Biomedical Instrumentation I
Chapter 7 Bioelectric Amplifiers from
Introduction to Biomedical Equipment Technology
By Joseph Carr and John Brown
Part 2
Differential Amplifiers
•
R2
R1 A
E1
Vinput
Infinite Input impedance thus current passes from
R3 to R4 and from R1 to R2
-
R3
+
E2
Voutput
R4
Book Assumes: Vinput = E2-E1
And R1 =R3 and R2=R4
R1
E1
I1
E2
I3  I 4  0
I2
R3
I3
R2
A
Voutput
A
R4
I4
E1  A
I1 
R1
Voutput  A
I2 
R2
E2  A E2  A
I3 

R3
R1
0 A 0 A
I4 

R4
R2
I1  I 2  0
E1  A Voutput  A

0
R1
R2
 Voutput  A 
E1  A

 
R1
 R2 
E1R 2  AR2  AR1  Voutput R1
E1R 2  Voutput R1  AR 2  AR1
E2  A 0  A

0
R1
R2
E2  A
 0  A
 

R1
R
2


E 2 R 2  AR 2  AR1
E 2 R 2  AR2  AR1
E 2 R 2  E1R 2  Voutput R1
Voutput R1  R 2( E 2  E1)
Voutput
E 2  E1

R2
R1
Advantages of Differential
Amplifier
• In differential mode you can cancel noise
common to both input signals
R2
1V
E1
R1 A
-
2V
R3
3V
E2
+
Voutput
R4
Instrumentation Amplifier
E1
+
E1
-
-
R2
R1
Vinput
+
R3
R6
E2
E2
•
•
R5
R4
Voutpt
R7
-
R 2  R3; R 4  R 6; R5  R7
+
Voutput  2 R 2  R5 

 1

Vinput  R1
 R 4 
Give you high gain and high-input impedance.
Composed of 2 amplifiers in noninverting format and a 3rd amplifier as a
differential amplifier
Derivation of Gain for Instrumentation
Amplifier step 1
E1
R1
I1
R3
E3
I2
E2
+
E2
R1
E1
I1
R3
E2
I2
E3
I1  I 2
E1  E 2 E 2  E 3

R1
R3
R3E1  R3E 2  R1E 2  R1E 3
R1E 3  R1E 2  R3E1  R3E 2
R3
R3
E3  E 2 
E1 
E2
R1
R1
 R3  R3
E 3  E 2
 1 
E1
 R1  R1
Derivation of Gain for Instrumentation
Amplifier step 2
E1
+
E4
-
E1
R2
I1
I2
R1
E2
R1
E2
I1
R2
E1
I2
E4
I1  I 2
E 2  E1 E1  E 4

R1
R2
R 2 E 2  R 2 E1  R1E1  R1E 4
R1E 4  R1E1  R 2 E 2  R 2 E1
 R2 
 R2 
E 4  E1  
E 2  
 E1
 R1 
 R1 
 R2   R2 
E 4  E11 

E 2
R1   R1 

Derivation of Gain for Instrumentation
Amplifier step 3
E4
R5
R4
-
I4
+
I6
E3
I4 
I5
I5 
I7
Voutput
R6
R7
Book Assumes R4 =R6 and R5=R7
R4
E4
I4
E3
R5
A
Voutput
I5
R6
I6
A
R7
I7
0
E4  A
R4
A  Voutput
R5
E3  A E3  A
I6 

R6
R4
A0 A0
I7 

R7
R5
I 4  I5
I6  I7
E 4  A A  Voutput

R4
R5
E 4 R5  AR5  AR4  VoutputR 4
E 4 R5  VoutputR 4  AR5  AR4
E3  A A  0

R4
R5
E 3R5  AR5  AR4
E 3R5  AR5  AR4
E 3R5  E 4 R5  VoutputR 4
R5( E 3  E 4)  VoutputR 4
 R5 
Voutput  E 3  E 4 
 R4 
Derivation of Gain for Instrumentation
Amplifier step 4
R2   R2 

E 4  E11 

E 2
R1   R1 

Step1
 R3  R3
E 3  E 2
 1 
E1
 R1  R1
Step2
 R5 
Voutput  E 3  E 4 
 R4 
Step3
 R2   R2 
Book Assumes R3 =R2
E 3  E 2
 1  
 E1
 R1   R1 
  R 2 
 R2     R2 
 R 2   R5 
Voutput   E 2
 1  E1
    E11 
  E 2
 

R
1
R
1
R
1
R
1
R
4


  


 

 
  R2
R2 
R 2  R5 
 R2
Voutput   E 2
1
1
  E1


R
1
R
1
R
1
R
1
R
4








 2 R 2  R5 
Voutput  E 2  E1
 1

R
1
R
4



Example of Instrumentation
Amplifier
• Find the gain of the previous
instrumentation amplifier if R2 = 10K;
R1=500; R4 = 10K ; R5 = 100K 
 2 R 2  R5 
Voutput  E 2  E1
 1

