2. Introduction to Operational Amplifier Design
Table of Contents
1 Preface ..................................................................................................... 1-1
2 Introduction............................................................................................... 2-1
3 The Ideal Voltage Feedback Op Amp .............................................................. 3-1
3.1 Ideal Characteristics .............................................................................. 3-2
3.2 Ideal Model with Feedback ...................................................................... 3-3
3.3 Inverting or Non-inverting ...................................................................... 3-5
3.4 About the Transfer Function .................................................................... 3-6
3.5 About Input Impedances ........................................................................ 3-7
4 Linear Amplifiers and Attenuators.................................................................. 4-1
4.1 Voltage Follower.................................................................................... 4-1
4.2 Inverting Amplifier or Attenuator ............................................................. 4-2
4.3 Non-Inverting Amplifier .......................................................................... 4-3
5 Summation Amplifier................................................................................... 5-1
5.1 Inverting Summation Amplifier ................................................................ 5-1
5.2 Non-inverting Summation Amplifier.......................................................... 5-3
6 The Integrator............................................................................................ 6-1
6.1 Inverting Integrator ............................................................................... 6-1
6.2 Non-inverting Low-pass Filter .................................................................. 6-3
7 The Differentiator ....................................................................................... 7-1
7.1 The Inverting Differentiator..................................................................... 7-1
7.2 Non-inverting High-pass Filter ................................................................. 7-2
8 Practical Considerations ............................................................................... 8-1
8.1 Device Parameters................................................................................. 8-1
8.2 Power Supplies ..................................................................................... 8-2
8.3 Offsetting and Stabilizing........................................................................ 8-3
8.4 Impedance Matching and Phase Compensation .......................................... 8-5
Appendix A. ..Commonly Used Terms ................................................................ A-1
Appendix B. ..Bibliography ............................................................................... B-1
ii
3. Introduction to Operational Amplifier Design
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List of Circuits
Circuit 4-1 Voltage Follower .........................................................................................4-1
Circuit 4-2 Inverting Amplifier or Attenuator...................................................................4-2
Circuit 4-3 Non-inverting Amplifier ...............................................................................4-3
Circuit 5-1 Inverting Summation Amplifier .....................................................................5-1
Circuit 5-2 Non-inverting Summation Amplifier ..............................................................5-3
Circuit 6-1 Inverting Integrator ....................................................................................6-1
Circuit 6-2 The Non-inverting Integrator ........................................................................6-3
Circuit 7-1 Inverting Differentiator ................................................................................7-1
Circuit 7-2 Non-inverting Differentiator..........................................................................7-2
iii
4. Introduction to Operational Amplifier Design
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List of Figures
Figure 3.1-1 Loop analysis example ..............................................................................2-1
Figure 3.1-1 Ideal operational amplifier .........................................................................3-1
Figure 3.1-2 Ideal op amp input impedances ..................................................................3-1
Figure 3.2-1 Ideal op amp feedback model 1 ..................................................................3-3
Figure 3.2-2 Ideal op amp feedback model 2 ..................................................................3-4
Figure 3.3-1 Non-inverting examples ............................................................................3-5
Figure 3.3-2 Loop equations for Fig 3.3-1 ......................................................................3-5
Figure 3.4-1 Black box diagram ....................................................................................3-6
Figure 3.4-2 Cascaded transfer functions .......................................................................3-6
Figure 3.5-1 Zero impedance source .............................................................................3-7
Figure 3.5-2 Source with internal impedance ..................................................................3-7
- +
Figure 4.2-1 E and E loops for Circuit 4-2 ...................................................................4-2
- +
Figure 4.3-1 E and E loops for Circuit 4-3 ...................................................................4-3
-
Figure 5.1-1 E loop for Circuit 5-1 ...............................................................................5-1
Figure 5.1-2 Thevenized input for Circuit 5-1..................................................................5-2
+
Figure 5.2-1 E loop for Circuit 5-2...............................................................................5-3
-
Figure 5.2-2 E loop for Circuit 5-2 ...............................................................................5-3
Figure 5.2-3 Thevenized input for Circuit 5-2..................................................................5-4
- +
Figure 6.1-1 E and E loops for Circuit 6-1 ...................................................................6-1
+
Figure 6.2-1 E loop for Circuit 6-2...............................................................................6-3
- +
Figure 7.1-1 E and E loops for Circuit 7-1 ....................................................................7-1
+
Figure 7.2-1 E loop for Circuit 7-2 ..............................................................................7-2
Figure 8.2-1 Filtering the Power Supply .........................................................................8-2
Figure 8.3-1 Avoiding common mode noise ....................................................................8-3
Figure 8.3-2 DC baseline shift example..........................................................................8-3
- +
Figure 8.3-3 E loop with Vs = 0 and E loop with bias VB ................................................8-4
Figure 8.4-1 Impedance matching ................................................................................8-5
Figure 8.4-2 Canceling reactance..................................................................................8-6
iv
5. 1 Preface
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1 Preface
The scope of this course is the design of basic voltage feedback operational amplifier circuits. Using the
ideal op amp model and solving for the currents and voltages at each terminal we get the transfer
function as a Laplace Transform. This course provides a practical way of going from paper design to
prototyping working circuits.
This course is intended for professional electrical engineers. The course-taker should be familiar with
the Laplace and inverse Laplace Transforms and basic AC network analysis.
