Laboratory Experiment 7_3101

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Transcript of Laboratory Experiment 7_3101

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Laboratory Experiment 7Feedback Amplifiers

By: Josh Stevens #4279185 Completed 11/27/2012

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I.) Abstract  

Conducting an experiment in the design, construction, implementation, stability 

and testing of feedback amplifiers was the inspiration for this lab. Feedback 

amplifiers were designed to provide desired gains by manipulating the values

of the resistive elements used in their feedback; these were then loaded with aresistance and the effects were observed. The series- shunt feedback as well as

the shunt-shunt feedback exhibits lower internal output resistances while the

series-series and shunt-series exhibit higher internal output resistances. This is

important because the series-shunt and shunt-shunt will have a gain that is less

responsive to resistive loads when compared to the shunt-series and series-

series feedback amplifiers. Also, cascaded implementations of basic feedback 

amplifiers were used to test frequency response as well as stability. The stability 

issues of the cascaded amplifiers were solved using dominant pole

compensation through the implementation of capacitive elements. Using this

method is important because it shows real world implementation of frequency 

response manipulation.

II.) Introduction 

Previous work:

This lab focuses on feedback amplifiers using a form of feedback 

known as negative feedback. Negative feedback was invented by

engineers in order to provide a circuit with one or more of the

properties of  gain de-sensitivity , reduction in nonlinear distortion, or

reduction in the the effect of noise. These properties are highly desiredand thus negative feedback is extremely useful and common in circuit 

implementation.

There are four general types of negative feedback, each with slightly

different effects on the amplifier. The four types, as well as their

respective feedback equations and effects are outlined below.

Series – Shunt: 

Closed Loop Gain:  

 

Feedback Factor:

 

Input and Output Resistances:

 

 

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Shunt – Shunt: 

Closed Loop Gain:  

 

Feedback Factor:

 

Input and Output Resistances:

 

 

 

Series – Series: 

Closed Loop Gain:  

 

Feedback Factor:  

Input and Output Resistances:

 

 

Shunt – Series: 

Closed Loop Gain:  

 

Feedback Factor:  

Input and Output Resistances:

 

 

Overview:

Now knowing the four various methods of implementing negativefeedback, this lab proceeds to examine the effects of load resistances on the

varying types of feedback. In theory, the negative feedback circuits are

implemented using the equations shown above and they will produce the

aforementioned effects on the circuit. More time is spent on the series-shunt 

feedback method because the additional frequency information examined

therein will be used in a later portion of the lab.

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Once the analysis of load resistance effects on mid-band gain is

complete, the experiment continues through the use of cascading series-

shunt feedback circuits in order to examine frequency response as well as

stability.

When the examinations are complete, dominant pole frequency

compensation is used to alter frequency response.

III.) Results

Part 1) Analyzing the four-feedback topologies

 Analyzing the Series-Shunt: 

The basic circuit showing the series-shunt style of feedback is shown

in Figure 7.1 below:

Figure 7.1 

The goal is to design this amplifier for a mid-band no load voltage gainof 15v/v and then test the effects of varying load resistances (RL) on

this gain.

Using the equations for feedback analyzed in the introduction above,

the feedback resistors R1 and R2 can be solved for analytical values as

follows:

 

for A >>1 yields : 

 

Feedback Factor:

 

Thus:   

Choose R2 = 1K and solve for R1 = 14K

A source resistance (RS) is generically chosen to be 1K and

analysis is conducted.

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According to the introduction section the equations state that 

the RO of a series-shunt should be decreased which means that 

when placed in parallel with resistive loads, the loads should

not have an overwhelming effect on the gain. i.e. for larger and

larger loads the Ro dominates.

The actual results are shown in Table 7.1 below.

Table 7.1

RL (ohms) Vin Vout   Af = Vout /Vin

0 1.10 15.8 14.36

1K 1.10 16.4 14.91

100K 1.10 16.2 14.73

1M 1.10 15.8 14.36

Thus, theory suggested right in that as the smaller load affectedthe gain somewhat, but as load increased the effect of it on gain

became almost negligible and the gain returned to its unloaded

value. Theory is also correct in that the resistive values chosen

for the feedback produced a gain very near 15V/V.

The lab then tasked to analyze the low-frequency input resistance.

This was done using a multimeter to measure the voltage across the

source resistance. Using this to calculate the current into the circuit 

along with the voltage at the circuit input RIN was determined.

