PH-315 A. La Rosa Portland State University

NEGATIVE FEEDBACK and APPLICATIONS Amplifiers circuits, Transfer function, Active low pass filters

I. PURPOSE

II. THEORETICAL CONSIDERATIONS

II.1 Negative feedback

II.2 Operational Amplifiers with “Infinite” and Finite Gain

II.2.A The concept of open loop gain and close loop gain

II.2.B Op-amps with infinite open loop gain: The Golden Rules of Operational Amplifiers

III. EXPERIMENTAL CONSIDERATIONS Applications of negative feedback with operational amplifiers

III.1 Amplifiers Circuits f3dB frequency (test in the frequency domain) Slew-rate (test in the time domain) Transfer function (analysis in the Laplace domain)

III.2 Active Filters

III.2.A Exploiting the low output impedance and high input impedance of op-amps

III.2.B Active pass filters: Filters with steeper roll-off

I. PURPOSE:

To use various types of negative feedback, using operational amplifiers, to build

a) gain-controlled amplifier, non-inverted amplifier, integrator, differentiator; and

b) active low p[ass RC filters. Students must i) provide an analysis of the circuit in the time domain, and ii) evaluate the corresponding transfer function (Laplace domain analysis) of the circuits.

II. THEORETICAL CONSIDERATIONS

FEEDBACK

In control systems, feedback consists in comparing the output of the system with the desired output and making a correction accordingly.1

II.1 Negative feedback Negative feedback is the process of coupling a portion of the output back into the input, as a way to cancel part of the input. This process, it turns out, has the effect of reducing the gain of the amplifier, but, in exchange, it improves other characteristics including freedom from distortion and nonlinearity, flatness in the frequency response, and predictability. In fact, as more negative feedback is used, the resultant amplifier’s characteristics becomes less dependent on the characteristics of the original open-loop amplifier

Caution: We have learned from our RC filter experiment that a phase lag exists between the input and output voltages, which increases as more components are added into the circuit. If a negative feedback-loop circuit were to accumulate large enough phase lag (i.e. greater than 180o), then positive feedback occurs (the circuit ends up being an oscillator.)

-

+ = A (v+ - v-

)

v+

v-

vout =

i-

Amplifying region

~ 120 V

(v+ - v- )

vout

+VCC /A

-VCC /A

-VCC

+VCC

i+

Differential input voltage

Ri > 10 M

i- , i+ ~ 20 nA for the LM358AP

~ pA for op amps with FET-input types

Fig. 1 Operational amplifier.

II.2 OPERATIONAL AMPLIFIERS WITH “INFINITE and FINITE GAIN

See additional information contained in the file “Operational Amplifiers (background)” posted on the website of this course.

Before performing the experiments listed below, review the data sheet of the op amp you

are using. In particular, check the slwe rate (it is 0.3 V/s for the LM358AP). This parameter

(that tell you how fast tye op-amp responds to an input signal) will allow you to estimate

the frequency bandwidth response of your circuit.

II.2.A The concept of open loop gain and close loop gain

0

20

40

60

80

100

100

101

102

103

104

105

106

107

Low-pass_unity-gain-Bandwidth

Gain (in dB)

Volt

age G

ain

(in

dB

)

Frequency Unity-gain

bandwidthfT

f3dB

Open loop

voltage gain

-

+

A =

vin vout

vout

vin

Open loop voltage gain

A

A ()

fT Unity-gain bandwidth

Fig. 2 Typical open loop voltage gain A as a function of frequency.

When integrated with an external network, the gain of the network is less than the open-loop gain of the amplifier. (Even though the gain is lower, the performance in terms of stability is much better).

0

20

40

60

80

100

100

101

102

103

104

105

106

107

Low-pass_unity-gain-Bandwidth

Gain (in dB)

Vo

lta

ge G

ain

(in

dB

)

Frequency Unity-gain

bandwidthfT

f3dB

Open loop

voltage gain

Ideal close

loop gain:

ACL=103

Close loop bandwidth

Open loop

voltage gain

Close loop

voltage gain

vin

ACL = vout

vin

Rf = 103 k Ro =

1 k vout

A

ACL A ()

ACL ()

Fig. 3 Open loop A and close loop ACL voltage gain as a function of frequency. Notice in the

graph that ACL < A.

