Amplifier design system and diode analysis
Miss Doody
BIOE3300-Lecture_Series_4_BiosignalAmplifiers_SP20201.pptxBiosignal Amplifiers
BIOE 3300 – Biomedical Electronics
(Chapters 11 & 14)
1
Biomedical Instrumentation
General Architecture
Biomedical
Amplifiers
Human Interface Subsystem
Signal
Processing
User Interface
Human
Interaction
Main Classes of Ideal Amplifiers
Voltage Amplifier
Amplifies an input voltage by a given gain factor (AV)
Vout=AVVin
Current Amplifier
Amplifies an input current by a given gain factor (AI)
Iout=AIIin
Transconductance Amplifier
Senses the input voltage and forces an output current proportional (G) to this voltage to flow through the load.
Iout=GVin
Transresistance Amplifier
Senses the input current and forces an output voltage proportional (R) to this current to appear across the load.
Vout=RIin
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Class Impedance Characteristics
Regardless of the amplifier class, a core objective of an amplifier system is to ensure that the output signal is a properly scaled version of the source signal. This can be achieved, in part, through the impedance characteristics of the amplifier.
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Input Voltage Amplifiers desire to have a high input impedance
Output Voltage Amplifiers desire to have a low output impedance
Input Current Amplifiers desire to have a low input impedance
Output Current Amplifiers desire to have a high output impedance
Ideal Voltage Amplifier Model
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Ideal Voltage Amplifier Model Understanding Input Resistance (Impedance)
Ri is the input resistance of the amplifier
Rs is the source resistance
vs is the source voltage
vi is the input voltage to the amplifier
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Voltage Amplifier Preferred Input Characteristics
Ri is the input resistance of the amplifier
Rs is the source resistance
vs is the source voltage
vi is the input voltage to the amplifier
In this problem,
vs is the source signal (i.e., heart) that we wish to amplify.
vi is the input voltage to the amplifier.
We have no control over Rs as this is the inherent resistance of the body.
Therefore, we can only design around Ri (amplifier input resistance).
The objective is to ensure that vi=vs (no distortion; the signal we want to amplify makes it to the input of the amplifier.
Voltage Amplifier Preferred Input Characteristics
Ri is the input resistance of the amplifier
Rs is the source resistance
vs is the source voltage
vi is the input voltage to the amplifier
The objective is to ensure that vi=vs (0% distortion; the signal we want to amplify makes it to the input of the amplifier.
Lets assume we design the amplifier with a very low input resistance (i.e., Ri0)
In this case, what would vi be equal to?
vi=0; we lost all of our signal vs
(100% Distortion); (not desired)
Lets assume the other extreme such that we design the amplifier with a very high input resistance (i.e., Ri )
In this case, what would vi be equal to?
Since ii=0A then vi=vs (i.e., there can’t be a voltage difference over RBODY as 0 current is flowing).
(Thus, 0% distortion; objective achieved)
Ideal Voltage Amplifier Model Input: Ri
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Voltage Amplifier Preferred Output Characteristics
RO is the output resistance of the amplifier
RL is the load resistance
vs is the source voltage
vi is the input voltage to the amplifier
Avo is the gain of the amplifier
iO is the current leaving the amplifier
vo is the amplifier’s output voltage
The objective is to ensure that
vo=Avovi =Avovs (0% distortion; the output is a properly amplified version of the source signal.)
Lets assume we design the amplifier with a very high output resistance (i.e., Ro )
In this case, what would vo be equal to?
vo=0 as io=0 we lost all of our signal. Avovi does not make it out of the amplifier
(100% Distortion); (not desired)
Lets assume the other extreme such that we design the amplifier with a very low output resistance (i.e., Ro 0)
In this case, what would vo be equal to?
As can be seen, vo=Avovi & assuming Ri then vo=Avovi =Avovs
(0% distortion; objective achieved)
Ideal Voltage Amplifier Model Input: Ri Output: Ro 0
As shown, to minimize distortion, voltage amplifiers must be designed with a high input resistance and low output resistance (i.e., Ri & Ro 0)
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Ideal Current Amplifier Model
Ai is the current gain of the amplifier
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Ideal Current Amplifier Model Understanding Input Resistance (Impedance)
Ri is the input resistance of the amplifier
Rs is the source resistance
is is the source current
ii is the input current to the amplifier
The objective is to ensure that ii=is (0% distortion; the source signal we want to amplify makes it into the input of the amplifier.
