Biomedical Instrumentation I

Chapter 7 Bioelectric Amplifiers fromIntroduction to Biomedical Equipment Technology

By Joseph Carr and John BrownPart 2

Differential Amplifiers• Infinite Input impedance thus current passes from

R3 to R4 and from R1 to R2 R2

0

02

0

4

01

2

3

22

1

1

43

21

4

3

2

1

II

IIR

A

R

AI

R

AE

R

AEI

R

AVI

R

AEI

output

-+ Voutput

R1 A

R3

Voutput

I4I3

R1 R2

I1 I2

R4

Vinput

E1

E2

R4R3

E1

E2

Book Assumes: Vinput = E2-E1 And R1 =R3 and R2=R4

1

2

12

)12(21

12122

1222

1222

2

0

1

2

02

0

1

2

12121

11221

21

1

021

1

R

R

EE

V

EERRV

RVRERE

ARARRE

ARARRE

R

A

R

AER

A

R

AE

ARARRVRE

RVARARRE

R

AV

R

AE

R

AV

R

AE

output

output

output

output

output

output

output

A

A

Advantages of Differential Amplifier

• In differential mode you can cancel noise common to both input signals

R2

-+ Voutput

R1 A

R3

R4E2

E11V

3V

2V

Instrumentation Amplifier

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

+

-R2

Vinput

E1

E1R5

-+

R4

R7R6

+

-E2

R1

E2

R3 Voutpt

4

51

1

22

75;64;32

R

R

R

R

Vinput

Voutput

RRRRRR

Derivation of Gain for Instrumentation Amplifier step 1

+

-E2

R1

E2

R3

E1

E3

E3R1 R3

I1 I2

E1 E2

I1

I2

11

31

1

323

21

31

1

323

23132131

312123133

32

1

21

21

ER

R

R

REE

ER

RE

R

REE

ERERERER

ERERERERR

EE

R

EE

II

Derivation of Gain for Instrumentation Amplifier step 2

+

-R2

E1

E1

R1

E2

E4

R1 R2

I1 I2

E2 E1

I1I2

21

2

1

2114

11

22

1

214

12221141

411112222

41

1

12

21

ER

R

R

REE

ER

RE

R

REE

ERERERER

ERERERERR

EE

R

EE

II

E4

Derivation of Gain for Instrumentation Amplifier step 3

R5

-+

R4

R7R6

Voutput

E4

E3

Voutput

I7I6

R4 R5

I4 I5

E4

R7R6E3

Book Assumes R4 =R6 and R5=R7

0

I4I5

I6 I7

76

54

7

6

5

4

5

0

7

04

3

6

35

4

4

II

IIR

A

R

AI

R

AE

R

AEI

R

VAI

R

AEI

output

4

543

4)43(5

45453

4553

45535

0

4

3

45454

4455454

4

R

REEV

RVEER

RVRERE

ARARRE

ARARRER

A

R

AE

ARARRVRE

RVARARRER

VA

R

AE

output

output

output

output

output

output

A

A

Derivation of Gain for Instrumentation Amplifier step 4

4

543R

REEVoutput

21

2

1

2114 E

R

R

R

REE

11

31

1

323 E

R

R

R

REE

Book Assumes R3 =R2

4

51

1

2212

4

5

1

21

1

21

1

21

1

22

4

5

1

22

1

211

1

211

1

22

11

21

1

223

R

R

R

REEV

R

R

R

R

R

RE

R

R

R

REV

R

R

R

RE

R

RE

R

RE

R

REV

ER

R

R

REE

output

output

output

Step1

Step2

Step3

Example of InstrumentationAmplifier

• Find the gain of the previous instrumentation amplifier if R2 = 10K; R1=500; R4 = 10K ; R5 = 100K

41010

1001

5.0

10*2

12

4

51

1

2212

K

K

K

K

EE

V

R

R

R

REEV

output

output

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?