II.2.B Op-amps with infinite open loop gain: The Golden Rules of Operational Amplifiers When implemented as part of a negative feedback external network, the behavior of the op amp can be predicted (for many practical purposes) based on two simple rules: 2

Rule I The output attempts to do whatever is necessary as to produce that the external feedback brings the differential

input voltage close to zero (v+ - v- ≈ 0). (1)

Rule II The inputs draw no current (i- , i+ ≈ 0).

Basically we are saying that the open loop gains A is infinite. This is of course an abstraction, but allows obtaining a quick grasp of the functioning of many widely used circuits (as we will see in the experimental section below).

III. EXPERIMENTAL CONSIDERATIONS Applications of negative feedback with operational amplifiers

III.A Amplifiers Circuits

III.B Active Filters

III.1 Amplifiers Circuits

III.1.A The Inverting Amplifier

-

+

Vout Vin

Rf = 10 k

Ro = 1 k

VCC = 12V

-VCC = -12V

Io

If

v+

v-

Fig. 4 DC inverted amplifier circuit. (Keep in mind that we will be using also ac input voltage, as in Fig.5 below)

Since v+ is grounded, then Rule I implies v- = 0. Accordingly,

the current through Ro is equal to Io =Vin / Ro; and

the current through Rf is then equal to If = -Vout / Rf.

Since no current flows into the op amp inputs (Rule II) we should have

Io = If

that is,

Vin / Ro= -Vout / Rf,

or simply,

in

o

f

out VR

RV (2)

TASKS:

Analysis in the time domain a) Construct the circuit shown in Fig. 4, and verify that indeed the voltage gain is equal to

of RR / .

b) Testing the operational amplifier in the frequency domain.

Using a sinusoidal wave (~ 50 mV amplitude) as an input.

Make two Bode plots, corresponding to two different gains (10 and 100 for example). Evaluate the corresponding f3dB for these two cases. (i.e. evaluate if the f3dB value

depends on the gain.)

For a given fixed frequency, plot the input and output voltages observed in your oscilloscope (a digital picture would be fine).

c) Testing the operational amplifier in the time domain. This step is to get familiar with the concept of slew-rate [see also the “slew-rate” section in the file “Operational amplifier

(background material)” posted online in the website of this course]

c1) Using first a square-wave input of 50 mV amplitude of low frequency, monitor with the highest precision possible the high to low transition of one edge of the output square wave. Estimate the rate of voltage per unit time at which the signal decays. Let’s call it,

S experimental = decay rate _______ (3)

Check if the value of S experimental compares with the slew rate value quoted in the datasheet.

On the other hand, for a given value of the slew-rate S (see footnote1), we expect the maximum amplitude Bmax, at which a sine-wave output of frequency f remains

undistorted, to be given by S= 2f Bmax.

That is, the amplitude B of a sine-wave B Sin(t) will remain undistorted if,

B f2

S

(4)

c2) Change you input to a sinusoidal wave, and use a fixed frequency f that is lower than, but close to, the 3dB found in part b above. Check the amplitude of the output signal has the expected value predicted by expression (2) above (for the gain value, 10 or 100, you are currently using).

1 Its value will depend on the current value of the circuit gain.

Now take different (increasing) values of the input voltage amplitude. Verify if the amplitude B of the output-signal deviates very much from its expected value (predicted by expression (2) ) when the input-signal amplitude is selected too large.

Check if this deviation start happening when B starts exceeding the value Bmax=S/

2f. (For this test, compare the results for B obtained when choose S to be equal to the slew rate you obtain from the data sheet of the op-amp and the one given by expression (3)). Make this evaluation for the two cases, gain 10 and 100.

-

+

vout

vin

Rf = 10 k

Ro = 1 k

VCC = 12V

-VCC = -12V

io

if

v+

v-

Fig. 5 AC inverted amplifier circuit. (It is the same circuit as in Fig. 4; it just emphasizes the ac input and output voltages.

Analysis in the Laplace domain

d) We analyzed above the functioning of the inverted amplifier circuit in the time domain. This time, instead, we ask you to also analyze the circuit in the Laplace domain; evaluate the transfer function of this circuit.

Suggestion: Proceed with a time-domain analysis first, and then evaluate the corresponding Laplace transformation of the quantities involved.