Lets assume we design the amplifier with a very high input resistance (i.e., Ri )
In this case, what would ii be equal to?
ii=0; we lost all of our signal is
(100% Distortion); (not desired)
Lets assume the other extreme such that we design the amplifier with a very low input resistance (i.e., Ri0 )
In this case, what would ii be equal to?
Since Ri0 no current will flow though Rs, therefore, ii=is
(Thus, 0% distortion; objective achieved)
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Ideal Current Amplifier Model Input: Ri 0
Ai is the current gain of the amplifier
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Current Amplifier Understanding output impedance
RO is the output resistance of the amplifier
RL is the load resistance
is is the source current
ii is the input current into the amplifier
Ai is the gain of the amplifier
iO is the current leaving the amplifier
The objective is to ensure that
io=Aiii =Aiis (0% distortion; the output is a properly amplified version of the source signal.)
Lets assume we design the amplifier with a very low output resistance (i.e., Ro 0)
In this case, what would io be equal to?
io=0 we lost all of our signal. Aiii does not make it out of the amplifier (all of this current flows through Ro)
(100% Distortion); (not desired)
Lets assume the other extreme such that we design the amplifier with a very low output resistance (i.e., Ro )
In this case, what would io be equal to?
As can be seen, io=Aiii & assuming Ri 0 then io=Aiii =Aiis
(0% distortion; objective achieved)
Ideal Current Amplifier Model Input: Ri 0 Output: Ro
As shown, to minimize distortion, current amplifiers must be designed with a low input resistance and high output resistance (i.e., Ri 0 & Ro )
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Ideal Transconductance Amplifier Model Input: Ri Output: Ro
Transconductance Amplifier
Senses the input voltage and forces an output current proportional (G) to this voltage to flow through the load.
Iout=GVin
These amplifiers behave as voltage amplifiers with respect to the input and current amplifiers with respect to the output. Therefore, using the same reasons as before, to minimize distortion, transconductance amplifiers must be designed with a high input resistance and high output resistance (i.e., Ri & Ro )
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Ideal Transresistance Amplifier Model Input: Ri 0 Output: Ro 0
Transresistance Amplifier
Senses the input current and forces an output voltage proportional (R) to this current to appear across the load.
Vout=RIin
These amplifiers behave as current amplifiers with respect to the input and voltage amplifiers with respect to the output. Therefore, using the same reasons as before, to minimize distortion, transresistance amplifiers must be designed with a low input resistance and low output resistance (i.e., Ri 0 & Ro 0 )
18
Class Impedance Characteristics
Regardless of the amplifier class, a core objective of an amplifier system is to ensure that the output signal is a properly scaled version of the source signal. This can be achieved, in part, through the impedance characteristics of the amplifier.
19
Input Voltage Amplifiers desire to have a high input impedance
Output Voltage Amplifiers desire to have a low output impedance
Input Current Amplifiers desire to have a low input impedance
Output Current Amplifiers desire to have a high output impedance
Biopotential Amplifiers: Ideal Impedance Characteristics
To amplify a voltage potential from the
body it is desired to have an amplifier with a high input resistance
Voltage Amplifier
Transconductance Amplifier
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Power relationship
Biocurrent Amplifiers: Ideal Impedance Characteristics
To amplify a current from the body, it is desired to have an amplifier with a low input resistance
Current Amplifier
Transresistance Amplifier
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Basic Amplifier Concepts
Regardless of the class of amplifier, certain basic concepts hold
Input/Output Relationship
Output=(Gain Factor)*Input
Cascade Effect
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Input/Output Relationship
Positive Gain
Non-inverting
Negative Gain
Inverting
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Operational Amplifiers
The most common integrated circuit used for voltage amplification is the operational amplifier (op-amp).
The op-amp is a high-gain differential amplifier
These are active components, therefore they require a external power supply to operate/function
There are many different versions of op-amps available with varying specifications (LM741, OP07, OP27, LM747, OP37, ….)
Price ranges vary from $0.15 to $10’s
Price mainly depends on performance specifications
e.g. noise, speed, input resistance, etc…
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Common Voltage Amplifier Configurations
Inverting
Non-Inverting
Summing
Differential including Instrumentation Amplifier
Buffer or Voltage Follower
Differentiating
Integrating
Bridge
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Operational Amplifiers
Simplified Schematic
MODEL
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Ideal Voltage Amplifier Model
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Schematic of Common Op-amp
Common IC packages
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Characteristics of Ideal Op-Amps
Infinite gain for the differential input signal (A=)
Zero gain for the common-mode input signal
vo=0 when v1=v2 (no offset voltage)
Infinite input impedance
Zero output impedance
Infinite bandwidth (no frequency response limitations or phase shift)
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Engineering design with op-amps is usually facilitated by treating these devices as “ideal”.
