ENGR 201 β Electrical Fundamentals I
SECTION 4: OPERATIONAL AMPLIFIERS
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Introduction2
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Amplification
Amplification Multiplication of electrical signals Voltage or current Scaling factor: gain
Performed by amplifiers
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Amplification β Why?
Often want to increase the amplitude of small electrical signals Sensor outputs, e.g.: Strain gauges Pressure sensors Flow meters Temperature sensors Photo-detectors, etc.
Wireless communication signals Audio signals
Larger signals are easier to measure Utilize the full dynamic range of the measurement system Higher accuracy
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Amplification β Why?
Amplifiers are also useful for impedance conversion Make a high-resistance source look like a low resistance Make a low-resistance load look like a high resistance Buffering a high- resistance source from a low-
resistance load
A single amplifier circuit can provide amplification and impedance conversion
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Amplifier Fundamentals6
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Amplifier Characteristics
Key amplifier characteristics: GainMay be designed for voltage, current, or power gain
Input resistance Output resistance
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Amplifier Characteristics - Gain
Voltage gain Ratio of output to input voltage
π΄π΄ππ =π£π£πππ£π£ππ
Current gain Ratio of output to input current
π΄π΄ππ =ππππππππ
Power gain Ratio of output to input power
πΊπΊ =ππππππππ
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Amplifier Characteristics β Input Resistance
Input resistance, π π ππ Equivalent resistance seen looking into the amplifier
Amplifier loads the source Source appears to be connected to a resistance of π π ππ Possible voltage division between source resistance
and input resistance
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Amplifier Characteristics β Output Resistance
Output resistance, π π ππ ThΓ©venin equivalent resistance of the amplifier output
From the perspective of the load, amplifier output is the source Modeled as ThΓ©venin equivalent circuit, with
resistance, π π ππ Possible voltage division between π π ππ and the load
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Amplifier Characteristics β Cascaded Amplifiers
Input to the second amplifier is the output from the first
π£π£πππ π‘π‘ = π΄π΄π£π£π β π£π£ππ π‘π‘
Output of the second amplifier is the output of the cascade
π£π£ππ π‘π‘ = π΄π΄π£π£π£ β π£π£πππ π‘π‘ = π΄π΄π£π£π β π΄π΄π£π£π£ β π£π£ππ π‘π‘
Overall gain is the product of the individual gains
π΄π΄π£π£ = π΄π΄π£π£π β π΄π΄π£π£π£
Consider two amplifiers connected in cascade
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Amplifier β Equivalent Circuit
Amplifier equivalent circuit:
π π ππ: input resistance π π ππ: output resistance π΄π΄π£π£πππ£π£: open-circuit voltage gain
Note that, in general, due to loading:
π΄π΄π£π£πππ£π£ =π£π£π΄π΄π£π£ππβ π΄π΄π£π£ =
π£π£πππ£π£π π
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Opamp Fundamentals13
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Operational Amplifiers - Opamps
How do we get the amplification we need? Different requirements for different applications
High gain/low gain β Inverting/non-inverting gain Accuracy Adjustability
Building amplifiers out of transistors is difficult, inconvenient Chip makers could make unique integrated circuits (ICs) for
all possible applications Impractical, not economical
Instead: operational amplifiers (opamps) General-purpose amplifier ICs Gain set by a few external components
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Operational Amplifiers - Opamps
High-gain amplifiers built from many transistors, resistors, and capacitors
Integrated circuits (ICs) All components fabricated on a single semiconductor (e.g.
Si) chip Gain set by a few external resistors
www.chiplook.com
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Ideal Opamps
Schematic symbol:
outputnon-inverting input
inverting input
Equivalent circuit: Differential input, π£π£ππππ
Difference between the two input voltages
π£π£ππππ = π£π£+ β π£π£β
Common-mode input, π£π£πππ£π£ππ
π£π£πππ£π£ππ =π£π£+ + π£π£β
2
Open-loop voltage gain, π΄π΄π£π£πππ£π£ Gain of the opamp without
feedback
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Ideal Opamp Characteristics
Ideal opamp characteristics:
1) Infinite input resistances β at π£π£+ and π£π£β No current flows into either input terminal
2) Infinite open-loop gain Any differential input will result in an infinite output
3) Zero output resistance Immune to the effect of loading
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Infinite Gain
Ideal opamps have infinite gain How do we get a gain of 2 or 9 or -3?
