The electrocardiogram

Op-amps, instrumentation amps, filters

This is your data

Time (sec)

V (a

rb)

EKG

The EKG is a powerful diagnostic tool. Regularly used by cardiologists.

Disclaimers

• We are not real doctors. Neither are you. Do not try to interpret your EKG.

Safety

Safety

• We will review this in lab, but basically– 100 K resistors between you and breadboard.– Unplug your laptop while collecting data.

Privacy

• Your EKG could be considered private medical information under Federal HIPAA laws.

• If using this data makes you at all uncomfortable – then use one of the instructors as your subject.

The circuit

C 3

1uF

+ 5 V

R 5

1M

+ 2.5 V

DAQ(A0+)

R 6

1k

R 8

2K

R 7

10K

R 1

100K

Arm

ArmC 1

1 uF

+ 2.5 V

U 3 A

L M C 6 48 4

1

3

2

411

O U T

+

-

V+

V-

Foot

+ 2.5 V

U 1 A

A D 6 2 3 A N

4

3

6

2

5

81

7-V

S

+I N

O U TP U T

-I N

R E F

+R G-R G

+VS

R 2

100K C 2

0.1 uF

R 3

10K

+ 5 V

R 4

100K

AD623 – Instrumentation amplifier

The Op-Amp

Dynamic model of an op-amp

out

Subtract Saturation

1s

Integrator

1e6

Gainin -

in +

out

in +

in -

The op-amp follower

3

21

411

-

+

U1A

TL084

+15V

-15V

OutIn

Dynamic model of the follower

out

Subtract Saturation

1s

Integrator

1e6

Gainin -

in +

out

in +

in -

Simplification for op-amp circuit analysis

• If the op-amp is wired with negative feedback, and - • Changes are “slow” (less than ~100 kHz), then –

– Assume no current flow through into the inputs. – The two input voltages are equal.

What does this circuit do?

What does this circuit do?

What does this circuit do?

The instrumentation amplifier

Wikipedia

AD623

Data sheet

Let’s look at the filters now…

C

R21

V in p u t

Vout

Impedance (Z)General form of Ohm’s Law V = I Z

CjjCZ

ejCeti

etv

dttdv

Cti

tjtj

tj

1

)(

)( :response AC

)()(

VV

IV

VI

V

Example: capacitor

RC circuit (via impedances)

C

R21

V in p u t

RZ 1

CjZ

1

2

21

2ZZ

ZVV inout

Voltage divider

Vout

Do you remember how to plot Vout/Vin ?

w = logspace(0,4,100);j = sqrt(-1);R = 10e3;C = 1e-6;

Z1 = R;Z2 = 1./(j*w*C);

loglog(w,abs(Z2./(Z1+Z2)));

Resistors in series and parallel

Impedances in series:

Impedances in parallel:

21 ZZZ

21

111

ZZ

Z

Can you derive these relationships?

What about two filters in series?

C 3

1uF

+ 5 V

R 5

1M

+ 2.5 V

DAQ(A0+)

R 6

1k

R 8

2K

R 7

10K

R 1

100K

Arm

ArmC 1

1 uF

+ 2.5 V

U 3 A

L M C 6 4 8 4

1

3

2

411

O U T

+

-

V+

V-

Foot

+ 2.5 V

U 1 A

A D 6 2 3 A N

4

3

6

2

5

81

7-V

S

+I N

O U TP U T

-I N

R E F

+R G-R G

+VS

R 2

100K C 2

0.1 uF

R 3

10K

+ 5 V

R 4

100KVin Vout

The circuit

C 3

1uF

+ 5 V

R 5

1M

+ 2.5 V

DAQ(A0+)

R 6

1k

R 8

2K

R 7

10K

R 1

100K

Arm

ArmC 1

1 uF

+ 2.5 V

U 3 A

L M C 6 48 4

1

3

2

411

O U T

+

-

V+

V-

Foot

+ 2.5 V

U 1 A

A D 6 2 3 A N

4

3

6

2

5

81

7-V

S

+I N

O U TP U T

-I N

R E F

+R G-R G

+VS

R 2

100K C 2

0.1 uF

R 3

10K

+ 5 V

R 4

100K