After the initial design, then the “non-ideal” or “real-world” properties are usually considered.
At this point, the “brand” or “model” of op-amp is determined to achieve desired response or the design configuration may need to altered if the “non-ideal” properties cause a problem
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Ideal Op-Amp Circuit Analysis
Based on Two Basic Rules
Rule 1: When the op-amp output is in its linear range, the two input terminals are at the same voltage
Rule 2: No current flows into either input terminal of the op-amp
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Circuit Analysis for Op-amp Networks
Operational amplifiers are almost always used with negative feedback, in which part of the output signal is returned to the negative input in opposition to the source signal.
This provides a stabilized output through an ability to control or set the gain of the configuration.
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Summing Point Constraint
In a negative feedback system, the ideal op-amp output voltage attains the value needed to force the differential input voltage and input current to zero. We call this fact the summing-point constraint.
In other words, if negative feedback is present, you can assume
V+ = V-
I+ = I-=0
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Circuit Analysis Approach for Op-Amp Networks
(1) Verify that negative feedback is present.
(2) Assume that the differential input voltage and the input current of the op amp are forced to zero. (This is the summing-point constraint.)
(3) Apply standard circuit-analysis principles, such as Kirchhoff’s laws and Ohm’s law, to solve for the quantities of interest.
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Standard Voltage Amplifier Configurations
Inverting
Non-Inverting
Summing
Differential including Instrumentation Amplifier
Buffer or Voltage Follower
Differentiating
Integrating
Bridge
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Inverting Amplifier
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Resistors
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Choosing resistors for amplifier design
Prefer Values between 1k to 100k
Why > 1 k?
Helps avoid unnecessarily large currents from flowing in the circuit
Avoids damage to op-amp
Human/Animal subject considerations
Power issues
Why < 100 k?
Avoids unnecessarily small currents
Results in noise issues
Very large resistances lead to instability due to leakage currents over the surface of the resistors and circuit board. Stray pickup of undesired signals
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Be aware of tolerance
Tolerance provides an indication of how the actual resistance value may deviate from the stated value
For example, suppose a 1 k resistor has a gold tolerance band.
This gold band indicates a 5% tolerance
Therefore, although 1 k is the stated value, the actual value may range between 950 to 1050
1000 * 5% = 50 1 k + 50
Tolerance may lead to unexpected behavior in a design
Error in Amplifier Gain
Unusually low CMRR in differential amplifiers
(e.g. imbalance in resistors R1R3 R2R4 leading to a large CMG)
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Tolerance and Amplifier Gain
(a) Design an inverting amplifier with a K= -150 V/V using 5% tolerance resistors
(b) Determine the minimum and maximum possible gain
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Summing Amplifier
A summing amplifier accepts multiple input voltage signals which are amplified in an inverting manner then summed together into a single output.
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Example (Amplifier System Analysis)
Find vo in terms of v1 and v2
Answer: vo=4v1-2v2
vo1
A
B
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Non-Inverting Amplifier
vin(t)
vout(t)
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Voltage Follower or Buffer
vin(t)
vout(t)
What is the purpose of a buffer?
What is the simplest configuration that has the same input-output relationship?
Does a buffer behave in the same manner?
Why is an amplifier needed?
Often in design, we need to cascade multiple configurations to achieve a desired outcome. In certain cases, however, subsequent stages can affect the behavior of prior stages (LOADING EFFECTS)
Used for stage isolation (minimize loading effects)
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High Pass Filter to Inverting Amplifier Configuration Example
Types of Biopotential Measurements
Two Types
Monopolar Recordings – Single site biopotential measurements (i.e., made with 1-channel/single input type amplifiers)
Bipolar Recordings – Dual site differencing biopotential measurements (i.e., made with 2-channel/dual input type amplifiers)
Most common (e.g., utilized for ECG, EMG, EEG, EOG,etc…)
Regardless, whether monopolar or bipolar biopotential measurements, the amplifier used to interface to the body must have an extremely high input impedance to avoid distortion (ideally Rin∞)
Standard Differential Amplifier (Dual-Input)
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Instrumentation Amplifier Bipolar Biopotential Amplifier
Excellent Bio-potential Amp
-High input impedance
-High Gain Possible
-Able to distribute over two stages
-Good Noise Rejection
1
2
3
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Good Biopotential Amp
High input impedance
Good Noise Rejection
High Gain Possible
High Bandwidth
Example: EEG Amplifier Design
Design a bipolar EEG voltage amplifier with a gain of 2000 V/V.