Negative feedback
By enclosing the opamp in a feedback loop we can create an amplifier with useable gain
Before applying feedback to opamp circuits, weβll first introduce the concept of feedback
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Introduction to Feedback19
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Feedback
Feedback A process in which a portion of the output of some system is
fed back to the input of that system
Positive feedback The addition of a portion of the output to the input Generally has a destabilizing effect
Negative feedback The subtraction of portion of the output from the input Generally has a stabilizing effect All opamp amplifiers we will encounter in this course
employ negative feedback
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Feedback
Feedback is everywhere: HVAC Cruise control Robotics Amplifiers
Toilets Ovens Autonomous vehicles Etc.
A very important concept in many engineering fields, particularly controls and electronics
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Signal Flow Diagrams
Signal flow diagrams Block diagrams Show the flow of signals β energy, information, etc. β through a system Used for all types of engineering systems: electrical, mechanical, etc.
Amplifier, with open-loop gain π΄π΄π£π£πππ£π£, enclosed in a feedback loop:
Summing junctionAmplifier
Indicates negative feedback
Feedback network with a gain of π½π½
Amplifier is in a closed-loop configuration
Feedback gain π½π½, determines how much output is fed back
Difference between input and feedback is amplifier input
π½π½ determines overall gain Closed-loop gain
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Closed-Loop Gain
Open-loop gain Gain of amplifier without feedback
π΄π΄π£π£πππ£π£ =π£π£πππ£π£ππππ
Closed-loop gain Gain of the closed-loop amplifier
π΄π΄π£π£π£π£π£π£ =π£π£πππ£π£ππ
Calculating the closed-loop gain:
π£π£ππ = π£π£ππππ β π΄π΄π£π£πππ£π£ = π£π£ππ β π½π½π£π£ππ π΄π΄π£π£πππ£π£π£π£ππ + π½π½ β π£π£ππ β π΄π΄π£π£πππ£π£ = π£π£ππ β π΄π΄π£π£πππ£π£π£π£ππ 1 + π½π½ β π΄π΄π£π£πππ£π£ = π£π£ππ β π΄π΄π£π£πππ£π£
π΄π΄π£π£π£π£π£π£ =π£π£πππ£π£ππ
=π΄π΄π£π£πππ£π£
1 + π½π½ β π΄π΄π£π£πππ£π£
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Closed-Loop Gain
E.g., an ideal opamp
limπ΄π΄π£π£π£π£π£π£ββ
π΄π΄π£π£π£π£π£π£ = limπ΄π΄π£π£π£π£π£π£ββ
π΄π΄π£π£πππ£π£1 + π½π½ β π΄π΄π£π£πππ£π£
limπ΄π΄π£π£π£π£π£π£ββ
π΄π΄π£π£π£π£π£π£ = limπ΄π΄π£π£π£π£π£π£ββ
π΄π΄π£π£πππ£π£π½π½ β π΄π΄π£π£πππ£π£
=1π½π½
Feedback gain, alone, determines the closed-loop gain For example, to get a gain of four, feed back one quarter of the
output signal
π΄π΄π£π£π£π£π£π£ =π£π£πππ£π£ππ
=π΄π΄π£π£πππ£π£
1 + π½π½ β π΄π΄π£π£πππ£π£
Consider the case of infinite open-loop gain
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Summing Point Constraint
For π΄π΄π£π£πππ£π£ = β, the differential input is
π£π£ππππ = π£π£ππ β π½π½π£π£ππ = π£π£ππ β π½π½1π½π½ π£π£ππ
π£π£ππππ = π£π£ππ β π£π£ππ = 0
A very important result, the summing-point constraint:
The input to an infinite-gain amplifier, enclosed in a negative feedback loop, is zero!
Any non-zero input would yield infinite output
Along with the properties of ideal opamps, the summing-point constraint will be essential for analyzing opamp circuits
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Opamps as Feedback Systems
Any amplifier enclosed in a negative feedback loop: Output scaled by π½π½ and fed back Feedback signal subtracted from
the input at a summing junction If π΄π΄π£π£πππ£π£ = β, then π£π£ππππ = 0
Ideal opamp enclosed in a negative feedback loop: Output scaled by π½π½ and fed back Differential input is a built-in
summing junction π£π£ππππ = 0, so π£π£+ = π£π£β A virtual short between input
terminals
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Example Problems27
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An amplifier with a gain of 4 is used to amplify the output of a sensor.The amplifier has an input resistance of 1 kΞ© and an output resistanceof 100 Ξ©. The sensor has an open-circuit voltage of 1 V, and an outputresistance of 50 Ξ©. The amplifier drives a 5 kΞ© load. What is theamplifierβs output voltage?
