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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.

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Presentation on theme: "Biomedical Instrumentation I Chapter 7 Bioelectric Amplifiers from Introduction to Biomedical Equipment Technology By Joseph Carr and John Brown Part 2."— Presentation transcript:

1 Biomedical Instrumentation I Chapter 7 Bioelectric Amplifiers from Introduction to Biomedical Equipment Technology By Joseph Carr and John Brown Part 2

2 Differential Amplifiers Infinite Input impedance thus current passes from R3 to R4 and from R1 to R2 R2 - + Voutput R1 A R3 Voutput I4I4 I3I3 R1R2 I1I1 I2I2 R4 Vinput E1 E2 R4 R3 E1 E2 Book Assumes: Vinput = E2-E1 And R1 =R3 and R2=R4 A A

3 Advantages of Differential Amplifier In differential mode you can cancel noise common to both input signals R2 - + Voutput R1 A R3 R4 E2 E1 1V 3V 2V

4 Instrumentation Amplifier Give you high gain and high-input impedance. Composed of 2 amplifiers in noninverting format and a 3 rd amplifier as a differential amplifier + - R2 Vinput E1 R5 - + R4 R7 R6 + - E2 R1 E2 R3 Voutpt

5 Derivation of Gain for Instrumentation Amplifier step E2 R1 E2 R3 E1 E3 R1R3 I1I1 I2I2 E1E2 I1I1 I2I2

6 Derivation of Gain for Instrumentation Amplifier step R2 E1 R1 E2 E4 R1R2 I1I1 I2I2 E2 E1 I1I1 I2I2 E4

7 Derivation of Gain for Instrumentation Amplifier step 3 R5 - + R4 R7 R6 V output E4 E3 V output I7I7 I6I6 R4R5 I4I4 I5I5 E4 R7 R6 E3 Book Assumes R4 =R6 and R5=R7 0 I4I4 I5I5 I6I6 I7I7 A A

8 Derivation of Gain for Instrumentation Amplifier step 4 Book Assumes R3 =R2 Step1 Step2 Step3

9 Example of Instrumentation Amplifier Find the gain of the previous instrumentation amplifier if R2 = 10K  ; R1=500  ; R4 = 10K  ; R5 = 100K 

10 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?

11 Solution 1

12 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 V input1 is 1V. V out1 = -2V. –Draw Schematic –Find R2 & R1.

13 Solution R2 Vinput E1 R5 - + R4 R7 R6 + - E2 R1 E2 R3 Voutpt Vout1 Vin1 IR2

14 Solution 2 cont

15 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

16 Example of a Low pass Filter 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? - + Voutput Vinput R A C R Voutput 0 ICIC IRIR Vinput Cf

17 Example of a Low pass Filter - + Voutput Vinput R A C R Voutput 0 ICIC IRIR Vinput Cf

18 Low Pass Active Filters = Integrator Attenuates High frequency where cutoff frequency is  =RfCf - + Voutput Vinput Ri A Rf Cf Rf Ri Voutput 0 I Rf Ii Vinput Cf I Cf

19 High Pass Active Filters=Differentiator 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. - + Voutput Vinput A Rf Ci Rf Voutput 0 I Rf Ii Vinput Cf

20 High Pass Active Filters - + Voutput Vinput A Rf Ci Rf Voutput 0 I Rf Ii Vinput Cf

21 High Pass Active Filters Attenuates High frequency where cutoff frequency is 1/(2  ) =1/ 2  RiCi - + Voutput Vinput Ri A Rf Ci Rf Ri Voutput 0 I Rf Ii Vinput Cf

22 Band Pass Active Filters Attenuates High frequency and low frequencies where cutoff frequency is  =RfCf - + Voutput Vinput Ri A Rf Cf Rf Ri Voutput 0 I Rf Ii Vinput Cf I Cf Ci


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