(EEG signals can range from .1mV to 1 mV in amplitude)
For this application, a differential amplifier would be needed (i.e. bipolar measurements) with a suitable input impedance (need high impedance).
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Instrumentation Amplifier
For this design, we need to determine R1, R2, R3, & R4 to realize an overall gain of K=2000 V/V.
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Design
It is best to distribute the overall gain between the two stages
Note: too much gain on stage 1 may cause saturation after the initial stage due to slight DC biopotentials and/or biasing issues inherent in the fabrication of the op-amp
Too much gain on any single stage may also cause problems meeting the 1kΩ to 100kΩ preferred resistive component design requirement
One possible design
K1=50 K2=40
K1*K2=50*40=2000 V/V
50
Design
To design K1=50, choose an R2
e.g. R2=100k
Compute an R1
R1=2R2/(K1-1)=2(100k)/(50-1)
R1=4.08k
To design K2=40, choose an R4
e.g. R4=100k
Compute an R3
R3=R4/K2=100k/40=2.5k
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Realistic Design
Note: resistor tolerance will cause ideal gain to deviate
Another consideration, in regards to gain, is that resistors are only commercially available in certain values
e.g. an R1=4.08k may be difficult to find
Using a 4 k resistor would cause the overall gain to be
K=51*40=2040 V/V
If a more precise gain is desired, a potentiometer (i.e. variable resistor) could be used for R1
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Noise Rejection and Differential Amplifiers
One main problem in the measurement of bio-potentials is the presence of noise (especially, 60 Hz power line noise -see figure next slide)
(1) direct coupling to surface of body
(2) Poorly shielded electrode leads
For a differential amplifier, this noise is present at both inputs.
Signals of this type are known as common-mode signals.
Common to both inputs
These signals can mask the signal of interest, however, through proper design practices common-mode signals can be attenuated considerably.
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Noise Rejection and Differential Amplifiers
The ability of an amplifier to attenuate common-mode signals is known as Common-Mode Rejection Ratio (CMRR)
Units: decibels (dB)
Ideally, we want CMG0
A more accurate expression for vo
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Noise Rejection and Differential Amplifiers Example
Find the minimum CMRR for an ECG amplifier if:
the differential gain is 1000
the desired differential input signal has a peak amplitude of 1 mV
the common-mode signal is a 100 V peak 60Hz sine wave
And it is desired that the output contains a peak common-mode contribution that is 1% or less of the peak output caused by the differential signal
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Noise Rejection and Differential Amplifiers Example
Solution: to compute CMRR we need DMG and CMG
Desired peak ECG signal is: (1mV)(1000)=1V
To meet the stated specification, the common-mode signal must have a peak value less than 1V*1%=0.01 V
Based on this, the CMG=Vocm/Vicm=0.01V/100V=10-4
DMG=1000
CMRR=20*log10(DMG/CMG)=20*log10(1000/10-4)=140dB
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For ECG amplifiers needed CMRR is between 90-130dB
Common Mode Noise & CMRR
ECG (K=1000 V/V) CMRR 65dB
ECG (K=1000 V/V) CMRR 125 dB
What determines the CMRR for a Differential Amplifier?
(1) The inherent CMRR for the model of op-amp being used. 741: 90dB
OP07: 100-120dB
OP27/37: 126dB
(2) imbalance in resistances. For maximum CMRR R1=R3 and R2=R4
slight differences between resistances can affect CMRR significantly)
(3) Frequency of differential and common-mode signal
The model of op-amp determines the maximum attainable CMRR
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CMRR and the Instrumentation Amplifier
For instrumentation amplifiers, the overall CMRR is solely determined by the second stage alone.
CMG for first stage is unity
Vo1
Vo2
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Practical Measurement of CMRR
Measure the Differential mode gain (DMG)
Ground one input and apply a sinusoidal input of known amplitude to the second input
DMG=Vo/Vin
Measure the Common mode gain (CMG)
Apply a high amplitude sinusoid signal of known amplitude to both inputs
CMG=Vo/Vin
Compute CMRR
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Practical Considerations (i.e. non-ideal characteristics) Operational Amplifiers
In practice, op-amps can behave quite differently from ideal under certain operating conditions
Known as Nonlinear effects
Clipping
Related to the physical achievable output voltage for an op-amp based amplifier. In practice, it is bound between an upper and lower limit determined in part by the DC power supply rails.