After completing the course, there is a quiz consisting of 16 multiple choice questions. On completion,
4 professional development hours will count towards satisfying PE licensure renewal requirements.
Navigating the course is facilitated by hyperlinked table of contents on each page or the tags in the
bookmark pane.
1-1
6. 2 Introduction
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2 Introduction
The voltage feedback differential amplifier (“op amp” as it is called) is used in a wide variety of
electronic applications such as: linear amplifier/attenuator, signal conditioner, signal synthesizer,
computer, or simulator.
A practical way to approach designing and implementing an op amp circuit is to start with the ideal
model and get an expression that relates the output to the input, regardless of the input. This is
[1]
accomplished by working with the loop equations in the frequency or s domain .
In summary, the key to getting the transfer function is that the voltages at the input terminals of a
closed-loop op amp circuit mirror each other. Also, no current flows into or out of an input terminal.
om
When a signal is applied to either or both terminals, the output will adjust itself to meet these
constraints.
.c
ng
Using the figure to the left :
ri
+ -
if V2 and Z2 = 0 then E = 0 ∴E = 0
ee
02 ngin
We know that no current flows into an input
+ -
terminal so I = I = 0 ∴I1 = If and I2 = 0
-0 e
- -
V1 − E E − Vo = − Vo
04 ed
V1
I1 = = and If =
Z1 Z1 Zf Zf
E .c
V1 Vo
so I1 = If gives us =−
Z1 Zf
se ww
Vo Z
ur /w
Figure 3.1-1 Loop analysis example hence =− f
V1 Z1
o :/
C tp
The closed loop transfer function is in the frequency domain, A v (s) =
Vo (s)
V1(s)
Z
=− f .
Z1
ht
Vo Z
For the remainder of the course, we' ll use the shorter notation as, A v = =− f .
V1 Z1
Also note : in AC network analysis, impedances are represented as phasors and do not vary with
time but with frequency. So there is no time domain representation or time variation of an impedance.
In the following sections, the same method is used for application specific circuits where the voltages
and impedances are arbitrary.
[1] To get the output in the time domain vo(t) we would have to multiply Vi(s) by Av(s) and then take the inverse Laplace
-1
Transform ; vo(t)=L [ Vi(s)·Av(s) ].
2-1
7. 3 The Ideal Voltage Feedback Op Amp
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3 The Ideal Voltage Feedback Op Amp
The voltage feedback op amp is a discrete device that has 2 input terminals and one output terminal.
Without feedback, the output is the difference between the input voltages, multiplied by the open-loop
gain (transfer function) of the op amp.
+
E the non - inverting terminal
-
E the inverting terminal
Eo the output terminal
om
Avol is the open loop gain (transfer function) of
.c
Figure 3.1-1 Ideal operational amplifier the op amp itself, without feedback
ng
+ -
The key is in maintaining E = E otherwise
ri
The open l oop equati ons :
the output will saturate:
ee
⎛ + −⎞
⎜ E − E ⎟ ⋅ A vol ≡ Eo where A vol = ∞ + -
⎝ ⎠ If we apply a voltage at E and ground E
02 ngin
⎛ + −⎞ Eo ≡ 0 + −
E o will saturate to positive supply voltage.
⎜E − E ⎟ ≡ so E −E ≡ 0
⎝ ⎠ A vol
-0 e
+ − - +
∴ E ≡E If we apply a voltage at E and ground E
04 ed
E o will saturate to negative supply voltage.
E .c
se ww
In the open loop model, each input terminal has infinite impedance so no current can flow into an
input terminal even with a voltage source or a ground applied. The output terminal has zero output
ur /w
impedance.
o :/
C tp
ht E
+ +
has Zin = ∞ so I
+
=0 E
− −
has Zin = ∞ so I
−
=0 Eo has Z out ≡0
Figure 3.1-2 Ideal op amp input impedances
The ideal characteristics are summarized in Table 3-1 below
3-1
8. 3 The Ideal Voltage Feedback Op Amp 3.1 Ideal Characteristics
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3.1 Ideal Characteristics
Summary Of Ideal Characteristics
Zin = ∞ the input impedance at each terminal is infinite
Zout = 0 the output impedance is zero
±
I =0 no current flows into either of the input terminals
Avol = ∞ the open loop gain is infinite
om
Bandwidth = ∞ the bandwidth is infinitely wide
.c
No temperature drift
ng
+ -
Eo = 0 the output voltage is zero when E = E
ri
Table 3-1 Summary of ideal characteristics
ee
02 ngin
-0 e
04 ed
E .c
se ww
ur /w
o :/
C tp
ht
3-2
9. 3 The Ideal Voltage Feedback Op Amp 3.2 Ideal Model with Feedback
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3.2 Ideal Model with Feedback
With feedback, all or a portion of the output is tied to either or both input terminals. The difference
between the voltages at the input terminals is still equal to zero and again no current flows into either
of the input terminals.
The voltage at each input terminal is treated as a node and is determined by the loop equations. The
terminal voltages are equated and we get a transfer function or closed loop gain which provides the
output over the input as a ratio (Laplace Transform).
