Bài giảng Electronic circuits - Chapter 1: Op-Amp - Trịnh Xuân Dũng
Dr. Dung Trinh
HCMUT / 2017
Electronic Circuits
Chapter 1: Op-Amp
Dr. Dung Trinh
1
Dr. Dung Trinh
HCMUT / 2017
Content
Introduction
Inverting & Non-Inverting Amplifier
Current-to-Voltage Converter
Instrumentation Amplifier
Integrator and Differentiator
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Dr. Dung Trinh
HCMUT / 2017
Introduction
The integrated circuit operational amplifier evolved soon after development of the
first bipolar integrated circuit.
The μA-709 was introduced by Fairchild Semiconductor in 1965 and was one of the
first widely used general-purpose op-amps. The new classic μA-741, also by
Fairchild, was introduced in the late 1960s.
In the ideal op-amp:
. The open-loop gain Aod
approaches infinity
. The common-mode output
Circuit symbol of Op-Amp
signal is zero.
. Input resistance Ri is infinite.
. Output resistance Ro is zero.
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Dr. Dung Trinh
HCMUT / 2017
Inverting Amplifier
One of the most widely used op-amp circuits is the inverting amplifier
. The closed-loop gain:
푣표
퐴푣 = = −
푣푖
푅2
푅1
Inverting op-amp circuit
. The input resistance:
푅푖 = 푅1
Example 1: Using the inverting Op-amp amplifier
to design the circuit such that the voltage gain is
Av = −5. Assume the op-amp is driven by an ideal
sinusoidal source, 푣푠 = 0.1sin휔푡 (푉), that can
supply a maximum current of 5휇퐴.
Inverting op-amp equivalent circuit
푅1 = 20푘Ω
푅2 = 100푘Ω
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HCMUT / 2017
Inverting Amplifier – Finite Gain
. We have:
푣퐼 − 푣1
푣퐼 − 푣푂
푖1 =
푖2 =
푅1
푅2
. The output voltage is: 푣푂 = −퐴표푑푣1
푣퐼
퐴표푑
푣푂
퐴표푑
푣퐼 +
푣푂 +
푣퐼 − 푣1
푖1 =
푣푂
=
= 푖2 = −
. We obtain:
. Then:
푅1
푅1
푅2
푅2
1
퐴푣 =
= −
1
퐴표푑
푅2
푅1
푣퐼
푅1
1 +
1 +
Example 2: Consider an inverting op-amp with 푅1 = 10푘Ω and 푅2 = 100푘Ω. Determine
the closed-loop gain for: Aod = 102, 103, 104, 105, and 106. Calculate the percent deviation
from the ideal gain.
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HCMUT / 2017
Summing Amplifier
. Using superposition theorem to analysis the
summing amplifier, we obtain:
푅퐹
푅1
푅퐹
푣퐼1 + 푣퐼2 + 푣퐼3
푅2 푅3
푅퐹
푣푂 = −
. If 푅1 = 푅2 = 푅3, then:
푅퐹
푣푂 = −
푣퐼1 + 푣퐼2 + 푣퐼3
푅1
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Dr. Dung Trinh
HCMUT / 2017
Non-Inverting Amplifier
We have:
0 − 푣퐼
푣퐼 − 푣푂
푖2 =
푖1 =
푅2
푅1
푣표
푅2
푅1
Because 푖1 = 푖2, then: 퐴푣 = = 1 +
푣푖
Non-inverting op-amp circuit
In voltage follower circuit: 푅2 = 0
퐴푣 = 1
푅푖 = ∞
푅표 = 0
Example 3: Derive the closed-loop gain of non-
inverting amplifier which has a finite differential
gain of Aod.
Voltage follower op-amp
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HCMUT / 2017
Current-to-Voltage Converter
In some situations, the output of a device or
circuit is a current. An example is the output of a
photodiode or photo-detector. We may need to
convert this output current to an output voltage.
푣푂 = −푖2푅퐹 = −푖푆푅퐹
Current-to-voltage converter
푣퐼
푖2 = 푖1 =
Voltage-to-current converter:
푅1
. Current i2 is independent of the load impedance
or resistance R2.
. NOT practical as the load need to be at ground
Simple voltage-to-current converter
potential.
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HCMUT / 2017
Voltage-to-Current Converter
푣퐼 − 푖퐿푍퐿 푖퐿푍퐿 − 푣푂
At the inverting terminal:
=
푅1
푅퐹
푣푂 − 푖퐿푍퐿
푖퐿푍퐿
At the non-inverting terminal:
= 푖퐿 +
푅3
푅2
From these two equations, we obtain:
푅퐹 푖퐿푍퐿 − 푣퐼
푖퐿푍퐿
= 푖퐿 +
푅1
푅3
푅2
Voltage-to-current converter
푅퐹푍퐿
푅1푅3
푍퐿
푅2
푅퐹
푅1푅3
푖퐿
− 1 −
= 푣퐼
Then:
푅퐹
1
푅퐹
푅1푅3
푣퐼
If
:
=
푖퐿 = −푣퐼
= −
푅1푅3 푅2
푅2
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Dr. Dung Trinh
HCMUT / 2017
Voltage-to-Current Converter
Example 4: Let 푍퐿 = 100Ω , 푅1 = 10푘Ω, 푅2 = 1푘Ω, 푅3 = 1푘Ω, and 푅퐹 = 10푘Ω.
If 푣퐼 = −5푉, determine the load current iL and the output voltage vO.