Slew Rate
Related to the rate of change in output voltage with respect to time. In practice, this is bound by an upper limit.
Bandwidth
Related to the range of frequencies in which the gain of the amplifier will behave as ideal.
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Practical (i.e. non-ideal) Operation Amplifiers
Clipping
The output voltage of a real op-amp is limited to an operational range. This range is dependent on the internal design of the op-amp and the external DC power supply.
When the output voltage tries to exceed these limits, clipping occurs.
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Example: Clipping
Listed in Specification Sheet
For a given op-amp model
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Nonlinear Effects
Slew Rate (SR) Units: Voltage/time
Describes another limitation of practical op-amps where the magnitude of the rate of change of the output voltage is limited.
In other words, SR is how fast an op-amp can change its output voltage
Very important consideration for amplifiers with time varying input signals
If the slew rate is exceeded, the output voltage will be distorted
741: 0.5 V/s OP27: 2.8 V/s OP37: 17 V/s OP07:0.1-0.3 V/s
For sinusoidal signals, the minimal slew rate needed such that no output distortion occurs is:
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Example: Slew rate
--Maximum rate of change is
0.5 V/us for a 741 op-amp
Vs(t)=2.5sin(105t)
Ideal
Actual
(741 op-amp)
If we want to achieve the ideal response for this configuration, what could be
done?
Use a different op-amp (e.g. OP37 SR=17 V/us)
What is the minimum slew rate needed to avoid distortion?
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Example 14.6
Nonlinear Effects
Gain Bandwidth Product (GBWP)
Sometimes referred to unity gain (Gain =1) small signal bandwidth
Determined by the model of the op-amp
For a single op-amp configuration, the GBWP is constant
Allows determination of the range of input frequencies for which the designed gain of the amplifier will behave as ideal
This is referred to as the amplifier bandwidth
A input signal which has a frequency that exceeds the bandwidth will not be amplified to the full extent.
Gain
Frequency
Ideal
Non-Ideal
Units: (Hz)
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Example #1
Determine the bandwidth of a non-inverting amplifier designed for a gain of 50 that is built with a
741 op-amp GBWP = 1 MHz
OP-27 op-amp GBWP = 8MHz
741
Bandwidth=GBWP(741) /50=1MHz/50=20kHz
OP-27
Bandwidth=GBWP(OP-27) /50=8MHz/50=160kHz
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Example #2
A standard differential amplifier is designed using a 741 op-amp (GBWP=1MHz) with an overall gain K=1000 V/V
What is the bandwidth?
Bandwidth=GBWP/1000=1MHz/1000=1KHz
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Example #3
An instrumentation amplifier is designed using 741 op-amps (GBWP=1MHz) with a 1st stage gain of 50 and a 2nd stage gain of 20 for an overall gain K=(50)(20)=1000 V/V
Would the bandwidth=GBWP/1000=1kHz?
NO!
Why?
The gain is distributed across multiple op-amp stages
Stage 1: Bandwidth=GBWP/50=1MHz/50=20KHz
Stage 2: Bandwidth=GBWP/20=1MHz/20=50KHz
Therefore Stage 1 limits the bandwidth of the instrumentation amplifier to 20KHz
However the bandwidth is 20 times larger compared to the standard differential amplifier case seen in Example #2!
This is a benefit of distributing large gains over multiple stages
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Other ‘Op-Amp’ Configurations of Interest
Integrators
Differentiators
Modified Bridge Circuits
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Integrators
Integrators produce an output voltage signal that is proportional to the running time integral of the input voltage signal.
Note: In a running time integral, the upper limit of integration is t .
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Integrator
Purpose: In a general sense, integrators are used to determine the area under the curve, which can be used to estimate signal energy.
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Example
If the input into an integrator is given by:
And R=10kΩ and C=0.1uF, sketch the output. (Assume an ideal op-amp)
Solution:
Output:
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Differentiator
Differentiators produce an output voltage signal proportional to the time derivative of the input voltage signal.
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Differentiator
Purpose: Differentiators are used to determine the rate of change within a signal. (e.g., an application could be to trigger an alarm based on a rate event, such as ventricular tachycardia.)
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Application Example
Suppose a system outputs a voltage signal proportional to velocity (Vv) of a sprinter
How could you design a system to take this voltage signal as an input and output a voltage signal proportional to displacement (x)?