According to the theory outlined in the introduction the Rin should be

high.

VT = 14V IT = 17.536uA (this is corrected from the RMS

value of the multimeter by dividing by the

sqrt 2.)

Rin = VT/IT = 798.35K 

This value is high, just as theoretically expected.

As a precursor to future portions of the lab, the series-shunt was

analyzed to determine its f 3db frequency as well as the dominant pole

in the open loop response.

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The below Table 7.2 shows the frequency response:

Table 7.2

 f (hertz) V in V out  V out  /V in 100 1.10 15.8 14.36

1K 1.10 15.8 14.36

5K 1.10 15.8 14.36

10K 1.14 15.8 13.86

15K 1.14 15.8 13.86

20K 1.14 14.4 12.63

24.4K 1.14 12.8 11.2 =f 3db

25K 1.14 12.6 11.05

50K 1.14 6.8 5.96

100K 1.14 3.8 3.33

15.8 * 0.707 = 11.17 Therefore from table f 3db = 24.4KHZ 

Now, proceed to find the location of the dominant pole (f p) in open

loop response.

 

 

 Analyzing the Shunt - Series: 

The basic circuit showing the shunt-series style of feedback is shownin Figure 7.2 below:

Figure 7.2 

The goal is to design this amplifier for a mid-band short circuit 

current gain of 100A/A and then test the effects of varying load

resistances (RL) on this gain.

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Using the equations for feedback analyzed in the introduction above,

the feedback resistors R1 and R2 can be solved for analytical values as

follows:

 

for A >>1 yields : 

 

Feedback Factor:

 

Thus: 

 

Choose R2 = 1K and solve for R1 = 101K

An added challenge of a shunt-series amplifier is providing a current 

source to the circuit. Without a current source at disposal, a current 

source was implemented by placing the voltage source in parallel with

a very large (1M

resistor. This method of source implementation is

shown again in the shunt-shunt case.

Thus, IIn=Vin/1M 

IOut =VO/RL 

With current source intact, analysis using varying load resistors was

conducted. The introduction section of this report details the

expected circuit resistance RO to be increased when using this method

of feedback. With an increased resistance we should see results where

the load resistance has more of an impact on the gain of the circuit.

The actual results of the analysis are shown in Table 7.3 below:

Table 7.3 

RL (ohms) VIn IIn=Vin/1M VO IOut =VO/RL  Af = IOut /IIn 

10k 1.10 1.1uA 1.26 0.126mA 114.545

100k 1.10 1.1uA 11.4 0.114mA 103.636

1M 0.26 0.26uA 19.8 0.0198mA 76.151

The final input when using a load of 1M had to be adjusted to avoid

clipping.

The results of the gain agree with the theory in that the gain is

relatively stable with a gain around 100A/A, but the load resistance

does have an observable effect on the gain. This resistance effect 

agrees with stated theory as well.

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 Analyzing the Series - Series: 

The basic circuit showing the series-series style of feedback is shown

in Figure 7.3 below:

Figure 7.3 

The goal is to design this amplifier for a mid-band short circuit 

transconductance of 1mA/V and then test the effects of varying loadresistances (RL) on this transconductance.

Using the equations for feedback analyzed in the introduction above,

the feedback resistor RF can be solved for analytical values as follows:

 

for A >>1 yields : 

 

Feedback Factor:  

Thus: 

Rf = 1K 

A source resistance (RS) is generically chosen to be 1K and analysis is

conducted.

The IOut =VO/RL 

According to the introduction section the equations state that the RO 

of a series-series should be increased when using this method of 

feedback. With an increased resistance we should see results where

the load resistance has more of an impact on the gain of the circuit.

The actual results of the analysis are shown in Table 7.4 below:

Table 7.4

RL (ohms) VIn  VO  IOut =VO/RL   Af = IOut /VIn 

10k 1.10 11.6 1.16mA 1.276 mA/V

100k 220mV 21.8 0.218mA 0.991 mA/V

1M 26.8mV 18.4 0.0184mA 0.687 mA/V

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The input when using a load of 100K and 1M had to be adjusted to

avoid clipping.

The results of the transconductance agree with the theory in that the

transconductance is relatively stable with a magnitude of nearly

1ma/V. Also agreeing with theory, the load resistance does have an

observable effect on the transconductance.