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An amplifier is enclosed in a negative feedback loop with a feedback gain of 0.5.Determine the closed-loop gain for an amplifier with an open-loop gain of:
a) 10b) 100E3
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A 1 V input is applied to an opamp enclosed in a negative feedback loop with afeedback gain of 0.2. Determine the differential input voltage for an opamp with anopen-loop gain of:
a) 10b) 100E3
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Opamp Amplifiers34
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Analysis of Ideal Opamp Circuits
Key properties for analysis of opamp circuits:
1) Infinite input resistance β no input current
2) Virtual short between inverting and non-inverting input terminals (as long as there is negative feedback)
3) Infinite open-loop gain
If there is negative feedback, 1 and 2 are sufficient
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Opamp Amplifiers
Two basic opamp amplifier configurations:
Non-inverting amplifier: Inverting amplifier:
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Non-Inverting Amplifier37
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Non-Inverting Amplifier β Gain
Non-inverting amplifier gain:
Negative feedback, so the summing-point constraint applies
π£π£β = π£π£+ = π£π£ππ
Zero opamp input current, so KCL at the inverting terminal gives:
πππ = πππ£
π£π£ππ β π£π£β
π π π£=π£π£β
π π π
Applying the summing-point constraintπ£π£ππ β π£π£πππ π π£
=π£π£πππ π π
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Non-Inverting Amplifier β Gain π£π£ππ β π£π£πππ π π£
=π£π£πππ π π
Solving for the amplifier output
π£π£ππ = π£π£ππ1π π π
+1π π π£
π π π£
Dividing both sides by π£π£ππ gives the non-inverting amplifier gain:
π΄π΄π£π£ =π£π£πππ£π£ππ
=π π π + π π π£π π π
Note that this is the inverse of the feedback path gain
π΄π΄π£π£ =1π½π½
=1π π π
π π π + π π π£
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Non-Inverting Amplifier β Gain
π΄π΄π£π£ =π£π£πππ£π£ππ
=π π π + π π π£π π π
Gain determined entirely by the relative value of two external resistors Any gain value is possible Resistor tolerance sets amplifier gain tolerance
Gain is positive (non-inverting) As input goes up, output goes up
Gain can never be less than one Unity gain for π π π£ = 0 Ξ©
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Non-Inverting Amplifier β π π ππ and π π ππ
Output resistance, π π ππ: Output is the output from an ideal opamp
π π ππ = 0 Ξ©
Input resistance, π π ππ: Input connected directly to
input terminal of ideal opamp
π π ππ = β
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Inverting Amplifier42
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Inverting Amplifier β Gain
Inverting amplifier gain:
Negative feedback, so the summing-point constraint applies
π£π£β = π£π£+ = 0
Zero opamp input current, so KCL at the inverting terminal gives:
πππ = πππ£π£π£πππ π π
=βπ£π£πππ π π£
Gain of the inverting amplifier:
π΄π΄π£π£ =π£π£πππ£π£ππ
= βπ π π£π π π
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Inverting Amplifier β Gain
π΄π΄π£π£ =π£π£πππ£π£ππ
= βπ π π£π π π
Again, gain determined entirely by external resistor values
Gain is negative (inverting) As input goes up, output goes down
Gain can be any value π΄π΄π£π£ < 1 for π π π > π π π£
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Inverting Amplifier β π π ππ
By definition, input resistance is
π π ππ =π£π£ππππππ
= π£π£πππ π πππππ
π π ππ = π π π
Input is a resistance, π π π, to (virtual) ground
Input resistance, π π ππ: Inverting terminal is a virtual
ground, so
ππππ =π£π£πππ π π
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Inverting Amplifier β π π ππ
Consider a non-ideal opamp with non-zero output resistance: π π 3 is driven by a dependent
source Determining π π ππ is a bit trickier
Output resistance, π π ππ: Output is the output of an
ideal opamp, so
π π ππ = 0 Ξ©
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Inverting Amplifier β π π ππ To determine a terminal
resistance in the presence of dependent sources: Set all independent sources
to zero Here, ground the input, π£π£ππ
Apply a test current, πππ‘π‘, to the terminal of interest
Analyze the circuit to determine the voltage at that terminal
Resistance at that terminal is given by
π π ππ =π£π£πππππ‘π‘
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Inverting Amplifier β π π ππ
The output is zero, independent of πππ‘π‘ (and independent of π π 3), so
π π ππ = 0 Ξ©
Feedback around an infinite gain amplifier forces the closed-loop output resistance to zero, even if the output resistance of the amplifier itself is non-zero
Still a virtual ground at the inverting terminal, so
πππ = πππ£ =π£π£πππ π π
=0 πππ π π
= 0
πππ£ = βπ£π£πππ π π
= 0
π£π£ππ = 0
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Example Problems49
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Determine the gain of the following circuits.
