程序代写代做代考 Part 7: Operational Amplifier or Op-Amp

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
 Given􏰀􏰃2􏰍10􏰎,input􏰉􏰏 􏰃2M,andoutput􏰉􏰇 􏰃50.
 Find􏰁􏰇⁄􏰁􏰏 and􏰐when􏰁􏰑 􏰃2V.
vvvvv s111o
vv vAv v200kv 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
ivv 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:
o50 20k 120k 50 50 1 v  1  1  A 2Av
Substituting􏰁 andrearranging: 􏰆
o5020k150 150s Avo2Avs andvo2
20k 50 v1 1 v 1 AAv
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: ii
12
vvvv i11o
RR 1f
vo Rf vi R
Negative
1

Find vo.
Example 7.2
KCL at node a:
vavo 6va 40k 20k
vo 3va 12 Notetoothat􏰁􏰘 􏰃􏰁􏰙 􏰃2:
vo 32126V

Prove the following two expressions.
Exercise
v vRR o R o R1 3  3 
i i 1RR 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:
vavo 6va 10k 4k
Notetoothat􏰁􏰘 􏰃􏰁􏰙 􏰃4: 4vo 64
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: iiii
vvvv 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 10k210k18V o  5k 2.5k 
iovo vo 4.8mA 10k 2k

Difference Amplifier
 Also: RRR R1RRR
v  2 1 4 v  2 v  2 1 2 v  2 v o R RR2R1R1RR2R1
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: vvvv
oR21 1
1aao RR
12
RR v  2 1v  2 v
oRaR1 11
vavbR4 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 R1RR2 R1 1341
One possible set of values is thus
􏰉􏰆 􏰃10k,􏰉􏰄 􏰃50k,and􏰉􏰞 􏰃􏰉􏰟 􏰃10k
R2 5 and 5 115 3 R 1R R
134
The second expression leads to:
R3 1 R4

Prove that
R 2R v21 3vv
Op-amp 􏰀􏰞: vo  R2 vo2 vo1 R
Op-amps 􏰀􏰆 and 􏰀􏰄: Two ways to calculate 􏰐:
iv v v v  1 2 o1 o2
R4 2R3 R4
 2R
v v1 3vv o2 o1  R42 1
Considering op-amp 􏰀􏰞 again:
R 2R v21 3vv
1
Example 7.6
oRR21 14
va v1 vb v2
oRR21 14

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 tvid 0 RC 0
votvo0 1 tvid RC 0
If􏰁􏰇 0 􏰃0:
vot 1 tvid RC 0

Find 􏰁􏰇 if its initial value is zero. 􏰁􏰆 􏰃 10􏰠􏰜􏰡2􏰢 mV
􏰁􏰄 􏰃 0.5􏰢 mV KCL at node a:
12o v  v 2dv
Example 7.7
3M 100k dt vt1tvd 1 tvd
o 6 0 1 0.2 0 2
v t15sin2t50.25t2
o
6
v t0.833sin2t1.25t2 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 votRCdvi
dt
RC  5k 0.2 1ms
Also note
vi 2000t, 0t2ms 82000t, 2t4ms
Therefore
􏰁􏰏 􏰢

vo 2V, 0t2ms 2V, 2t4ms
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 AAvAAAv
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 112k20100mV a  3k
Second op-amp:
v 110k100350mV o  4k
Current 􏰐􏰇 flows through 10 k and 4 k in series:
i  vo 25A
Example 7.9
o
14k

Find vo.
v1 = 1 V
Op-amp A:
va 6k13V Op-ampC:
v2 = 2 V
Op-ampB:vb 8k24V 4k
 8.667 V
Example 7.10
2k
10k 10k  vo  5k 315k4

End