III.1.B The Noninverting Amplifier

Rule I implies v- = Vin

At the same time, v- is part of a voltage divider: v

- = [Vout /( R1+ R2)] R1

Equating these two expressions, we obtain,

inout VR

RV 1

1

2

(4)

+ Vout Vin

R2 = 20 k

R1 = 2 k

-

Fig. 6 Non-inverting amplifier.

Implementation: a) Analysis in the time domain. Construct the circuit shown in Fig, 6, whose input is, first, a DC

voltage, and verify that the voltage gain is indeed equal to 1 + (R2 / R1). Then, using a square wave (~50 mV amplitude) as an input, make two Bode plots,

corresponding to two different gains (10 and 100 for example). Check if the f3dB value depends on the gain.

For a given fixed frequency, plot the input and output voltages observed in your oscilloscope (a digital picture would be fine).

b) Analyze the circuit in the Laplace domain. Evaluate the transfer function of this circuit.

Suggestion: Proceed with a time-domain analysis first, and then evaluate the corresponding Laplace transformation of the quantities involved.

III.1.C Differential Amplifier

Implementation:

a) Apply the golden rules to demonstrate that the output voltage of the circuit shown in Fig. 5 is

given by,

) ( 12

1

2 VVR

RVout (5)

b) Build the circuit shown in Fig, 7 and verify if the output voltage varies according to the

expression given in part a) above.

-

+

Vout V1

R2

R1

R2

V2

R2 = 20 k

R1 = 2 k

R1

Fig. 7 Differential amplifier.

III.1.D Integrator

Since v + is grounded, the input v- acts as a virtual ground. The current IR passing through the

resistor is then given by,

iR = vin /R

The current through the capacitor is given by,

)C( outC 0dt

d(q)

dt

di v

dt

d C outv

Rule II implies that IR = IC ,

dt

dC

R

outin vv

This implies,

t

intout )dt'(t'RC

1(t) vv (6)

-

+

vout

vin

q

R = 1 k

VCC = 12V

-VCC = -12V

IR

IC

C= 0.01 F

-q

Fig. 8 Integrator circuit.

Implementation:

a) Analysis in the time domain. Implement an integrator circuit shown in Fig. 8.

Test your circuit using a square signal of 1 kHz at the input.

Since charging effects can cause serious offsets, a parallel resistor Rp may be needed (to

prevent any long term voltage shift at the input). Try different values for Rp. (100K, 1 M).

See Fig. 9.

Investigate the effect of changing the various parameters.

Plot the input and output voltages observed in your oscilloscope (a digital picture would be fine).

-

+

Vout Vin R = 1 k

VCC = 12V -VCC = -12V

C= 0.01 F

Rp

Fig. 9 Integrator.

c) Analyze the circuit in the Laplace domain; evaluate the transfer function of this circuit.

Suggestion: Proceed with a time-domain analysis first, and then evaluate the corresponding Laplace transformation of the quantities involved.

III.1.E Differentiator

The circuitry is similar to the integrator but with the R and C reversed.

The current through the capacitor is given by,

)( int)( CVdt

dq

dt

dIC

dt

dVC in

The current through the resistor is given by,

IR = (0-Vout /R

Rule II implies that IR = IC ,

R

V

dt

dVC outin

This implies,

dt

dVRCtV in

out )( (7)

Implementation:

a) Time domain analysis. Implement a differentiator circuit. Test the circuit with triangle waves at the input. Plot the input and output voltages observed in your oscilloscope (a digital picture would be fine).

-

+

Vout Vin q

R = 1 k

VCC = 12V -VCC = -12V

IC

IR C= 0.01 F

-q

Fig. 10 Differentiator

b) Analysis in the Laplace domain.

Analyze the circuit in the Laplace domain; evaluate the transfer function of this circuit.

III.2 Active Filters III.2.A Exploiting the low output impedance and high input impedance of op-amps In a previous laboratory you worked with two first-order low pass filters connected in a cascade arrangement, as shown in Fig. 11.

R

C

vin R

C vout

Fig. 11 Two low-pass RC filters.

The output voltage of this circuit is given by,

2

CCC

C

C

C

)z(/z1

1

)z(

z

)z(

z

R R

R

Rinout

vv (8)

where Cz is the capacitor impedance.