In the following subsections, we’ll look at the feedback models : the non-zero reference and the zero
reference feedback models. As the names imply, the non-zero reference model is characterized by the
±
voltage at either input terminal, E ≠ 0 , and the zero-reference model is where the voltage at either
om
±
input terminal is referenced to 0 volts or ground; E
.c
=0.
ng
3.2.1 The non-zero reference model
ri
ee
+
The figure to the left models a source Vi connected to the E
02 ngin
terminal through a series impedance Z1 and feedback k Eo
− Z2
-0 e
tied to E ; with k being a voltage divider =
Z2 + Zf
04 ed
The voltage at each input terminal balances out to k Eo
E .c
+ − +
So E = E = k Eo and we see that I = I1 .
se ww
+
+ V −E Vi − k Eo
ur /w
But I = 0 , giving us I1 = i = =0
Z1 Z1
o :/
Eo 1 Z2 + Zf Z
Hence = = = 1+ f
C tp
Vi k Z2 Z2
Figure 3.2-1 Ideal op amp feedback model
Eo ⎛ Z ⎞
ht
1 The transfer function A v = = ⎜1 + f ⎟
⎜
Vi ⎝ Z2 ⎟
⎠
3.2.1.1 The input impedance
V
The input impedance seen by Vi is Zin = i
Iin
Vi − k Eo ⎛ 1 − kA v ⎞
The input current Iin = I1 , so Iin = Since kA v = 1, Vi ⋅ ⎜
⎜ Z ⎟=0
⎟
Z1 ⎝ 1 ⎠
With Iin = 0, Zin = ∞, so the input impedance seen by Vi is infinite or an open circuit.
3-3
10. 3 The Ideal Voltage Feedback Op Amp 3.2 Ideal Model with Feedback
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3.2.2 The zero-reference model
The figure to the left models a source Vi connected
−
to the E terminal through a series impedance Z1 ,
+
with E tied to ground.
The voltage at each input must balance to 0 volts;
±
thus E =0
om
− −
We can see that I1 + If = I ; with I = 0 , I1 = −If
.c
- -
Vi − E Eo − E -
ng
Substituting I1 = , If = and E = 0
Z1 Zf
ri
Figure 3.2-2 Ideal op amp feedback model 2
Vi E
=− o
ee
into I1 = −If gives us
Z1 Zf
02 ngin
Eo Z
So the transfer function A v = =− f
Vi Z1
-0 e
04 ed
E .c
3.2.2.1 The input impedance
se ww
ur /w
V V V
The input impedance seen by Vi is Zin = i ; with Iin = I1 = i , Zin = i = Z1
Iin Z1 Vi
o :/
Z1
C tp
ht
3-4
11. 3 The Ideal Voltage Feedback Op Amp 3.3 Inverting or Non-inverting
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3.3 Inverting or Non-inverting
The determining factors for whether the output is inverted or not, are the circuit configuration and the
loop equations for the terminal currents and voltages. With voltage feedback op amp circuits, which
terminal the signal is connected to, by itself, does not determine whether the output is inverted or not.
For example, the 2 circuits below are both inverting amplifier/attenuators circuits. The only difference
is the op amp input terminals are reversed but both provide an output that is a linear amplifier or
attenuator with 180° phase shift or inversion:
.c om
ring
ee
02 ngin
-0 e
Figure 3.3-1 Non-inverting examples
04 ed
E .c
E
-
=0 ∴E
+
=0 E
+
=0 ∴E
-
=0
se ww
+ + - -
I1 + I f = I I = 0 ∴ I1 = − I f I1 + I f = I I = 0 ∴ I1 = − If
ur /w
V V Vi Vo
I1 = i If = o I1 = If =
R1 Rf R1 Rf
o :/
C tp
V
R1
V
∴ i = − o and A v = − R f
Rf R1
∴
Vi
R1
= −
Vo
Rf
R
and A v = − f
R1
ht
Figure 3.3-2 Loop equations for Fig 3.3-1
3-5
12. 3 The Ideal Voltage Feedback Op Amp 3.4 About the Transfer Function
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3.4 About the Transfer Function
The transfer function gives you the output over the input expressed as a ratio. To get the output for a
specific input, multiply the Laplace Transform of the input by the transfer function. To convert to the
[2]
time domain, take the inverse Laplace Transform . Using Laplace Transform tables available on the
internet or in printed textbooks, is a very useful tool in op amp circuit design. Some online references
are listed in Appendix B.
Working in the s domain (using Laplace Transforms) is advantageous and readily gives any transients
that might exist
Vo (s)
so Vo (s) = Vi (s) ⋅ A v (s)
om
A v (s) ≡
Vi (s)
.c
to get the time domain representation
−1
[ Vi (s) ⋅ A v (s) ]
ng
v o (t) = L or
ri
ee
by using convolut ion :
Figure 3.4-1 Black box diagram
t
∫ v o (t) = v i (τ)⋅av (t − τ)dτ
02 ngin
0
-0 e
04 ed
When cascading circuits, the overall transfer
function is the product of all the transfer
functions in the cascade and the overall
E .c
transfer function can be viewed as a single
se ww
ratio or Laplace transform
ur /w
V (s)
A v (s) o2 = A v1(s) ⋅ A v 2 (s)
o :/
Vi (s)
C tp
Figure 3.4-2 Cascaded transfer functions
ht
In the time domain, you could not express the transfer function as a ratio. You would have to solve
differential equations for the terminal voltages and currents and use the convolution integral to get the
output as a function of time, because superposition does not apply.
