit shown in Fig. 9.
[1]
A. S. Sedra and K. C. Smith, Microelectronic circuits, Fifth edition. Saunders, Philadelphia,
2004
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ECE306
Lab 2, Feedback amplifiers
04.12.2018
Appendix
Inverting op-amp configuration gain, Rin and RO discussion.
The inverting op-amp is really a trans-impedance amp even though the gain is often specified with the
voltage amp gain equation. Let’s look closely at what is going on.
Rf
100K
4
5
Vminus
2
V1
U1
Vo
V2
1K
7
1
3
Ro
6
LF356
+
1K
-
Rin
GND
R1
1K
GND
Vplus
Figure A1
The positive input terminal of the op-amp is at ground potential because there is no current going into
or out of it and thus no voltage drop across R1. This tells us that the negative input terminal will also
be at ground potential, the same voltage as the positive terminal, because the negative feedback tries
to keep the negative terminal at the same voltage as the positive terminal. If you don’t believe this
statement then glance at pages 64 – 76 [1]. This is true when the op-amp is operating linearly, when
the output is not clipping, and we are assuming this. So the input impedance is simply Rin since the
input voltage sees is a resistor to ground. So the input voltage V1 divided by the input resistance Rin
is the input current Iin, Eq. A1.
Iin 
V1
Rin
(A1)
Solving for V1 gives:
V1Iin Rin
Substituting A2 into Eq. 2 yields:
VO
R
 f
IinRin
Rin
(A3)
multiply both sides by Rin and you get Eq. 3.
VO
 Rf
Iin
(3)
(A2)
The output impedance of the basic amp is RO. We have a voltage source, VO, with near 0Ω output
impedance driving the output node, V2, through RO. This one we can get by inspection.
The above discussion is for the mid band of the op-amp circuit. Note that you will always be in the
mid band of the inverting op-amp circuit in these experiments.
A and β determination for the inverting op-amp configuration.
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Lab 2, Feedback amplifiers
04.12.2018
We will use the technique described in Fig 8.20 of [1], shown in appendix. The basic amp must be
loaded by R11, R22 and RS to calculate the gain. RS is Rin. We will assume the output impedance of
the op-amp is 0Ω and the open loop gain of the op-amp is 100,000V/V. The A gain is expressed in
terms of VO/Iin. β we know is -1/Rf = -1/100,000 from b in Fig. 8.20. This makes R11 and R22 = 100KΩ
(Fig. 8.20).
In the shunt-shunt topology of Fig. 8.XX the A circuit shows the input parameters Vi, Ii and Ri. The
output parameters are VO and RO. VO is generated by a voltage source with a gain of A*Ii. We know
from the gain of the op-amp that 1V in yields -100,000V out when operated open loop. 1V in yields
1/Ri = 1/1000 = 1mA Ii for our circuit. So A*Ii must be -100,000 for Ii = 0.001. A = -100,000/0.001 = 100M.
Figure 8.20 [1] Finding A and β for shunt-shunt.
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ECE306
Lab 2, Feedback amplifiers
Figure 8.11 [1] Finding A and β for series-shunt.
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04.12.2018
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