Solution 1

VVVVVV

V

VV

V

out

outout

10)12(*1012

1012

KKR

K

R

R

R

V

V

f

f

in

f

in

out

404*10

410

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+

-R2

Vinput

E1

E1R5

-+

R4

R7R6

+

-E2

R1

E2

R3 Voutpt

Vout1

Vin1

IR2

7504

3

4

)2(12

2

11

mA

V

mA

VV

I

VVR

R

outin

Solution 2 cont

KRR

KR

K

K

K

R

R

R

R

R

Vinput

Voutput

RRRRRR

5.111

5.11

11

5.12

10

20

1

101

1

)750(220

204

51

1

22

75;64;32

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 • 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 0.2F capacitor. Find the output voltage after 1 second if the input voltage is a constant 0.5V?

VV

dtF

V

VdtVRC

V

output

output

ic

t

inputoutput

5.205.02.0

1

0)5.0(10*2*10

1

1

1

076

0

-+ Voutput

Vinput

RA

C

R Voutput0

ICIR

VinputCf

Example of a Low pass Filter

-+ Voutput

Vinput

RA

C

R Voutput0

ICIR

VinputCf

t

inputoutput

inputoutput

R

inputR

output

VicdttVinputRC

tVoutput

CRjjVinput

jVoutput

R

jVCjjV

R

jV

Cj

jV

IIcR

jVI

Cj

jVIc

0

)(1

)(

1

)(

)(

1

)(0

1

0)(

)(0

1

0)(

Low Pass Active Filters = IntegratorAttenuates High frequency where

cutoff frequency is =RfCf

-+ Voutput

Vinput

RiA

Rf

Cf

RfRi Voutput0

IRfIi

Vinput

Cf

ICf

1

1

11

1

00

1

0

0

0

1

0

CRfj

Rf

RiV

V

Ri

V

Rf

CRfjV

Rf

CjV

Ri

V

Rf

V

Cj

V

IiIRfIcfRi

VIi

Rf

VIRf

Cj

VIcf

input

output

inputoutputoutput

inputoutputoutput

input

output

output

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 =0.5F and Vin is a constant slope of 400 V/s.

-+ Voutput

Vinput

A

Rf

Ci

Rf Voutput0

IRfIi

VinputCf

VV

SVFV

dt

VdRfCiV

output

output

inputoutput

20

40010*510

)(

75

High Pass Active Filters

-+ Voutput

Vinput

A

Rf

Ci

Rf Voutput0

IRfIi

VinputCf

CRfj

Cj

Rf

jV

jV

Cj

jV

Rf

jV

IiI

Cj

jVIi

Rf

jVI

input

output

inputoutput

Rf

input

outputRf

1

1

00

1

0

0

High Pass Active FiltersAttenuates High frequency where

cutoff frequency is 1/(2) =1/ 2RiCi

-+ Voutput

Vinput

RiA

Rf

Ci

RfRi Voutput0

IRfIi

VinputCf

CjRi

Rf

V

V

CjRi

V

Rf

V

IiI

CjRi

VIi

Rf

VI

input

output

inputoutput

Rf

input

output

fR

1

1

00

1

0

0

Band Pass Active FiltersAttenuates High frequency

and low frequencies where cutoff frequency is =RfCf

-+ Voutput

Vinput

RiA

Rf

Cf

RfRi Voutput0

IRfIi

Vinput

Cf

ICf

11

111

11

1

1

00

1

0

1

0

0

1

0

CfRfj

Rf

CiRij

Cij

V

V

CiRij

CijV

Cij

CiRij

V

CijRi

V

Rf

CfRfjV

Rf

CfjV

CijRi

V

Rf

V

Cfj

V

IiII

CijRi

VIi

Rf

VI

Cfj

VI

input

output

inputinputinput

outputoutput

inputoutputoutput

Rfcf

input

outputRf

outputcf

Ci

Ci