 R 4 
 R1
Voutput
 2 *10 K  100 K 

 1
  410
E 2  E1  0.5K
 10 K 
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
Vout
Vout
 10 
V 2  V1
2V  1V
Vout  10 * (2V  1V )  10V
Rf
Vout R f

 10 
Vin
Rin
4 K
R f  10 * 4 K  40 K
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.
E1
Solution 2
+
E1
-
-
R2
Vin1
R1
Vinput
IR2
R3
+
R6
E2
E2
R5
Vout1 R4
Voutpt
R7
+
Vin1  Vout1 1V  (2V )
3V
R2 


750
I R2
4mA
4mA
Solution 2 cont
R 2  R3; R 4  R6; R5  R7
Voutput  2 R 2  R5 

 1
  20
Vinput  R1
 R 4 
 2(750)  10 K 
20  
 1

 R1
 1K 
20
1.5 K
2
1
10
R1
1.5 K
1
R1
R1  1.5 K
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
•
•
•
•
•
•
C
R
-
A
Vinput
+
Voutput
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
Voutput
Vinput
R
IR
0
Cf
Voutput
1 t

Vinput dt  Vic

0
RC
1
1
 6
(0.5)dt  0
7

10  * 2 *10 F 0
Voutput 
IC
1
0.05  2.5V
0.2
Example of a Low pass Filter
C
R
-
A
Vinput
Vinput
+
R
0
Voutput
Cf
Voutput
Voutput ( j )  0
Ic 
1
j C
0  Vinput ( j )
IR 
R
Ic  I R
Voutput ( j )  0 0  Vinput ( j )

1
R
j C
Vinput  j 
 j C 
Voutput  j 
  
R
 1 
Voutput ( j )
1

Vinput ( j )
Rj C
t
IR
IC
1
Voutput(t )  
Vinput (t )dt  Vic

RC 0
Low Pass Active Filters = Integrator
Cf
Attenuates High frequency where
cutoff frequency is =RfCf
Icf 
Rf
Ri
IRf 
-
A
Vinput
Ii 
+
Voutput
Cf
Vinput
Ri
Ii
0
Rf
IRf
ICf
Voutput
Voutput  0
1
jC
Voutput  0
Rf
0  Vinput
Ri
Icf  IRf  Ii
Voutput  0 Voutput  0 0  Vinput


1
Rf
Ri
j C
Vinput
 j C
 jCRf  1 
1 
  Voutput 
  
Voutput 

Rf 
Rf
Ri
 1


Voutput  1 

Rf



Vinput
Ri  jCRf  1 
High Pass Active Filters=Differentiator
Rf
Ci
-
A
Vinput
+
Voutput
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 opamp differentiator when Rf = 100K and C
=0.5F and Vin is a constant slope of 400
V/s.
Voutput   RfCi
Vinput
Cf
0
Ii
Rf
IRf
Voutput

d (Vinput )

dt

Voutput   10 5  5 *10 7 F 400V
Voutput  20V
S

High Pass Active Filters
Rf
Ci
I Rf 
Ii 
-
A
Vinput
+
Voutput
I Rf
Voutput  j   0
Rf
0  Vinput  j 
1
jC
 Ii
Voutput  j   0
Rf
Voutput  j 
Vinput  j 
Vinput
Cf
0
Ii
Rf
IRf
Voutput


0  Vinput  j 
1
j C
 Rf
  RfjC
1
j C
High Pass Active Filters
Rf
Ci
Ri
-
A
Vinput
+
Voutput
Attenuates High frequency where
cutoff frequency is 1/(2)
=1/ 2RiCi
Voutput  0
IR f 
Rf
0  Vinput
Ii 
1
Ri 
j C
I Rf  Ii
Voutput  0
Rf
Voutput
Vinput
Cf
Ri
Ii
0
Rf
IRf
Voutput
Vinput

0  Vinput

1
Ri 
j C
 Rf
Ri 
1
j C
Band Pass Active Filters
Cf
I cf 
Rf
Ci
Ri
I Rf 
-
A
Vinput
+
Voutput
Cf
Ri
Vinput Ci
Ii
0
Rf
IRf
ICf
Voutput
Voutput  0
1
jCf
Voutput  0
Attenuates High frequency
and low frequencies
where cutoff frequency
is =RfCf
Rf
0  Vinput
Ii 
1
Ri 
jCi
I cf  I Rf  Ii
Voutput  0 Voutput  0
0  Vinput


1
1
Rf
Ri 
jCf
jCi
 jCf
 jCfRf  1 
1 
  Voutput 
 
Voutput 

1
Rf
Rf




Vinput
 Vinput
 Vinput  jCi 



1
Rij Ci  1
Rij Ci  1
Ri 
jCi
jCi
Voutput
Vinput


 jCi 
Rf


Rij Ci  1  jCfRf  1 