-1
[2] The time domain representation of Av (that is av(t) = L [Av(s)] ) is actually the unit impulse response of the circuit. The
output vo(t) after applying an arbitrary input vi(t) is found by convolving vi(t) with av(t) { vo(t) ≡ vi(t)*av(t) } .
3-6
13. 3 The Ideal Voltage Feedback Op Amp 3.5 About Input Impedances
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3.5 About Input Impedances
An input source can be either a directly connected zero impedance source, or a thevenin equivalent
source with an internal impedance. We need to be aware if a source does have internal impedance.
This needs to be considered in determining both the input impedance seen by the voltage source and
the transfer function of the circuit.
3.5.1 Directly connected source
Consider a directly-connected source, with zero output impedance, connecting to an op amp input
terminal, through a series impedance Z1 :
The input to the circuit is Vi = Vg .
om
Vo
In general, the transfer function A v gives us
.c
Vg
ng
±
When E = 0 , the input impedance seen by Vg is
ri
±
Vg Vg − E Vg − 0 Vg
ee
Zin = ; with Iin = = , Zin = = Z1 .
Iin Z1 Z1 Vg
02 ngin
Figure 3.5-1 Zero impedance source Z1
-0 e
⎛ ± ⎞
± Vi Vi − E
⎜ Vi ⎟
04 ed
When E ≠ 0 , the input impedance seen by Vi is Zin = ; with Iin = , Zin = Z1 ⎜ ⎟
Iin Z1 ⎜ V − E± ⎟
⎝ i ⎠
E .c
* It is interesting to note that in a case like this, we have a capability of synthesizing a voltage
se ww
±
dependent input impedance that varies with Vi and E .
ur /w
o :/
3.5.2 Source with built-in internal impedance
C tp
Consider the case where the input is the end of a cable span or an input from a previous circuit stage
in the cascade:
ht
Now we have a source Vg with internal impedance Z g .
⎛ Zin ⎞
We have to modify our calculatio ns because V = Vg ⎜ ⎟.
i ⎜ Zin + Z g
⎝
⎟
⎠
Substituti ng into the transfer function A v , we get
Vo V ⎛ Zin + Z g ⎞ Vo ⎛ Zg ⎞
= o ⎜ ⎜ ⎟ = ⎜
⎟ V ⎜1 + Z ⎟
⎟
Vi Vg ⎝ Zin ⎠ g ⎝ in ⎠
the total impedance seen by Vi is Zin
the total impedance seen by Vg is Z g + Zin
Figure 3.5-2 Source with internal impedance
3-7
14. 4 Linear Amplifiers and Attenuators 4.1 Voltage Follower
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4 Linear Amplifiers and Attenuators
The most basic circuit configurations are linear amplifiers and attenuators. Ideally, they provide flat
gain or loss across the device-rated bandwidth.
4.1 Voltage Follower
The voltage follower circuit provides unity gain with no inversion. This circuit is used to isolate a high
impedance source input and provide a buffered output.
For Circuit 4 - 1, the terminal voltages are :
Vo = Eo
om
E+ = Vi
.c
-
E = Eo
ng
- V
E+ E
, Vi = Vo giving us o = 1 .
ri
Since =
Vi
ee
Circuit 4-1 Voltage Follower So A V = 1 is the transfer function for Circuit 4.1.
02 ngin
-0 e
4.1.1 The Input impedance
04 ed
E .c
The input impedance seen by source Vi is Zin =
Vi
where Iin =
Vi − E
+
se ww
Iin R1
ur /w
+ Vi
Since E = Vi , Iin = 0 , making Zin = =∞
0
o :/
So the input impedance seen by Vi for Circuit 4 - 1 is an open circuit.
C tp
ht
4-1
15. 4 Linear Amplifiers and Attenuators 4.2 Inverting Amplifier or Attenuator
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4.2 Inverting Amplifier or Attenuator
The inverting configuration provides flat gain or attenuation with constant 180° phase shift.
For Circuit 4 - 2, the terminal voltages are :
Vo = Eo
E+ =0
-
E = E+ =0
* the terminal voltages balance out to zero
.c om
Circuit 4-2 Inverting Amplifier or Attenuator
ng
-
ri
From Figure 4.2 - 1 , for the E loop we have :
ee
- -
Vi − E V −E
I1 = and I f = o
02 ngin
R1 Rf
− −
With I1 + I f = I and I = 0 , we have I1 = - I f
-0 e
-
E
04 ed
Since the terminal voltage at = 0,
Vi V V R
I1 = and I f = o and with I1 = - I f , o = − f
E .c
R1 Rf Vi R1
se ww
R
So A v = − f is the transfer function for Circuit 4.2
ur /w
- + R1
Figure 4.2-1 E and E loops for Circuit 4-2
o :/
C tp
ht
4.2.1 The Input impedance
Vi Vi
The input impedance seen by source Vi is , Zin = ; where Iin = I1 =
Iin R1
V
Substituting Iin into the expression for Zin gives us Zin = i = R1
Vi
R1
So the input impedance seen by Vi for Circuit 4 - 2 is R 1
4-2
16. 4 Linear Amplifiers and Attenuators 4.3 Non-Inverting Amplifier
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4.3 Non-Inverting Amplifier
The non-inverting configuration provides flat gain with no phase shift.