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Determine the output voltage, Vo.
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Determine the gain of the following circuit.
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Determine the output voltage, Vo.
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Determine the LED bias current, Io.
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Unity-Gain Buffer57
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Unity-Gain Buffer
Start with a non-inverting amplifier, and let the feedback path gain go to unity
π½π½ β 1
Inverting terminal is connected directly to the outputπ£π£β = π£π£ππ
Virtual short at the input terminalsπ£π£β = π£π£+ = π£π£ππ = π£π£ππ
Closed-loop gain is unity
π΄π΄π£π£ =π£π£0π£π£ππ
= 1
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Unity-Gain Buffer β Buffering
What good is an amplifier with unity gain? Impedance conversion A buffer between a high-resistance source and a lower-resistance
load Eliminates loading effects
Consider the following scenario: Sensor with π π π‘π‘π‘ = 10 ππΞ© drives a load of π π πΏπΏ = 1 ππΞ©
Signal measured at the load is attenuated
π£π£πΏπΏ = π£π£π π π π πΏπΏ
π π π π + π π πΏπΏ= π£π£π π
1 ππΞ©11 ππΞ©
π£π£πΏπΏ =1
11π£π£π π
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Unity-Gain Buffer β Buffering
To prevent signal attenuation due to loading: Add a unity-gain buffer between the source and the load
Buffer input is the load for the source π π ππ = β
Buffer output is the source for the load π π ππ = 0 Ξ©
Now, full sensor signal appears across the load
π£π£πΏπΏ = π£π£+ = π£π£π π
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Summing & Difference Amplifiers61
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Summing Amplifier
KCL at the inverting node:
πππ΄π΄ + πππ΅π΅ + πππΆπΆ = πππππ£π£π΄π΄π π π΄π΄
+π£π£π΅π΅π π π΅π΅
+π£π£πΆπΆπ π πΆπΆ
= βπ£π£πππ π πΉπΉ
Output is the (inverted) weighted sum of all of the inputs
π£π£ππ = βπ π πΉπΉπ£π£π΄π΄π π π΄π΄
+π£π£π΅π΅π π π΅π΅
+π£π£πΆπΆπ π πΆπΆ
For the special case of π π π΄π΄ = π π π΅π΅ = π π πΆπΆ = π π ππ:
π£π£ππ = βπ π πΉπΉπ π ππ
π£π£π΄π΄ + π£π£π΅π΅ + π£π£πΆπΆ
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Difference Amplifier
This looks a bit like a combination of a non-inverting and an inverting amplifier
Analyze by applying superposition First, set π£π£π£ = 0
A non-inverting amplifier with a voltage divider at the input
π£π£ππ οΏ½π£π£1= π£π£+
π π π + π π π£π π π
π£π£+ = π£π£ππ π 4
π π 3 + π π 4
π£π£ππ οΏ½π£π£1= π£π£π
π π 4π π 3 + π π 4
π π π + π π π£π π π
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Difference Amplifier
Next, set π£π£π = 0 No current through π π 3 and π π 4
π£π£+ = 0 ππ
An inverting amplifier
π£π£ππ οΏ½π£π£2
= βπ£π£π£π π π£π π π
Summing the contributions from each output:
π£π£ππ = π£π£ππ οΏ½π£π£1
+ π£π£ππ οΏ½π£π£2
π£π£ππ = π£π£ππ π 4
π π 3 + π π 4π π π + π π π£π π π
β π£π£π£π π π£π π π
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Difference Amplifier
Restricting resistor values: π π 3 = π π π and π π 4 = π π π£
π£π£ππ = π£π£ππ π π£
π π π + π π π£π π π + π π π£π π π
β π£π£π£π π π£π π π
π£π£ππ = π£π£ππ π π£π π π
β π£π£π£π π π£π π π
π£π£ππ =π π π£π π π
π£π£π β π£π£π£
Inverting and non-inverting amplifiers are special cases of the difference amplifier Inverting amplifier: π£π£π is grounded Non-inverting amplifier: π£π£π£ is grounded
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Example Problems66
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Determine the output voltage, Vo.