Velocity=dx/dt
Therefore an integrator could be used
How could you design a system to take this velocity voltage as an input and output a voltage proportional to acceleration (a)?
A=d2x/dt2=d(velocity)/dt
Therefore a differentiator could be used
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Bridge Amplifiers
Wheatstone bridge
Provides method for monitoring changes in sensor resistance
To work, this bridge circuit requires a differential measurement to be taken
How would this be achieved in lab?
A digital differential input multimeter would be used
Suppose we did not have a differential input multimeter, of the circuits discussed in class, how could we implement the differencing measurement?
Use a differential amplifier
What about input impedance?
Standard differential versus instrumentation
need instrumentation amp
Drawbacks:
(1) # of op-amps (three)
(2) only linear for small variations in sensor resistance (due to the Wheatstone configuration)
Better circuit is a modified bridge amplifier
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Modified Bridge Amplifier
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Design Example#0
You have been asked to design a system that will be used in an oscillometric based blood pressure monitor. For this, you are provided a pressure sensor whose resistance changes proportional to input pressure (For reference, the sensor specifications state that at 40 mmHg the sensor has a corresponding resistance of 10kΩ and at 160 mmHg the corresponding resistance is 50kΩ.) .
The designed circuit should output a voltage proportional to the pressure within a pneumatic cuff. More specifically, the output voltage should be between 0V to 10V corresponding to cuff pressures between 40 mmHg to 160 mmHg (e.g. 0V corresponds to 40mmHg and 10 V corresponds to 160mmHg).
Summarize Information Given
@P=40mmHg Rs=10kW
@P=160mmHg Rs=50kW
From the problem statement,
we want
0 < Vo < 10
40mmHg < P < 160mmHg
10kΩ < Rs < 50kΩ
Design: Use Modified Bridge Amplifier and Solve for RBAL, R1, and V
Always start with the BALANCE case (i.e., when Vo=0)
This occurs when RBAL=RS; therefore RBAL=10kΩ
When P=40 mmHg, Rs=10kΩ which indicates DR=0.
We need to solve for R1, so now consider when P=160 mmHg. At this pressure, Vo=10 and Rs=50kΩ
We know Rs=RBAL+DR,
therefore, DR=Rs-RBAL=50k-10k=40kΩ
At this point, we need to define our input voltage (V). Assume V=5;
Was this a good assumption?
No, resistances can’t be negative (-)
Ok, assume V again… V= - 5
So we have: V=-5, RBAL=10kΩ, R1=10kΩ
Under this design; 0 < Vo < 10 when 40mmHg < P < 160mmHg
EXAM #3 Material Begins Here
Larger System Designs
It is usually easiest to start from the system output and work back toward the input(s). This is often because it is difficult to anticipate the effects that later stages may have on the overall design.
In system design, there are often many correct design architectures that can achieve the same output; however, some approaches may be more optimal (e.g., require fewer stages).
Design Steps
(1) Layout a block diagram to develop a flow pathway and logic required to realize the design. When possible, try to use standard design configurations and keep in mind associated output/input relationship associated with these configurations (e.g., although differentiator configuration does differentiate an input signal, it will also invert it which needs to be taken account).
(2) When the block diagram is finished, evaluate the system starting with the input(s) to confirm it will achieve the overall output you expect.
(3) After the block diagram is validated, then assemble the schematic by connecting the appropriate standard configurations. (Be sure to use unique variable names for the components to avoid confusion).
(4) Calculate the components values to realize associated amplifier gains based on the details specified in the block diagram.
Design: Example #1
You are working for a medical device company that designs and markets medical instrumentation for hospital intensive care units (ICUs). A team of engineers at your company has extensively researched and developed a new algorithm to alert nurses and physicians that an ICU patient is in distress and in need of immediate medical attention. The new device has input signals that are provided by other medical devices already used in the ICU.
These device include: (1) a respirometer that outputs a voltage signal, vR(t), proportional to respiration rate
(2) a blood pressure monitor that outputs a voltage signal, vBP(t), proportional to
instantaneous blood pressure
(3) a pulse oximeter that outputs a voltage signal, vOX(t), proportional to arterial blood
oxygenation
(4) a ECG system that outputs a voltage signal, vHR(t), proportional to heart rate
Note: vR(t), vBP(t), vOX(t), and vHR(t) are the inputs to the system you are to design.
The developed algorithm takes each of these as inputs and outputs a voltage signal, vA(t), which can be used to trigger an audible alarm. The output voltage signal is governed by the following equation:
Using ideal op-amps, design a system to realize the needed output..