For Circuit 4 - 3, the terminal voltages are :
Vo = Eo
E+ = Vi
- R1
E = kVo ; where k is a voltage divider =
R1 + R f
- V
Since E+ E
, kVo = Vi ; readily giving us o =
=
1
Vi k
om
1 ⎛ R ⎞
= ⎜1 + f ⎟ .
.c
So by inspection, we see that A v = ⎜
k ⎝ R1 ⎟
⎠
ng
Circuit 4-3 Non-inverting Amplifier
ri
−
Analysis of the E loop gives us the same result :
ee
- -
E Vo − E
From Fig 4.3 - 1 we see that I1 = and If =
02 ngin
R1 Rf
− −
-0 e
With If = I1 + I and I = 0 , If = I1
- + - +
04 ed
Since E = kVo , E = Vi and E = E ,
-
we substitute E = Vi , into the expressions for I1 and If
E .c
se ww
V Vo − Vi
- + giving us I1 = i and If =
E E R1 Rf
ur /w
Figure 4.3-1 and loops for Circuit 4-3
o :/
C tp
With If = I1 we get
Vi
R1
=
Vo − Vi
Rf
∴
⎛ 1
Vi ⋅ ⎜
⎜R + R
⎝ 1
1
f
⎞
⎟ ⋅ R f = Vo ; giving us
⎟
⎠
ht
Vo ⎛ R ⎞
Av = = ⎜1 + f ⎟ as the transfer function for Circuit 4 - 3.
Vi ⎜
⎝ R1 ⎟
⎠
4.3.1 The Input impedance
+
Vi V −E
The input impedance seen by source Vi is Zin = From Figure 4.3 - 1, Iin = I g where I g = i
Iin Rg
+ Vi − kVo V
Substituting E = kVo into Iin gives us Iin = With kVo = Vi , Iin = 0 making Zin = i = ∞
Rg 0
So the input impedance seen by Vi for Circuit 4.3 is an open circuit.
4-3
17. 5 Summation Amplifier 5.1 Inverting Summation Amplifier
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5 Summation Amplifier
The summation amplifier provides flat gain or loss and can be configured in an inverting or non-
inverting configuration. Typical uses are signal combiner, voltage comparator or summing junction.
5.1 Inverting Summation Amplifier
The inverting summation amplifier provides flat gain or loss across the rated bandwidth with constant
180° phase shift.
For Circuit 5 - 1,
the terminal voltages are :
om
Vo = Eo
.c
-
E+ = 0 so E = 0
ng
* the terminal voltages
ri
balance out to zero
ee
02 ngin
-0 e
04 ed
E .c
Circuit 5-1 Inverting Summation Amplifier
se ww
ur /w
-
From Figure 5.1 - 1, for the E loop we have :
IS = I1 + I 2 + I 3 where
o :/
C tp I1 =
V1 − E
R
−
, I2 =
V2 − E
R
−
, I3 =
V3 − E
R
−
ht
−
V −E
and I f = o
Rf
−
Since the terminal voltage at E = 0
V1 V V V
I1 = , I 2 = 2 , I 3 = 3 and I f = o
R R R Rf
− −
- With I = IS + I f and I = 0 , IS = −I f
Figure 5.1-1 E loop for Circuit 5-1
V1 V2 V3 V
With IS = −If , we have + + =− o
R R R Rf
Rf
Simplifyin g the expression for Vo , we get Vo = − ⋅ (V1 + V2 + V3 )
R
5-1
18. 5 Summation Amplifier 5.1 Inverting Summation Amplifier
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To get the transfer function as a single ratio, we have to thevenize the 3 sources into one
V1 + V2 + V3 R
equivalent source Vi with thevenin impedance Z eq ; giving us Vi = and Z eq =
3 3
as illustrated in the figure below.
-
Analyzing the E loop in Figure 5.1 - 2 we have :
- -
V −E V −E
Is = i and I f = o
R Rf
3
om
-
Substitutin g E = 0 into I s and I f we have,
.c
3 Vi V
Is = and I f = o
ng
R Rf
ri
− −
With I s + I f = I and I = 0 , Is = − If
ee
3Vi V Vo 3R f
∴
02 ngin
= − o = −
R Rf Vi R
-0 e
3R f
04 ed
So A v = − is the transfer function for Circuit 5 - 1.
Figure 5.1-2 Thevenized input for Circuit 5-1 R
E .c
se ww
5.1.1 The Input impedance
ur /w
By inspection, the input impedance seen by our thenevized source Vi is Zin = 0 because, as Figure 5.1 - 2
o :/
R -
illustrates, we are connecting a source with an internal impedance of directly to the E terminal.
C tp
3
ht
Vn
However, the input impedance seen by each of the sources V1 , V2 and V3 is Zn = ; where
In
−
Vn − E −
In = With E = 0 , Zn is going to be whatever impedance is in series with each source.
Zn
For Circuit 5 - 1, each of the sources Vn sees Zn = R as the input impedance.
5-2
19. 5 Summation Amplifier 5.2 Non-inverting Summation Amplifier
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5.2 Non-inverting Summation Amplifier
The non-inverting summation amplifier provides flat gain across the rated bandwidth with no phase
shift.