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Determine the output voltage, Vo.
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Design an opamp circuit to perform the following mathematical operation:
ππππ = 3πππ β 5πππ£
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Design an opamp circuit to perform the following mathematical operation:
ππππ = 5πππ β 3πππ£
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Comparators75
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Open-Loop Opamp Behavior
So far, weβve looked at opamp amplifier circuits Closed-loop configuration Negative feedback Output remains within the opampβs linear output range
Weβll now consider using an opamp open-loop βwithout feedback β or with positive feedback For ideal opamp, π΄π΄π£π£πππ£π£ = β For any non-zero input, π£π£ππ β Β±β But, π£π£ππ limits, or saturates, somewhere near the supply
voltages, ππππ,ππππππ In practice, the output of an open-loop opamp is always
saturated at Β±ππππ,ππππππ
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Comparators
Open-loop opampβs output determined by relative values of the two inputs, differential input
ππ+,ππβ Vid Vo
ππ+ > ππβ
ππ+ < ππβ
An open-loop opamp acts as a comparator Compares the two input voltages Sets the output based on which input is higher
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Comparators
Comparator Inputs
Comparator Output
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Comparators β Applications
Simple comparator application: thermostat Inverting input connected to a temperature sensor Non-inverting input connected to a variable reference voltage determined by
the temperature setpoint If ππππππππππ < πππ π π π π‘π‘
πππ‘π‘π π πππ‘π‘ < πππππ π ππ Output is high Heat turns on
If ππππππππππ > πππ π π π π‘π‘ πππ‘π‘π π πππ‘π‘ > πππππ π ππ Output is low Heat is off
Another example application: motion-sensing light One input from a motion sensor β variable analog voltage Other input is a threshold voltage set by sensitivity setting
Want light to turn on for people and cars, not birds or insects
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Comparators and Noise
Consider the following comparator circuit: Inverting input driven by
sinusoidal voltage Non-inverting input is
threshold voltage β connected to ground (0 ππ)
Output switches cleanly at each input threshold crossing
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Comparators and Noise
Now, the sinusoidal input is corrupted by noise Multiple input threshold
crossings each time π£π£π π π‘π‘ goes through 0 ππ
Multiple, unwanted, output transitions
Would like to be able to reject this noise at the input: Schmitt trigger
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Hysteresis82
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Hysteresis β Schmitt Trigger
Schmitt trigger employs hysteresis Characteristics of the circuit are
dependent on its previous state Looks like a non-inverting amplifier,
but it is not Positive feedback
A comparator with a threshold voltage that depends on the output
π£π£+ = π£π£πππ π π£
π π π + π π π£= π½π½π£π£ππ
Threshold voltage switches between two values as the output switches between Β±ππππ,ππππππ
π£π£+ = Β±π½π½ππππ,ππππππ
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Schmitt Trigger β Hysteresis voltage
Consider the case where the input is lowπ£π£ππ < π£π£+
Output will be highπ£π£ππ = +ππππ,ππππππ
The input then increases and exceeds the threshold voltage
π£π£ππ > π£π£+
The output will then switch lowπ£π£ππ β βππππ,ππππππ
The threshold voltage will switch lowπ£π£+ β βπ½π½ππππ,ππππππ
Threshold voltage switches away from the rising input Similar thing is true for falling input
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Schmitt Trigger β Hysteresis voltage
Output, and threshold voltage, always switches away from the input signal at the first threshold crossing
Hysteresis voltage: Magnitude of the threshold voltage
change
πππ‘π¦π¦π π π‘π‘ = 2π½π½ππππ,ππππππ
