Part 7: Operational Amplifier or Op-Amp
Topics Ideal Op-Amp
Inverting and Non-Inverting Amplifiers
Summing and Difference Amplifiers
Integrator and Differentiator
Cascading of Op-Amps
Textbooks:
– Chapter5andChapter6
Inverting terminal (-)
Non- inverting terminal (+)
Output
Op-Amp in General
Capable of realizing many mathematical operations like signal summation, amplification, integration and differentiation.
It is hence popularly used in analog circuits. Common pins found with op-amp are …
+ Power supply terminal
Dual in-line package (DIP)
– Power supply terminal
Inverting terminal (-)
Non- inverting terminal (+)
Output
Op-Amp in General
Input at the non-inverting terminal will appear with the same polarity at the output.
Input at the inverting terminal will appear inverted at the output. For simplicity, power supply terminals are usually not drawn.
+ Power supply terminal
Dual in-line package (DIP)
– Power supply terminal
Equivalent Circuit Model
To avoid internal complexity of op-amp, Thevenin’s theorem can be applied to model its input and output.
Output source voltage can be expressed as
where A is the open-loop gain when there isn’t any external feedback from output to input.
Only Thevenin’s resistance
Both Thevenin’s resistance and controlled source
Typical Values
High since op-amp is after all an amplifier
High to avoid drawing large input current and hence high losses Low so that with zero voltage drop across
Negative saturation
Linear range:
Positive saturation
Negative Feedback
Output being fed back to the inverting () terminal of the op-amp.
Ratio of output voltage to input voltage is then called the closed-loop gain.
Closed-loop gain is almost insensitive to the open-loop gain A.
Practical op-amp circuits therefore always have feedback paths.
For example …
Example 7.1
Given210,input 2M,andoutput 50.
Find⁄ andwhen 2V.
vvvvv s111o
vv vAv v200kv 1oodo 1
10k 2000k 20k 200v 301v 100v
20k 50 50 Substituting and solving give:
s1o s1o o
2v 3v 1v v 1.9999699
v 2vs vo 13
vs
Example 7.1
When 2 V, 3.9999398 V and 20.066667 V (from earlier red enclosed expression)
1o
ivv 0.19999mA
20k
Example 7.1
To what extent closed-loop gain ⁄ is affected by A?
Expressions derived earlier:
v 2vs vo and 13
v v v Av 1oo1
Since A is usually large:
o50 20k 120k 50 50 1 v 1 1 A 2Av
Substituting andrearranging:
o5020k150 150s Avo2Avs andvo2
20k 50 v1 1 v 1 AAv
150 150 vs
Almost insensitive of A and very close to earlier computed answer of ⁄ 1.9999699
Op-amp with the following features. – Infinite open-loop gain → ∞. – Infinite input resistance → ∞. – Zero output resistance 0.
Ideal Op-Amp
Ideal representation can greatly simplify computation without compromising accuracy significantly.
With an infinite , input currents are zero. With zero input currents, 0 implying .
Is output current also zero?
Remember …
Ideal Op-Amp
KCLrequires .
Therefore, 0doesnotimply 0.
Topics
Ideal Op-Amp
Inverting and Non-Inverting
Amplifiers
Summing and Difference Amplifiers
Integrator and Differentiator
Cascading of Op-Amps
Textbooks:
– Chapter5andChapter6
Grounded
Note also 0, implying:
Inverting Amplifier
• Inverting amplifier reverses polarity of input signal, while amplifying it.
• Closed-loop gain ⁄ depends only on external
components, and not inner open-loop gain of the op-amp.
To find relationship between input voltage and output voltage.
KCL at node 1: ii
12
vvvv i11o
RR 1f
vo Rf vi R
Negative
1
Find vo.
Example 7.2
KCL at node a:
vavo 6va 40k 20k
vo 3va 12 Notetoothat 2:
vo 32126V
Prove the following two expressions.
Exercise
v vRR o R o R1 3 3
i i 1RR ss12
Voltage Follower
→∞and 0 Simplify to a voltage follower
What happen if → ∞, 0 or both?
R
vo 1 f vi vi
1
R
Voltage Follower as Buffer What happen upon connecting the two stages?
Current may not be zero
Current = zero because of infinite op-amp input impedance
Second stage “affects” first stage by drawing current from it.
First stage does not “see” second stage because of the op- amp buffer.
Find vo.