For Circuit 5 - 2,
the terminal voltages are :
Vo = Eo
E+ = kE o
R1
om
k = ; a voltage divider
R1 + R f
.c
-
E = E+ = kE o
ring
Circuit 5-2 Non-inverting Summation Amplifier
ee
02 ngin
From Figure 5.2 - 1, we see that the voltage at the
-0 e E+ terminal is Eo through a voltage divider.
04 ed
E .c
So E+ = k Eo
+ +
se ww
Also , I f = I1 + I and with I = 0 , I1 = I f
ur /w
+
Figure 5.2-1 E loop for Circuit 5-2
o :/
−
From Figure 5.2 - 2, for the E loop we have :
C tp IS = I1 + I 2 + I3 where
ht
− − −
V1 − E V2 − E V3 − E
I1 = , I2 = , I3 =
R R R
−
V V V 3⋅E
so IS = 1 + 2 + 3 −
R R R R
- -
With I = IS and I = 0, IS = 0
−
V1 V2 V3 3 ⋅ E
∴ R
+
R
+
R
=
R
-
Figure 5.2-2 E loop for Circuit 5-2
5-3
20. 5 Summation Amplifier 5.2 Non-inverting Summation Amplifier
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- V1 V2 V 3 ⋅ kVo R1
Substituting E = kVo , we get + + 3 = Substituting k = we get,
R R R R R1 + R f
⎛ V1 V2 V3 ⎞ 3Vo ⎛ R 1 ⎞ ⎛ R ⎞
⎜
⎜ R + R + R ⎟ = R ⋅ ⎜R + R
⎜ ⎟ ⎟
⎟ Simplifying we get, Vo =
1
(V1 + V2 + V3 ) ⋅ ⎜1 + f ⎟
⎜
⎝ ⎠ ⎝ 1 f ⎠ 3 ⎝ R1 ⎟
⎠
To get the transfer function in the form of a ratio, we have to thevenize the 3 sources into one
V1 + V2 + V3 R
equivalent source Vi with thevenin impedance Z eq ; so Vi = and Z eq =
3 3
-
From Figure 5.2 - 3 below, we look at the loop equations for the E terminal.
.c om We have :
−
ng
Is = I =0
ri
−
V −E −
with Is = i , substitute E = kVo
ee
R
02 ngin
3
Vi − kVo ⎛ Vi − kVo ⎞
Expanding Is we have ; Is = = 3⋅⎜ ⎟ =0
-0 e
R ⎜ R ⎟
⎝ ⎠
04 ed
3
giving us Vi = kVo
E .c
⎛ R1 ⎞ Vo ⎛ Rf ⎞
Substituti ng k = ⎜
⎜ R + R ⎟ , we get V = ⎜1 + R ⎟ , so
⎟ ⎜ ⎟
se ww
⎝ 1 f ⎠ i ⎝ 1⎠
⎛ R ⎞
ur /w
A v = ⎜1 + f ⎟ is the transfer function for Circuit 5 - 2.
⎜
⎝ R1 ⎟
⎠
o :/
Figure 5.2-3 Thevenized input for Circuit 5-2
C tp
ht
5.2.1 The Input impedance
Vi ⎛ Vi − kVo ⎞
The input impedance seen by Vi is Zin = where Iin = Is = 3 ⋅ ⎜
⎜
⎟
⎟
Iin ⎝ R ⎠
Vi
Substituti ng kVo = Vi into Iin gives us Iin = 0 making Zin = =∞
0
The input impedance for Circuit 5 - 2 is an open circuit.
The input impedance seen by each source is going to be voltage - dependent.
⎛ R1 ⎞
We know that Vo =
1
(V1 + V2 + V3 ) , where k = ⎜
⎜R + R ⎟
⎟
3k ⎝ 1 f ⎠
5-4
21. 5 Summation Amplifier 5.2 Non-inverting Summation Amplifier
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- Vn
We also know that the voltage at E = kVo . The input impedance seen by each source is Zin n =
In
-
Vn − E
where In = and Zn is the series impedance of each source.
Zn
For example;
-
V1 − E
In Circuit 5 - 2, source V1 has a series impedance of R so the input current I1 =
R
- V1 −
1
(V1 + V2 + V3 ) 2V1 − (V2 + V3 )
om
= (V1 + V2 + V3 ) , I1 =
1
Substituting E 3 =
3 R 3R
.c
ng
V V1
so the input impedance seen by source V1 is Zin 1 = 1 = which simplifies to
I1 2V1 − (V2 + V3 )
ri
3R
ee
⎛ V1 ⎞
Zin 1 = 3R ⋅ ⎜ ⎟
⎜ 2V − (V + V ) ⎟
02 ngin
⎝ 1 2 3 ⎠
+ V3 ) (V2 (V + V3 )
So Zin 1 is an open circuit when V1 − , and is a negative impedance when V1 < 2
-0 e
2 2
A negative impedance means V1 is drawing current from the other sources and the output.
04 ed
E .c
To get a nominal value for Zin 1 , we have to know something about the harmonic content and magnitude
se ww
coefficients of the other sources so we can calculate an average.
ur /w
The same calculations apply to getting the input impedances seen by the other sources at the input.
o :/
C tp
⎛
Zin n = 3R ⋅ ⎜
Vn
⎜ 2V − ∑ V
⎝ n m
⎞
⎟
⎟
⎠
ht
5-5
22. 6 The Integrator 6.1 Inverting Integrator
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6 The Integrator
Typical uses of the integrator circuit are: noise reduction, simulation of a first order RC low-pass
circuit, low pass filter, calculating an integral or just phase compensation.