푖퐿 = 5푚퐴
푣표 = 6푉
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Dr. Dung Trinh
HCMUT / 2017
Difference Amplifier
An ideal difference amplifier amplifies only the
difference between two signals. It rejects any
common signals to the two input terminals.
For example, a microphone system amplifies an
audio signal applied to one terminal of a
difference amplifier, and rejects any 60 Hz noise
signal or “hum” existing on both terminals
Op-amp difference amplifier
푅4
푅3
푅2
푅1
푅2
푅1
푣푂 = 1 +
푣퐼2 −
푣퐼1
The output voltage:
푅4
푅3
1 +
푅2 푅4
푅2
If
If
= :
푣푂 =
푣퐼2 − 푣퐼1
푅1 푅3
푅1
푅2 푅4
≠ :
푣푂
퐴푑
1
퐴푐푚 =
퐶푀푅푅 =
푣푐푚 = 푣퐼2 + 푣퐼1
푅1 푅3
푣푐푚
퐴푐푚
2
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Dr. Dung Trinh
HCMUT / 2017
Difference Amplifier
Example 5: Consider the difference amplifier. Let 푅2/푅1 = 10 and 푅4/푅3 = 11.
Determine CMRR(dB).
퐴푑 = 10.042
퐴푐푚 = 0.0833
퐶푀푅푅 = 41.6푑퐵
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Dr. Dung Trinh
HCMUT / 2017
Instrumentation Amplifier
Obtain a high input impedance and a high
gain in a difference amplifier with
reasonable resistor values: DIFFICULT.
SOLUTION: insert a voltage follower
Problem: GAIN is not easily to change.
INSTRUMENTATION AMPLIFIER allows us to
change the gain by changing only a single
resistance value.
Instrumentation amplifier
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Dr. Dung Trinh
HCMUT / 2017
Instrumentation Amplifier
. The output of difference amplifier is:
푅4
푅3
푅4
푅3
푅2
푅1
푣푂 =
푣푂2 − 푣푂1
=
1 + 2
푣퐼2 − 푣퐼1
Problems:
1. The common mode gain Acm and the
differential gain Aid of the first stage are
equal. This means the common mode
signal will be amplified and the overall
CMRR will be reduced.
2. In order to change the overall gain, we
need to vary the values of two resistance.
This is not an easy task.
Instrumentation amplifier
Solution: Disconnect point X to the ground.
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Dr. Dung Trinh
HCMUT / 2017
Instrumentation Amplifier
. The current in resistor R1 and R2 is
푣퐼1 − 푣퐼2
푖1 =
푅1
. The output voltages of op-amps are:
푅2
푅2
푣푂1 = 푣퐼1 + 푖1푅2 = 1 +
푣푂2 = 푣퐼2 − 푖1푅2 = 1 +
푣퐼1 − 푣퐼2
푅1
푅1
푅2
푅1
푅2
푣퐼2 − 푣퐼1
푅1
Voltages and currents in instrumentation amplifier
. The output of difference amplifier is:
푅4
푅3
푅4
푅3
푅2
푅1
푣푂 =
푣푂2 − 푣푂1
=
1 + 2
푣퐼2 − 푣퐼1
The overall gain does not depend on the matching between the two resistors.
v01 and vO2 are equal if equal voltages appear at the negative terminal of A1 and A2
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Dr. Dung Trinh
HCMUT / 2017
Instrumentation Amplifier
Example 6: Consider the instrumentation amplifier circuit. Assume that 푅4 = 2푅3
so that the difference amplifier gain is 2. Determine the range required for resistor
R1 to realize a differential gain adjustable from 5 to 500. Assume that R1 is a variable
resistor varying from 푅1푓 to 푅1푓 + 100푘Ω
푅1푓 = 0.606푘Ω
푅2 = 75.5푘Ω
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Dr. Dung Trinh
HCMUT / 2017
Integrator and Differentiator
Op-amp integrator
Generalized inverting amplifier
푣퐼
푣푂 = −
Op-amp integrator:
푠푅1퐶2
푣푂 = −푣퐼푠푅2퐶1
Op-amp differentiator:
Op-amp differentiator
Reading: Microelectronics, Circuit Analysis and Design, D.A. Neamen, 4th edition, p621-670
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Dr. Dung Trinh
HCMUT / 2017
Integrator and Differentiator
Example 7: Find the output produced
by an integrator in response to an input
pulse of 1V height and 1ms width. Let
R = 10푘Ω and C = 10푛퐹 . If the
integrator is shunted by a 1푀Ω resistor.
How will the response be modified.
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Dr. Dung Trinh
HCMUT / 2017
Comparator
When 푣2 is slightly greater than 푣1:
The output is driven to a high
saturated state 푉퐻
When 푣2 is slightly less than 푣1:
The output is driven to a low
saturated state 푉
퐿
The transition region occurs when the difference input voltage in the range −훿, 훿
Example: if the open-loop voltage gain is 105 and the difference between the two
푉 −푉
10
퐻
퐿
stage is 푉퐻 − 푉 = 10푉 then 2훿 =
=
= 10−4 푉 .
퐿
5
퐺
10
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Dr. Dung Trinh
HCMUT / 2017
Comparator
For input bias current compensation
Using the superposition, we obtain:
푅1
푅2
푣+ =
푉푅퐸퐹 +
푣퐼
푅1 + 푅2
푅1 + 푅2
The ideal crossover voltage occur:
푅1
푣+ = 0 ↔ 푣퐼 = − 푉푅퐸퐹
푅2
For input bias current compensation
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