Example 7.3
KCL at node a:
vavo 6va 10k 4k
Notetoothat 4: 4vo 64
10 4 vo 1V
Topics
Ideal Op-Amp
Inverting and Non-Inverting
Amplifiers
Summing and Difference Amplifiers
Integrator and Differentiator
Cascading of Op-Amps
Textbooks:
– Chapter5andChapter6
Grounded
f123 RRR
Summing Amplifier
• Summing amplifier or summer reverses polarities of the inputs, while amplifying them.
• Individual closed-loop gain ⁄ , 1, 2 3 again does not depend on inner open-loop gain of the op-amp.
Output equals weighted sum of all inputs.
KCL at node a: iiii
vvvv RRRR
123
Since 0, KCL becomes: o123
v f v f v f v o R1R2R3
123
Find vo and io.
By KCL:
Example 7.4
For a summing amplifier:
v 10k210k18V o 5k 2.5k
iovo vo 4.8mA 10k 2k
Difference Amplifier
Also: RRR R1RRR
v 2 1 4 v 2 v 2 1 2 v 2 v o R RR2R1R1RR2R1
1 3 4 1 1 3 4 1 If⁄ ⁄: v R2 v v
Output amplifies difference, while rejecting common signals between inputs.
KCL at node a: vvvv
oR21 1
1aao RR
12
RR v 2 1v 2 v
oRaR1 11
vavbR4 v2 R3 R4
For a difference amplifier:
Example 7.5
Design a difference amplifier with output 3 5, where and are inputs.
v R 1 R R v R v 2122
o R1RR2 R1 1341
One possible set of values is thus
10k, 50k,and 10k
R2 5 and 5 115 3 R 1R R
134
The second expression leads to:
R3 1 R4
Prove that
R 2R v21 3vv
Op-amp : vo R2 vo2 vo1 R
Op-amps and : Two ways to calculate :
iv v v v 1 2 o1 o2
R4 2R3 R4
2R
v v1 3vv o2 o1 R42 1
Considering op-amp again:
R 2R v21 3vv
1
Example 7.6
oRR21 14
va v1 vb v2
oRR21 14
Topics
Ideal Op-Amp
Inverting and Non-Inverting
Amplifiers
Summing and Difference Amplifiers
Integrator and Differentiator
Cascading of Op-Amps
Textbooks:
– Chapter5andChapter6
Integrator
Output proportional to integral of input.
KCL at node a:
iR iC
vi Cdvo R dt
tdvo 1 tvid 0 RC 0
votvo0 1 tvid RC 0
If 0 0:
vot 1 tvid RC 0
Find if its initial value is zero. 102 mV
0.5 mV KCL at node a:
12o v v 2dv
Example 7.7
3M 100k dt vt1tvd 1 tvd
o 6 0 1 0.2 0 2
v t15sin2t50.25t2
o
6
v t0.833sin2t1.25t2 mV o
Differentiator
Output proportional to derivative of input.
KCL at node a:
iR iC
vo Cdvi R dt
vo t RC dvi dt
Cautious: Any high-frequency input noises will be amplified.
Differentiator is hence not as popular as integrator.
Given input below, sketch . Use votRCdvi
dt
RC 5k 0.2 1ms
Also note
vi 2000t, 0t2ms 82000t, 2t4ms
Therefore
vo 2V, 0t2ms 2V, 2t4ms
Example 7.8
Topics
Ideal Op-Amp
Inverting and Non-Inverting
Amplifiers
Summing and Difference Amplifiers
Integrator and Differentiator
Cascading of Op-Amps
Textbooks:
– Chapter5andChapter6
Cascading of Op-Amps
Cascaded connection means output of one op-amp stage fed as input of another stage.
Common approach applied to build complex circuits with more demanding requirements.
Overall gain is given by:
v Av AAvAAAv
o333223211
vo A AAA v overall 3 2 1
1
Find vo and io.
Circuit consists of two cascaded non-inverting amplifiers
First op-amp:
v 112k20100mV a 3k
Second op-amp:
v 110k100350mV o 4k
Current flows through 10 k and 4 k in series:
i vo 25A
Example 7.9
o
14k
Find vo.
v1 = 1 V
Op-amp A:
va 6k13V Op-ampC:
v2 = 2 V
Op-ampB:vb 8k24V 4k
8.667 V
Example 7.10
2k
10k 10k vo 5k 315k4
End