6.1 Inverting Integrator
The inverting integrator provides low-pass filtering with constant +90° phase shift. The roll-off starts
at DC. Typical use is direct integration of a time domain function, noise reduction, or low pass filter.
For Circuit 6 - 1, the terminal voltages are :
om
Vo = Eo
.c
E+ =0
ng
- +
E = E =0
ri
* the terminal voltages balance out to zero
ee
02 ngin
−
From Figure 6.1 - 1, for the E
-0 e
loop we have :
− −
V −E V −E
04 ed
Circuit 6-1 Inverting Integrator ⎛ −⎞
I1 = i , and I f = o = sC ⎜ Vo − E ⎟
R1 1 ⎝ ⎠
E .c
sC
−
se ww
Substituting E = 0 into I1 and I f gives us
ur /w
Vi
I1 = , and I f = sCVo
R1
o :/
− −
With I f + I1 = I and I = 0 , I f = −I1 so
C tp sC Vo = −
Vi
∴
Vo
=−
1
ht
R1 Vi sCR 1
1
- + Av = − is the transfer function for Circuit 6 - 1.
Figure 6.1-1 E and E loops for Circuit 6-1 sCR 1
6.1.1 The Input impedance
V V
The input impedance Zin seen by Vi is Zin = i in this case Iin = I1 = i
Iin R1
Vi Vi
With Iin = Zin = = R1
R1 Vi
R1
The input impedance seen by source Vi for Circuit 6 − 1 is R 1.
6-1
23. 6 The Integrator 6.1 Inverting Integrator
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6.1.2 Bandwidth Considerations
1
For Circuit 6 − 1, A v = − ; substituting s = jω gives us the value of A v(s) at s = jω.
sCR 1
We see at s = 0 (DC), A v = − ∞ which means the output of the circuit will be saturated at the negative
supply voltage, or 0 volts if we are using a single supply. We may want to offset this if we are working
with input signals that have DC components.
See Section 8.3.2 for offsetting the baseline.
π
j
1 2 1
We can rewrite A v =− as A v = A v ⋅ e where A v = ; noting that the constant 90°
jωCR 1 ωCR 1
om
phase shift is independent of ω. If we start the design at a fundamental frequency ω 0 , then we would
.c
1
select CR 1 to give us the desired gain magnitude at ω 0 ; i.e CR 1 =
ω0 ⋅ A v
ng
1
ri
Our - 3dB bandwidth is the ω at which A v (ω) = A v (ω 0 )
2
ee
A v (ω) ω 1
This gives us the direct relationship = 0 =
A v (ω 0 ) ω
02 ngin
2
So our - 3dB bandwidth occurs at ω = 2ω 0
-0 e
04 ed
E .c
se ww
ur /w
o :/
C tp
ht
6-2
24. 6 The Integrator 6.2 Non-inverting Low-pass Filter
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6.2 Non-inverting Low-pass Filter
The non-inverting low-pass filter provides filtering with phase shift that varies with the complex
transfer function.
For Circuit 6 - 2, the terminal voltages are :
Vo = E o
⎛ 1 ⎞
⎜ ⎟ ⎛ ⎞
E = Vi ⎜ sC 1 ⎟ = Vi ⎜ 1 ⎟
+
⎜ sCR + 1 ⎟
⎜R + ⎟ ⎝ 1 ⎠
⎜ 1 ⎟
⎝ sC ⎠
om
⎛ 1 ⎞
⎜
⎜ sCR + 1 ⎟ being a voltage divider
⎟
⎝ 1 ⎠
.c
- R2
E = kVo where k =
ng
R2 + R 3
ri
k being a voltage divider
Circuit 6-2 The Non-inverting Integrator
ee
- ⎛ 1 ⎞ Vo 1⎛ 1 ⎞ ⎛ R3 ⎞ ⎛ 1 ⎞
02 ngin
With E+ = E , Vi ⎜
⎜ sCR + 1 ⎟ = kVo
⎟ so = ⎜
⎜ sCR + 1 ⎟ = ⎜1 + R ⎟ ⋅ ⎜ sCR + 1 ⎟
⎟ ⎜ ⎟ ⎜ ⎟
⎝ 1 ⎠ Vi k⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 1 ⎠
-0 e
Vo ⎛ R ⎞ ⎛ 1 ⎞
Av = = ⎜1 + 3 ⎟ ⋅ ⎜
⎜ ⎟ is the transfer function for Circuit 6 - 2.