 Analyzing the Shunt - Shunt: 

The basic circuit showing the shunt-shunt style of feedback is shown

in Figure 7.4 below:

Figure 7.4 

As was seen in the shunt-series feedback, a current source was

implemented by placing the voltage source in parallel with a very

large (1M resistor.

Thus, IIn=Vin/1M 

The goal is to design this amplifier for a mid-band open circuit 

transresistance of 100K and then test the effects of varying load

resistances (RL) on this transresistance.

Using the equations for feedback analyzed in the introduction above,

the feedback resistor RF can be solved for analytical values as follows:

 

for A >>1 yields : 

 

Feedback Factor:

 

Thus:   

Rf = 100K 

According to the introduction section the equations state that the RO 

of a shunt-shunt should be decreased which means that when placed

in parallel with resistive loads, the loads should not have an

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overwhelming effect on the transresistance. i.e. for larger and larger

loads the Ro dominates.

The actual results are shown in Table 7.5 below.

Table 7.5

RL (ohms) VIn  IIn=Vin/1M  VO   Af = Vo /Is 

10K 1.10 1.1uA 112mV 101,818 V/A

100K 1.10 1.1uA 112mV 101,818 V/A

1M 1.10 1.1uA 116mV 105,455 V/A

Theory was correct in providing a stable gain at very near 100KΩ.

Also theory was correct in that the varying load resistances had a

minimal effect on the transresistance.

Part 2) Analyzing Frequency Response of 2 cascaded series-shunt  

Designing the circuit  

It is the goal of this portion of the lab to connect two identical series-

shunt amplifiers, as designed in the initial portion of this lab, and

provide overall feedback to make an overall voltage gain of 15V/V.

The overall feedback can be determined in the same manner in which

the original feedback values were determined for the series-shunt 

resistors.

 

for A >>1 yields : 

 

Feedback Factor:

 

Thus: 

 

Choose R2 = 1K and solve for R1 = 14K

Lacking the necessary resistors to implement another 14K, a

resistance of 15.1K was chosen. The results will be slightlyvaried from the ideal 14K resistor case, but the overall effect 

should not produce differences to drastic to keep this change

from being implemented.

A source resistance (RS) is generically chosen to be 1K.

As an added challenge to this implementation, it was realized that the

dc-offset voltage of the first op amp would be amplified by the second

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op amp and would cause saturation. This would yield incorrect data.

Thus, a potentiometer was used to provide compensation, in op amp

one, that would nullify the dc offset voltage.

Once this was complete, the circuit was assembled as shown in Figure

7.5 below.

Figure 7.5

The gain is tested at 200Hz and the gain of the circuit was 15.5V/V.Compared with the theoretical gain of 15V/V, this result is within

reasonable range to merit this circuit ’s implementation in the rest of 

the lab.

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Measuring the frequency response 

The next portion of the lab involved measuring the frequency

response of the compound amplifier assembled, and shown in Figure

7.5, above.

The frequency response is shown in the below table, Table 7.6.

Table 7.6 

 f (herz) VIn (mV)  V0 (V)   Af = Vout /Vin 

200 154 2.44 15.84

500 154 2.44 15.84

1K 154 2.44 15.84

5K 154 2.44 15.84

10K 160 2.52 15.75

20K 156 2.52 16.15

40K 158 2.56 16.1560K 156 2.60 16.67

80K 156 2.84 18.21

90K 158 3.04 19.49

100K 156 3.24 20.77 = f 0110K 156 3.04 18.21

150K 156 2.28 14.62 = f 2200K 154 1.68 10.91

500K 154 480mV 3.12

1M 154 160mV 1.04

The response is shown in the below graph as well.

Graph 7.1 

0

5

10

15

20

25

1 100 10000 1000000

Series1

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The frequency at which the maximum gain occurs, f 0 , is equal to

100KHz with a gain of 20.77V/V. This gain is identified and bolded in

the table.

V  f2 = 0.707 * 20.77   f  2 = 150kHz. This frequency is identified and 

bolded in the table as well.

These two frequencies are pulled from the table as they will be used

in a later portion of the lab to identify the Q, quality factor, of the

circuit.

Examining output for varying square wave frequencies 

The next portion of the lab tasked that the output of the Figure 7.5

circuit be examined for square waves of various frequencies. The

figures representing these frequency effects are shown in Figure 7.6

– 7. 11 below.

