Biomedical Instrumentation I Chapter 7 Bioelectric Amplifiers from Introduction to Biomedical...
-
Upload
janice-fairburn -
Category
Documents
-
view
288 -
download
4
Transcript of Biomedical Instrumentation I Chapter 7 Bioelectric Amplifiers from Introduction to Biomedical...
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?
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