ENTC 4350BIOMEDICAL 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.

'

'

─

+

2.2 k

4.7 k

22 k

RS

Ri

Rf

VOUT

RD

Inverting Amplifier Forcing V2 to 0 develops an inverting

amplifier with an output, VOUT of:

'

'

─

+

2.2 k

4.7 k

22 k

RS

Ri

Rf

VOUT

RD

V1

1111 102222 VkkV

RR

VVi

fOUT

.

i

fOUT R

RVV 11

Non-inverting Amplifier Forcing V1 to 0 develops a non-inverting

amplifier.

'

'

─

+

2.2 k

4.7 k

22 k

RS

Ri

Rf

VOUT

RD

V2

Applying Thevenin’s Theorem:

'

'

─

+

2.2 k

4.7 k

22 k

RS

Ri

Rf

VOUT

RD

V2

22 VRR

RVVDS

Dopen

DS

DSTH RR

RRR

The output of the non-inverting amplifier is:

'

'

─

+

2.2 k

22 k

RTh

Ri

Rf

VOUT

DS

D

RRRV2

i

fOUT R

RVV 122

i

fOUT R

RVV 122

The total output is the sum:

To balance the circuit, we set the coefficients to add to zero.

i

f

i

fOUTOUT R

RV

RR

VVV 1212

i

f

i

f

RR

RR

fi

f

fi

i

i

f

i

f

i

fi

i

f

i

f

i

i

i

f

i

f

RRR

RRR

RR

RR

RRR

RR

RR

RR

RR

RR

1212

12

1

2

2

12

VVRR

VRRRR

VRR

VRR

RRRRV

RR

RR

VRRR

RRRR

RRV

RR

VRR

RRR

RRR

VV

i

f

fi

fi

i

f

fi

f

fi

i

i

f

i

f

fii

f

fii

fi

i

f

i

f

fi

f

fi

fOUT

So the balanced condition yields

• and the differential gain Ad is

12 VVRR

Vi

fOUT

i

fD R

RA

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.

V Vwhen grounded is Vwhen

12

2

OUT

OUT

VV

CMRR

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

CMRRCMRRdb log20

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 voltage-divider 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.