Figure 7. 6 Figure 7. 7

 f = 100Hz f = 500Hz 

Figure 7. 8 Figure 7. 9

 f = 10KHz f = 50KHz 

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Figure 7.10 Figure 7.11

 f = 100KHz f = 1MHz 

As can be seen by these figures above, as the frequency increases the

output becomes more and more sinusoidal.

Determining the Q 

As a final portion to the circuit in Figure 7.5, the frequency response

was used to determine the Q of the circuit.

The equation for Q is as follows:

 

 f 0 and f  2 are taken from the table, Table 7.6 

 f 1 has a very low frequency, of a few hertz, and is thus

disregarded because it is much smaller than f  2. 

= 2/3

A Q of 2/3 is not a very good Q. The idea is to create a Q as large as

possible and this Q is not large.

Part 3) Analyzing Stability and Implementing Compensation 

Designing the circuit and analyzing stability  

The lab tasked the implementation of a circuit created from the

combination of 3 identical series-shunt amplifiers, as designed in the

initial portion of this lab, and provide overall feedback to make an

overall voltage gain with the range of 10V/V to 100V/V.

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The overall feedback can be determined in the same manner in which

the original feedback values were determined for the series-shunt 

resistors.

For a gain of 10:

  for A >>1 yields :   

Feedback Factor:

 

Thus: 

 

Choose R2 = 1K and solve for R1 = 9K

For a gain of 100:

 

for A

>>1 yields :

 

 

Feedback Factor:

 

Thus: 

 

Choose R2 = 1K and solve for R1 = 99K

Therefore, for a gain ranging between 10 and 100, one must vary the

resistor R1 between 9K and 99K.

The PSPICE network for the 10V/V gain is shown in Figure 7.12 

below.

Figure 7.12

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The circuit becomes unstable for higher gains; this becomes evident 

by the oscillating output produced by the PSPICE analyzer.

We wish to correct this unstable nature by adding a dominant pole.

This is the point of the remainder of the lab.

Compensation 

For the 3-stage amplifier shown in Figure 7.12 above, the lab tasks

the determination a dominant pole compensation to give stability

with a phase margin between 45 and 90 degrees.

(  ) 

for 45 degrees:

 

 f = 24.4 kHz  

for 90 degrees

 

 f = 14.1 kHz 

Therefore the frequency must be between 14.1KHz and 24.4KHz.

This can be implemented in the circuit by adding a resistor in serieswith a capacitor to ground in between the stages. The relationship

between this capacitor, resistor combination and frequency is shown

by:    

Choosing an R of 1K and a capacitor C = 10nF yields a frequency f =

15,915.5 Hz.

This implementation is shown in Figure 7.13 below.

Figure 7.13

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To visualize the effects of the compensation, plotted below are the

phase and mangnitude of the uncompensated 100 gain amplifier,

Figure 17.14, as well as the compensated phase and magnitude of the

100 gain amplifier, Figure 17.15.

Figure 17.14 ……… Uncompensated Amplifier

Figure 17.15 ……… Compensated Amplifier

As can be seen by the phase and magnitude plots, the dominant pole

compensation chosen eliminates the high resonant peak of the

original amplifier. This is done at the sacrifice of frequency range as

can be seen that the frequency range is significantly decreased with

the dominant pole compensation. This agrees perfectly with the

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theory of dominant pole compensation in that frequency range is

given up in favor of stability. The phase is also compensated; this is

shown by a decrease in slope.

IV.) Conclusion 

The results of this lab show importance in three relevant areas. For the initial

portion of this lab, experimenting with the various topologies of negative

feedback led to the understanding of the varying benefits of differing

implementations and the various responses to load resistances that result.

Despite the slight variances in output due to load resistances, the main result 

to focus on is the increased stability of gain provided by negative feedback.

The second relevant area of importance stems from the frequency response

of the cascaded op amps. Table 7.6 and Graph 7.1 do an excellent job of 

demonstrating the dependence of gain on frequency. This dependence is

extremely important when implementing real world circuits as it can have

drastic impacts on physical elements. This frequency dependence creating

instability introduces the necessity of the third portion of the results,

compensating an amplifier. The amplifier in this lab was compensated using

dominant pole analysis and the most notable result of this compensation was

the trade-off of an increase in stability of gain for reduced frequency range.