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Biomedical Instrumentation I. Chapter 7 Bioelectric Amplifiers from Introduction to Biomedical Equipment Technology By Joseph Carr and John Brown Part 2. Differential Amplifiers. R2. Infinite Input impedance thus current passes from R3 to R4 and from R1 to R2 . R1. -. E1. A. Vinput.

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biomedical instrumentation i

Biomedical Instrumentation I

Chapter 7 Bioelectric Amplifiers from

Introduction to Biomedical Equipment Technology

By Joseph Carr and John Brown

Part 2

differential amplifiers
Differential Amplifiers

R2

  • Infinite Input impedance thus current passes from R3 to R4 and from R1 to R2

R1

-

E1

A

Vinput

R3

+

Voutput

E2

R4

Book Assumes: Vinput = E2-E1

And R1 =R3 and R2=R4

R1

R2

A

E1

I1

I2

Voutput

E2

A

R4

R3

I4

I3

advantages of differential amplifier
Advantages of Differential Amplifier
  • In differential mode you can cancel noise common to both input signals

R2

1V

R1

-

E1

A

2V

R3

+

3V

Voutput

E2

R4

instrumentation amplifier
Instrumentation Amplifier

E1

+

R5

R4

  • Give you high gain and high-input impedance.
  • Composed of 2 amplifiers in noninverting format and a 3rd amplifier as a differential amplifier

-

E1

-

R2

+

R1

Vinput

Voutpt

R3

R6

R7

E2

-

+

E2

derivation of gain for instrumentation amplifier step 3

R5

R4

E4

I5

-

I4

+

I6

Voutput

I7

E3

R6

R7

Book Assumes R4 =R6 and R5=R7

R4

R5

E4

Voutput

I4

I5

E3

R7

0

R6

I7

I6

Derivation of Gain for Instrumentation Amplifier step 3

A

A

example of instrumentation amplifier
Example of InstrumentationAmplifier
  • Find the gain of the previous instrumentation amplifier if R2 = 10K; R1=500; R4 = 10K ; R5 = 100K 
problem 1
Problem 1
  • Design a differential amplifier where the feedback resistors are equal and the input resistors are equal. The gain should be equal to 10. One input voltage is 1 V and the second input voltage is 2 V. What is the output voltage?
  • If the input resistance is 4 K what is the feedback resistance?
problem 2
Problem 2
  • An instrumentation amplifier has a gain of 20. Using the schematic discussed earlier in the lecture, R5 = R7; R4=R6; R2 = R3.
  • If R5 = 10K and R4 = 1K. The current across R2 is 4 mA and Vinput1 is 1V. Vout1 = -2V.
    • Draw Schematic
    • Find R2 & R1.
solution 2
Solution 2

E1

+

R5

R4

Vout1

-

E1

-

R2

Vin1

IR2

+

R1

Vinput

Voutpt

R3

R6

R7

E2

-

+

E2

review for exam 1
Review for Exam 1
  • Review all Homework Problems
  • Review Wheatstone Bridge Lab & Amplifier Lab
  • Review Studio exercises (precision & accuracy and aliasing exercises)
  • Bring Calculators
  • Closed book
  • Equation sheet given previously will be given out at exam
example of a low pass filter
Example of a Low pass Filter

C

  • Vout = output potential in volts(v)
  • Vinput = input potential in volts(v)
  • R = input resistance
  • C =feedback capacitance
  • T = Time (sec)
  • Vic = initial conditions present at integrator output at t =0

Analog Integrator using a 1M resistor and a 0.2F capacitor. Find the output voltage after 1 second if the input voltage is a constant 0.5V?

R

-

A

Vinput

+

Voutput

R

Cf

Voutput

Vinput

0

IR

IC

example of a low pass filter1
Example of a Low pass Filter

C

R

-

A

Vinput

+

Voutput

R

Cf

Voutput

Vinput

0

IR

IC

low pass active filters integrator
Low Pass Active Filters = Integrator

Cf

Rf

Attenuates High frequency where cutoff frequency is =RfCf

Ri

-

A

Vinput

+

Voutput

Cf

ICf

Ri

Voutput

Rf

Vinput

0

Ii

IRf

high pass active filters differentiator
High Pass Active Filters=Differentiator

Rf

Voutput = differentiator output voltage (v)

Vinput = input potential in volts (v)

Rf = feedback resistor ohms ()

Ci = input capacitance farads (F)

Find the output voltage produced by an op-amp differentiator when Rf = 100K and C =0.5F and Vin is a constant slope of 400 V/s.

Ci

-

A

Vinput

+

Voutput

Cf

Voutput

Rf

Vinput

0

Ii

IRf

high pass active filters
High Pass Active Filters

Rf

Ci

-

A

Vinput

+

Voutput

Cf

Voutput

Rf

Vinput

0

Ii

IRf

high pass active filters1
High Pass Active Filters

Rf

Attenuates High frequency where cutoff frequency is 1/(2)

=1/ 2RiCi

Ci

Ri

-

A

Vinput

+

Voutput

Ri

Cf

Voutput

Rf

Vinput

0

Ii

IRf

band pass active filters
Band Pass Active Filters

Cf

Rf

Attenuates High frequency and low frequencies where cutoff frequency is =RfCf

Ci

Ri

-

A

Vinput

+

Voutput

Cf

ICf

Ri

Ci

Voutput

Rf

Vinput

0

Ii

IRf