Hysteresis voltage set for the amount of noise that is present Threshold must switch far enough away from the noisy input that
it will not be crossed multiple times by noise alone πππ‘π¦π¦π π π‘π‘ set greater than peak-to-peak noise on the input signal
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Comparators and Noise β Hysteresis
Feedback gain, π½π½, set to provide adequate hysteresis to reject the input noise Noisy input crosses threshold Threshold voltage switches away
from the input by the hysteresis voltage
πππ‘π¦π¦π π π‘π‘ > ππππ,π‘π‘π‘π‘
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Hysteresis Voltage
Hysteresis voltage Full peak-to-peak
swing of the threshold voltage
Twice the magnitude of the feedback signal
πππ‘π¦π¦π π π‘π‘ = 2π½π½ππππ,ππππππ
πππ‘π¦π¦π π π‘π‘ = 400 ππππ
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Hysteresis Voltage
Can also measure πππ‘π¦π¦π π π‘π‘ by performing a bidirectional DC sweep of the input Low-to-high, then high-to-low
Output traces a different path depending on direction of π£π£ππππ Vertical lines indicate
threshold voltages
πππ‘π¦π¦π π π‘π‘ given by difference between upper and lower thresholds
Vin increasing
V inde
crea
sing
πππ‘π¦π¦π π π‘π‘ = 400 ππππ
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Adjustable Hysteresis
If noise level is not known ahead of time Use a potentiometer to get adjustable hysteresis
π π π£ adjustable from 0 Ξ© to π π π£,ππππππ Hysteresis varies as π π π£ varies
πππ‘π¦π¦π π π‘π‘ = 2π π π£
π π π + π π π£ππππ,ππππππ
Max hysteresis at π π π£ = π π π£,ππππππ
πππ‘π¦π¦π π π‘π‘,ππππππ = 2π π π£,ππππππ
π π π + π π π£,ππππππππππ,ππππππ
Minimum hysteresis for π π π£ = 0 Ξ©
πππ‘π¦π¦π π π‘π‘,ππππππ = 20 Ξ©
π π π + 0 Ξ©ππππ,ππππππ = 0 ππ
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Schmitt Trigger β Example
Consider the following scenario: Test engine instrumented and running on a dynamometer Want an engine temperature warning light (LED) RTD (resistive temperature detector or resistive thermal device) installed
to measure coolant temperature. RTD is biased such that a 0 ππ output corresponds to ππππππππ Noise: ππππ,π‘π‘π‘π‘ β 100 ππππ Opamp available for use as a comparator:
πππ π π π π‘π‘π‘π‘π£π£π¦π¦ = Β±5 ππ
ππππ,ππππππ = Β± πππ π π π π‘π‘π‘π‘π£π£π¦π¦ β 500 ππππ
πΌπΌππ,ππππππ = Β±10 πππ΄π΄
πΌπΌπΏπΏπΏπΏπΏπΏ = 9 πππ΄π΄
K. Webb ENGR 201
91
Schmitt Trigger β Example
When ππ > ππππππππ RTD output exceeds 0 ππ Schmitt trigger output goes low: π£π£ππ β β4.5 ππ LED turns on: πΌπΌπΏπΏπΏπΏπΏπΏ = 9 πππ΄π΄ Opamp output sinks LED current and feedback network current Feedback network current must not exceed 1 πππ΄π΄
Determine π π π and π π π£ to: Set required
hysteresis voltage Limit feedback path
current
K. Webb ENGR 201
92
Schmitt Trigger β Example
Required hysteresis is at least the peak-to-peak noise voltage, ππππ,π‘π‘π‘π‘ β 100 ππππ Design for πππ‘π¦π¦π π π‘π‘ = 150 ππππ The opamp saturates at Β±4.5 ππ, so
πππ‘π¦π¦π π π‘π‘ = 2 β 4.5 πππ π π£
π π π + π π π£= 150 ππππ
π π π£π π π + π π π£
=150 ππππ
9 ππ= 16.67 Γ 10β3
This specifies the ratio between the feedback resistors
π π π£ = 16.95 Γ 10β3 β π π π
Absolute values will be selected to limit feedback path current
K. Webb ENGR 201
93
Schmitt Trigger β Example
When the LED is on, the feedback path current flowing into the opamp output is
πΌπΌππ =4.5 πππ π π + π π π£
< 1 πππ΄π΄
So, the total feedback path resistance is
π π π + π π π£ > 4.5 ππΞ©
Select a standard value for the larger resistor
π π π = 5.1 ππΞ©
Calculate π π π£ using the previously derived ratio
π π π£ = 16.95 Γ 10β3 β 5.1 ππΞ© = 86.4 Ξ©
Select the next larger standard value: π π π£ = 91 Ξ© Larger, so as not to decrease πππ‘π¦π¦π π π‘π‘
K. Webb ENGR 201
94
Schmitt Trigger β Example
The resulting circuit:
Verifying hysteresis:
πππ‘π¦π¦π π π‘π‘ = 9 ππ91 Ξ©
5.1 ππΞ© + 91 Ξ©
πππ‘π¦π¦π π π‘π‘ = 157 ππππ
Feedback path current:
πΌπΌππ =4.5 ππ
5191 Ξ©= 867 πππ΄π΄
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