Biomedical Instrumentation I
Chapter 7 Bioelectric Amplifiers from
Introduction to Biomedical Equipment Technology
By Joseph Carr and John Brown
Bioelectric Amplifiers
• Definition: An amplifier is used to process biopotential
– Adjust gain
– DC coupled: needed when signal is very slowly changing or is dc
i.e. O2level may change pressure mmHg per min or hour
– AC coupled: need to overcome electrode offset
– Frequency Response: range that amplifier can work over i.e.
ECG is 0.05 to 100 Hz
• Low Frequency Response or High Frequency Response:
frequencies where gain drops 3 dB below its mid frequency value.
– Gain Types of Amplifiers:
• Low Gain Amplifier: gain between 1 and 10 where unity or 1 is most
common (used for Isolation, buffering and possibly impedance
transformation between signal source and readout device)
• Medium Gain Amplifier: gain between 10 and 1000 used for ECG,
EMG etc.
• High Gain Amplifier: gain greater than 1000 used for EEG
Important Design Parameters
• Noise: want to keep noise as low as possible so keep
thermal and shot noise low by choosing proper
components and shielding wires
• Drift: want to keep drift at a minimum where drift is
spurious changes in output signal caused by change in
operating temperature rather than input signal chagnes
• High Input Impedance: Needed with all three types of
amplifiers because almost all bioelectric signal sources
have high source impedance of 103 to 107 Ώ
– Engineering Rule of thumb is to have input impedance at least 1
order of magnitude higher than source impedance
Operational Amplifier
• Circuit Diagram where power supply is bipolar (positive voltage to
ground and negative voltage to ground)
+V
Inverting
Noninverting
-
Inverting Input Produces an output signal
180o out of phase of the input signal
Output
+
-V
Noninverting Input Produces an output
Signal in phase with the input signal
Inverting and Noninverting input have
Same gain but different phase effects
• +V and –V can be DC or AC (note this circuit has no ground terminal
-> ground connection is between 2 power supplies)
+
+
-
+V
Ground
-V
Notice how –V is negative to ground
• Differential Signals: when E1 ≠ E2 then op amp will see a
differential input of E2 –E1 thus output ~ gain (E2-E1)
+V
E1
E3
E2
-
Output
+
-V
• Common Mode Signals: Signal Voltages are common or the
same to both inputs or E1 = E2 such that the input = 0 so the
output = 0
• Common Mode Rejection Ratio (CMRR): expression of how
close an op amp is to ideal where the common mode signal has no
effect on output terminal voltage
741 Properties of an Ideal
Operational Amplifier
1. Infinite Open Loop (no feedback) voltage
gain Avol = infinity
2. Zero Output impedance (Zo = 0)
3. Infinite Input impedance (Zi= infinity)
4. Infinite Frequency Response
5. Zero noise Contribution
6. Both Inputs follow each other in feedback
circuits
Implications of Ideal Properties on Operational
Amplifiers
• Infinite open loop gain : gain without any feedback; thus
the closed loop characteristics of the circuit are
determined entirely by the properties of the feedback
loop network and are independent of amplifying device
• Output impedance = 0 : ideal voltage source
• Input impedance = infinity : input terminals neither sink
nor source any current thus don’t load any circuit to
which they are connected
• Input tend to follow each other: thus can treat inputs as if
they were the same (if voltage is applied to noninverting
input then we treating inverting input as if it were the
same in other words have same voltage at other input)
Inverting Followers
R2
R1
A
Vinput
+
Voutput
Noninverting Input is grounded thus
inverting input is ground by
definition of property 6 (slide6)
and is called a virtual ground
Kirrchhoff’s Current Law = Σ of all
currents entering = Σ leaving the
summation
Ohm’s Voltage Law = V = IR
Point A = Summing Junction
R1 = Input Resistance
R2 = Feedback Resistance
Vinput
R1
I1
0
R2
Voutput
I2
Transfer Function =>
I1

0 V 

I2 
input
R1
(Voutput  0)
I1  I 2 
Voutput
Vinput
R2
 Vinput
R1
 R2

R1

Voutput
R2
Example of Inverting Amplifier
• What is the gain of the inverting amplifier if R2 =
120KΏ and R1 = 12KΏ
I1

0 V 

120KΏ
I2 
12KΏ
A
Vinput
+
input
R1
(Voutput  0)
I1  I 2 
Voutput
Voutput
Vinput
R2
 Vinput