R 2 ⎟ ⎜ sCR 1 + 1 ⎟
04 ed
Vi ⎝ ⎠ ⎝ ⎠
E .c
6.2.1 The input impedance
se ww
+
E
ur /w
To calculate the input impedance seen by Vi we analyze the loop.
o :/
Vi
The input impedance seen by source Vi is Zin =
C tp
Iin
For Circuit 6 - 2 , Iin = I1
ht
+
Vi − E
From Figure 6.2 - 1, we have I1 =
R1
Vi
Vi −
+ Vi sCR 1 + 1
+ Recall E = so I1 =
Figure 6.2-1 E loop for Circuit 6-2 sCR 1 + 1 R1
⎛ sC ⎞ Vi sCR 1 + 1 1
I1 simplifies to Vi ⋅ ⎜
⎜ sCR + 1 ⎟ so Zin =
⎟ = = R1 +
⎝ 1 ⎠ ⎛ sC ⎞ sC sC
⎜ sCR + 1 ⎟
Vi ⋅ ⎜ ⎟
⎝ 1 ⎠
1
The input impedance seen by source Vi for Circuit 6 - 2 is R 1 + .
sC
6-3
25. 6 The Integrator 6.2 Non-inverting Low-pass Filter
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6.2.2 Bandwidth considerations
Circuit 6 - 2 is a first order low - pass filter with a fundamental frequency of ω 0 = 0.
Vo ⎛ R ⎞ ⎛ 1 ⎞ ⎛ R ⎞
Previously, we found the transfer function to be A v = = ⎜1 + 3 ⎟ ⋅ ⎜ ⎟ ; let' s replace ⎜1 + 3 ⎟
Vi ⎜
⎝ R 2 ⎟ ⎜ sCR 1 + 1 ⎟
⎠ ⎝ ⎠
⎜
⎝ R2 ⎟
⎠
1 1 ⎛ 1 ⎞ - jθ
with so A v = ⎜
⋅⎜ ⎟ Rewriting A v in phasor form, and substituing s = jω gives us A v = A v ⋅ e
⎟
k k ⎝ sCR 1 + 1 ⎠
⎛ ⎞
1 ⎜ 1 ⎟ -1
where A v = ⋅⎜ ⎟ and θ = tan ωCR 1
k ⎜(ωCR 1 ) + 1 ⎟
2
⎝ ⎠
om
1
We see that at ω 0 , A v (ω 0 ) = and θ = 0 radians
.c
k
A v (ω)
ng
1 1
The - 3dB bandwidth is where A v (ω) = ⋅ A v (ω 0 ) or =
A v (ω 0 )
ri
2 2
⎛ ⎞
ee
1 1 ⎜ 1 ⎟ A v (ω) 1 1
Since A v (ω 0 ) = and A v (ω) = ⋅ ⎜ ⎟ , = =
k k ⎜ (ωCR 1 ) 2 ⎟
+ 1⎠ A v (ω 0 ) (ωCR 1 ) 2
+1 2
02 ngin
⎝
1 1 3
-0 e
The - 3dB point is where = or where ω =
(ωCR 1 )2 +1 2 CR 1
04 ed
E .c
se ww
ur /w
o :/
C tp
ht
6-4
26. 7 The Differentiator 7.1 The Inverting Differentiator
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7 The Differentiator
Typical uses of the differentiator circuit are: simulation of a first order RC high-pass circuit, high pass
filter, calculating a derivative or just phase compensation.
7.1 The Inverting Differentiator
The inverting differentiator provides high-pass filtering with constant - 90° phase shift. The roll-off
starts at DC. Typical use is direct differentiation of a time domain function, or high pass filter.
For Circuit 7 - 1, the terminal voltages are :
Vo = Eo
om
E+ =0
.c
-
E = E+ = 0
ring From Figure 7.1 - 1, for the E
−
loop we have :
ee
− −
IC + If = I ; with I = 0 , IC = − If where
02 ngin
Circuit 7-1 Inverting Differentiator − −
Vi − E Vo − E
-0 e
IC = and If =
1 R1
04 ed
sC
−
With the voltage at E = 0 ,
E .c V V
se ww
IC = i and If = o
1 R1
ur /w
sC
Substituting into IC = − If ,
o :/
Vo Vo
C tp
we get sCVi = − ∴ = −sCR1
R1 Vi
ht
Vo
Av = = −sCR1 is the transfer function
- + Vi
Figure 7.1-1 E and E loops for Circuit 7-1
for Circuit 7 − 1 .
7.1.1 The input impedance
V
The input impedance seen by Vi is Zin = i where Iin = IC
Iin
Vi
with Iin = sCVi Zin =
sCVi
1
The input impedance seen by source Vi for Circuit 7 - 1 is
sC
7-1
27. 7 The Differentiator 7.1 The Inverting Differentiator
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7.1.2 Bandwidth considerations
For Circuit 7 − 1, A v = −sCR1 ; substituting s = jω gives us the value of A v(s) at s = jω.
We see at s = 0 (DC), A v = 0 which means the output of the circuit will be 0 volts
π
−j
2
We can rewrite A v = − jωCR 1 as A v = A v ⋅ e where A v = ωCR 1 ; noting that the constant - 90°
phase shift is independent of ω. If we start the design at a fundamental frequency ω 0 , we would
Av
select CR 1 to give us the desired gain magnitude at ω 0 ; i.e CR 1 =
ω0
om
Our - 3dB bandwidth is the ω at which A v (ω) = 2 ⋅ A v (ω 0 )
.c
A v (ω) ω
This gives us the direct relationship = =2
A v (ω 0 ) ωo
ng
So our - 3dB bandwidth occurs at ω = 2ω 0
ri
ee
02 ngin
-0 e
04 ed
E .c
se ww
ur /w
o :/
C tp
ht
7-1
