ENTC 4350
BIOMEDICAL
INSTRUMENTATION I
BASIC DIFFERENTIAL
AMPLIFIER
Introduction

The differential amplifier can measure as
well as amplify small signals that are
buried in much larger signals.

There are two input terminals, labeled
() input, and (+) input.
Superposition

If E1 is replaced by a short circuit, E2
sees an inverting amplifier with a gain of
m.
• Therefore, the output voltage due to E2 is
 mE2.
 mE2

Now let E2 be short-circuited:
• E1 divides between R and mR to apply a
voltage of E1m/ (1+ m) at the op amp’s (+)
input.

This divided voltage sees a noninverting
amplifier with a gain of (m + 1).
• The output voltage due to E1 is the divided
voltage:
• E1m/(1 + m) times the noninverting amplifier gain,
(1 + m), which yields mE1.
mE1

Therefore, E1 is amplified at the output
by the multiplier m to mE1.
• When both E1 and E2 are present at the (+)
and () inputs, respectively.
• Vo is mE1  mE2.


The output voltage of the differential
amplifier, Vo, is proportional to the
difference in voltage applied to the (+)
and () inputs.
Multiplier m is called the differential gain
and is set by the resistor ratios.

When E1 = E2 the output voltage is 0.
• To put it another way, when a common (same)
voltage is applied to the input terminals, Vo =
0.
Lab 6_Differential Amplifier

The gain of the amplifier below can be
determined using the Superposition
Principle.
22 kW
Rf
'
'
2.2 kW
─
Ri
RS
+
4.7 kW
RD
VOUT
Inverting Amplifier

Forcing V2 to 0 develops an inverting
amplifier with an output, VOUT of:
VOUT 1  V1
Rf
Ri
 V1
22kW
 10V1
2.2kW
22 kW
Rf
V1
'
'
2.2 kW
─
Ri
RS
+
4.7 kW
RD
VOUT
VOUT 1  V1
Rf
Ri
Non-inverting Amplifier

Forcing V1 to 0 develops a non-inverting
amplifier.
22 kW
Rf
'
'
V2
2.2 kW
─
Ri
RS
+
4.7 kW
RD
VOUT

Applying Thevenin’s Theorem:
Vopen  V2
RD
 V2
RS  RD
RTH 
RS RD
RS  RD
22 kW
Rf
'
'
V2
2.2 kW
─
Ri
RS
+
4.7 kW
RD
VOUT

The output of the non-inverting amplifier
is:
Rf 


VOUT 2  V2  1 
R
i 

22 kW
Rf
'
'
V2
RD
RS  RD
2.2 kW
Ri
RTh
─
+
Rf 


VOUT 2  V2  1 
VOUT
R
i 


The total output is the sum:
Rf  
Rf 

    V1

VOUT 2  VOUT 1  V2   
Ri  
Ri 


To balance the circuit, we set the
coefficients to add to zero.
 
Rf
Ri

Rf
Ri
 
Rf
Ri

Rf
Ri
 Ri R f  R f
 
  
 Ri Ri  Ri
 Ri  R f  R f
 
 
 Ri  Ri
R f  Ri


Ri  Ri  R f

Rf

 Ri  R f

VOUT
 Rf
Rf
Rf  
Rf 
    V1

 V2 


 Ri  R f Ri  R f Ri 
Ri 

 
2
 Ri R f
 
R
Rf 
f



 V2

   V1
 Ri Ri  R f
Ri Ri  R f  
Ri 




 




R
 
R
f
i
   V1 
 V2 



Ri  R f 
  Ri  R f

  Rf 
 R f   Ri  R f 
  V1   
 V2 
V2  V1 
 




  Ri 
R
R

R
f 
 i   i

 Rf
 
 Ri

 


So the balanced condition yields
VOUT
 Rf
 
 Ri

V2  V1 

• and the differential gain Ad is
 Rf
AD  
 Ri




Common-mode rejection of 60 cycle
power line interference in medical
instrumentation which measures
difference potentials on the body is a
fundamental problem.
• Power-line interference may exceed the level
of the signal being measured.
• This bad news is often cancelled by the fact that
the interfacing signal appears equally intense at
both input terminals of the diff amp, and is
therefore called a common-mode signal.

If the diff amp is not perfectly balanced, as is
always the case in the real world, then the
common-mode signal input will cause an
output signal that then constitutes interference
with the desired amplified signal.
•
Since one of the functions of the diff amp is to reject
the common-mode signal, we define a figure of merit,
the common-mode rejection ratio (CMRR), which
measures how well the rejection occurs.

The common-mode rejection ratio
CMRR is defined as the magnitude of
the ratio of the differential voltage gain
Ad to the common-mode voltage gain Ac.

Ad equals VOUT divided by V1 when node
2 is grounded, and V1 is applied to node
1.
• Also, AC equals VOUT divided by V1 when
node 1 is connected to node 2, and V1 is
applied again.
CMRR 
VOUT when V2 is grounded
VOUT when V2  V1

In practice the CMRR is measured in the
following steps:
1. Ground V2, and apply a voltage V1 to the upper
terminal.
2. Measure the resulting VOUT.
3. Lift V2 from ground and short the two input leads,
then apply the same value of V1.
4. Measure the resulting VOUT.
5. To compute CMRR, divide the results of step 2 by the
result of step 4, and take the magnitude.

The CMRR is a voltage ratio, and
therefore in decibel units we may define
CMRRdb as
CMRRdb  20logCMRR
Common-Mode Voltage

The simplest way to apply equal
voltages is to wire inputs together and
connect them to the voltage source.

For such a connection, the input voltage
is called the common-mode input
voltage, ECM.

Now Vo will be 0 if the resistor ratios are
equal (mR to R for the inverting amplifier
gain equals mR to R of the voltagedivider network.)

Practically, the resistor ratios are
equalized by installing a potentiometer in
series with one resistor.

The potentiometer is trimmed until Vo is
reduced to a negligible value.
• This causes the common-mode voltage gain,
•
Vo/ECM to approach 0.
It is this characteristic of a differential amplifier
that allows a small signal voltage to be picked
out of a larger noise voltage.

It may be possible to arrange the circuit
so that the larger undesired signal is the
common-mode input voltage and the
small signal is the differential input
voltage.
• Then the differential amplifier’s output voltage
will contain only an amplified version of the
differential input voltage.