Voutput
R1
R2
 R 2  120 K


 10
R1
12 K
Note: R1 and R2 determine the magnitude of the Gain and the sign tells if the output
Is in phase (+) or 180o out of phase (-)
Non Inverting Amplifier
R2
R1
A
+
Vinput
Voutput
Noninverting Input is grounded thus
inverting input is ground by
definition of property 6 (slide6)
and is called a virtual ground
Kirrchhoff’s Current Law = Σ of all
currents entering = Σ leaving the
summation
Ohm’s Voltage Law = V = IR
I1
Point A = Summing Junction
R1 = Input Resistance
R2 = Feedback Resistance
0
R1
I1
Vinput
R2
I2
Voutput

V

I2 
input
 0
R1
(Voutput  Vinput )
I1  I 2 
R2
Vinput

V

output
 Vinput 
R1
R2
VoutputR1  Vinput R1  Vinput R 2
VoutputR1  Vinput R1  R 2 
Voutput
Vinput

R1  R 2  1  R 2
R1
R1
Non Inverting Amplifier
Calculate the voltage gain of the following noninverting amplifier when
R2 = 10KΏ and R1 = 2 KΏ

Vinput  0 
I1 
R1
(Voutput  Vinput )
I

R2
2
R2
Vinput Voutput  Vinput 
I1  I 2 

R1
R1
R2
A
VoutputR1  Vinput R1  Vinput R 2
-
Vinput
+
Voutput
VoutputR1  Vinput R1  R 2 
Voutput
Vinput

R1  R 2  1  R 2  1  10 K  6
R1
R1
2 K
Can this circuit ever be an attenuator? (meaning have the gain be less than 1)
Unity Gain Noninverting Followers
• Connect output directly to input using a wire.
Circuit is used for output buffering and
impedance matching between a high source
impedance and a low-impedance input circuit
A
Vinput
+
Voutput  Vinput
Voutput
Voutput
Vinput
1
Multiple Inverting Input Circuits
I1 
Rf
Vinput R1
I2 
R2
I3 
-
A
R3
If 
+
Voutput
0  V 
input
R1
(0  Vinput )
R2
0  Vinput 
V
R3
output
Vinput
R2
I1
0
R3 I2
I3
Rf
If
Voutput
Kirrchhoff’s Current Law
= Σ of all currents
entering = Σ leaving the
summation
 0
Rf
I1  I 2  I 3  I f 
R1
}
Ohm’s Law: I = V/R
 Vinput
R1

 Vinput
R2

  Vinput  Vinput  Vinput  Voutput

 


R2
R3 
Rf
 R1
  Vinput  Vinput  Vinput  Voutput

 


R
1
R
2
R
3
Rf


1
1  Voutput
 1
 Vinput  


Rf
 R1 R 2 R3 
Voutput
1
1 
 1
  Rf  


Vinput
R
1
R
2
R
3


 Vinput
R3

Voutput
Rf
Multiple Non Inverting Input Circuits
I1 
Rf
R1
I2 
R2
R3
If 
Vinput
0
I3 
-
A
+
Voutput
I1
Vinput
R3 I2
I3
input
 0
R1
(Vinput  0)
R2
Vinput  0
V
R3
output
Rf
If
Voutput
}
Ohm’s Law: I = V/R
Kirrchhoff’s Current Law
= Σ of all currents
entering = Σ leaving the
summation
 Vinput 
Rf
I1  I 2  I 3  I f 
R1
R2
V
Vinput
R1

Vinput
R2

Vinput
R3
 Vinput Vinput Vinput  Voutput Vinput

 



R2
R3 
Rf
Rf
 R1
 Vinput Vinput Vinput  Vinput Voutput

 



R
1
R
2
R
3
Rf
Rf


 1
1
1
1  Voutput
 
Vinput  


R
1
R
2
R
3
Rf
Rf


Voutput
 1
1
1
1 

 Rf  


Vinput
 R1 R 2 R3 Rf 

Voutput  Vinput
Rf
Homework Problems
• Read rest of Chapter 7 and begin Ch8
• Problems 1-10
• Problem 6 described on page 143. Where
the input resistances (“R1 or Rin”) must be
10 times larger than the source impedance
looking into the amplifier
• Problem 7 and 8 refer to figure 7-19 not
figure 7-20
Review
• Three key points on how to solve Op Amp
Gain Equations
• Derivation of Gain Equation
• Inverting Op Amps
• Non Inverting Op Amps