Notice, we cannot state that outv is simply the product of two voltage dividers

expressions, like R

R )z(

z

)z(

z

C

C

C

C

, because of the “loading” effects of the

second simple low-pass filter over the first simple low-pass filter. TASK: Using expression (8) prove that, for an harmonic input voltage of angular

frequency , one obtains,

RCj1 RCj

inout vv

2)(

1 (9)

One way to get around the loading effect that one circuit stage exerts on another is to use operational amplifiers in a “follower” configuration as shown in Fig. 12. Notice, it is simply a non-inverting amplifier, as shown in Fig. 6 above with R2=0 and R2=∞). TASKS: Build the circuit shown in Fig. 12.

Make a Bode plot and identify the corresponding f3dB frequency. Compare the result with the one corresponding to the circuit in Fig. 11 (obtained in your previous lab-4).

Evaluate vout using an analysis in the time domain (use the op-amp golden rules).

Evaluate the corresponding transfer function (analysis in the Laplace domain).

In both cases justify all your steps.

R

C

vin R

C

vout

-

+

IR

Fig. 12 Low pass filters with a decoupling amplifier.

Hint: The answers you should arrive are the following,

in

out v

v

R

R )z(

z

)z(

z

C

C

C

C

(10)

In the time domain )( /1z C Cj ; hence,

t

t

in

out )(

)(

v

v

RCj

RCj )(1

1

)(1

1

(11)

In the Laplace domain )s( /1zC C ; hence,

in

out s

s

)(

)(

V

V

RC

RC s)(1

1

s)(1

1

(12)

III.2.B Active pass filters: Filters with steeper roll-off

Simple RC filters produce gentle low low-pass gain characteristics with a roll-off of 6dB/octave at frequencies well beyond the -3dB frequency. Such filters are sufficient for many multiple purposes. Often, however, filters with flatter pass-bands and steeper roll-off are needed.

One solution may consists of cascading many RC filters, using buffer amplifiers (as indicated in Fig. 12) to avoid loading effects. It turns out, such an approach produces indeed steeper roll-off, but the curved “f3dB knee” in the gain vs frequency response does not disappear.

On the other hand, filters made with inductors and capacitors can have very sharp responses. In fact, by including inductors in the design, it is possible to create filters with any desired flatness of passband, combined with roll-off steepness. The only problem is that

inductors as circuit elements are often bulky and expensive (not to mention having significant series resistance. Hence the task becomes to design “inductorless filters” with the characteristics of ideal RLC filters.

Solution: Using op-amps as part of the filter design, it is possible to synthesize any RLC filter characteristic without using inductors.

A specific example of this statement is provided in the next example, where you are asked to compare the gain, as well as the transfer functions, corresponding to a RLC passive filter and an active RC filter.

TASKS: Consider the two circuits shown in Fig. 13. No need to build the RLC circuit. a) For the passive RLC filter (circuit on the left side of Fig. 13):

a1) Using the method of complex impedance in the time domain, show that,

RCj LCt

t

in

out

1-

1

2)(

)(

v

v (13)

a2) Using the method of complex impedance in the Laplace domain, show that,

RC LC in

out

1

1

2)(

)(

sssV

sV (14)

Hint: Use the expressions, in the Laplace domain, for the individual impedances of the R, C and L elements, as given in the “Transfer Function” notes posted in the website of this course.

R

C

vin vout

IR

L

R

C

vin R

C

vout

-

+

Fig. 13 Left: RLC passive filters. (No need to build this circuit) Right: Active RC pass filter. Build it and

make a Bode plot.

b) For the active RC filter shown on the right side of Fig. 13):

b1) Using the method of complex impedance in the time domain, show that,

in

out v

v

R

R )z(

z

)z(

z

C

C

C

C

RCj

RCj

)(1

1

)(1

1

(15)

b1) Using the method of complex impedance in the Laplace domain, show that,

in

out )(

)(

sV

sV

RC

RC s)(1

1

s)(1

1

RC RC 1) s2 s)(

122

(16)

Hint: Use the expressions, in the Laplace domain, for the individual impedances of the R, C and L elements, as given in the “Transfer Function” notes posted in the website of this course.

c) Compare the transfer functions for the passive RLC circuit (given by expression (14)) and the transfer function for the active RC filter (given by expression (17)). Include your comments on whether this active RC pass filter will behave similar to a passive RLC circuit.

d) Make a Bode plot for the active RC pass filter shown in Fig. 13.

1 (Chapter 4) Horowitz and Hills, “The Art of Electronics.” 2nd Ed.; Cambridge University Press (1990). 2 (Page 177) Horowitz and Hills,