Differential Amplifiers差分放大器英文版讲义

KNE222 University of Tasmania

G. Vertigan Page 1

2009 School Of Engineering

KNE222 Electronic Engineering

Differential Amplifiers

Introduction:

One of the most useful circuits in the field of instrumentation is the differential amplifier. This

device is designed to amplify the difference between two voltages and to provide a ground

referenced output voltage. Operational amplifiers are frequently used in this application, and a

common approach is shown in Figure 1.

Figure 1. A Simple Differential Amplifier

This circuit amplifies the difference between input signals v1(t) and v2(t). It can be analysed using the

Principle of Superposition by summing the contributions to vo(t) from v1(t) and v2(t) when acting

independent of each other.

Consider the case where v1(t) alone is applied, as shown in Figure 2.

Figure 2. Differential Amplifier with v1(t) acting alone.

Here v2(t) has been replaced by its internal impedance, which for a voltage source is zero Ohms. If

we define the voltage at the non-inverting input terminal as v+ and that at the inverting input as v-

then we can write:

)()(2121RRRtvv …(1) KNE222 University of Tasmania

G. Vertigan Page 2

2009 And since the amplifier has a very high open loop gain we know that to a good

approximationvv. Because the input impedance of the Op Amp is very high we can also write:

)()(2'1''10RRRtvv …(2)

Therefore by combining (1) and (2) we find:

'1'2'12121)()()(RRRRRRtvtvo …(3)

Similarly, when v2(t) acts on its own, the circuit can be redrawn as shown in Figure 3.

Figure 3. Differential Amplifier with v2(t) acting alone.

In this case 0vv, therefore summing the currents at the inverting input we can write:

0'2'12RvRvo

Thus: '1'22)()(RRtvtvo …(4)

Finally to obtain the complete expression for )(tvo we must sum these contributions to obtain:

'1'22'1'2'12121)()()()(RRtvRRRRRRtvtvo …(5)

And if R1’= R1 and R2’= R2 then:

1221)()()(RRtvtvtvo …(6)

Differential Mode and Common Mode Signals.

Equation (6) has shown that we can amplify the difference between two signals,)(1tvand )(2tv.

However from the point of view of our amplifier’s performance, we need to consider the relative

size of the differential voltage as opposed to the size of either )(1tvor)(2tv. To do this we define two

new terms, the Differential Mode Voltage and the Common Mode Voltage as follows:

(Differential Mode Voltage)/2 2)()(21tvtv KNE222 University of Tasmania

G. Vertigan Page 3

2009 And:

Common Mode Voltage 2)()((21tvtv

For example, take the trivial case where v1(t)= 3 volts and v2(t)= 1 volt. In this case the (Differential

Mode Voltage)/2 is 1 volt, and the Common Mode Voltage is 2 volts. When applied to our

differential amplifier the Differential Mode component must be amplified, however the Common

Mode component must be ignored. This can be a tall order when the Common Mode component is

large compared to the Differential component.

Figure 4 shows how the Differential Mode and Common Mode signals are presented to the amplifier.

Effectively the non-inverting input sees +VDM/2 while the inverting input sees -VDM/2, both

superimposed on the Common Mode component, VCM.

Figure 4. Common and Differential Mode Signals

The Differential Mode Gain (DMG)

It is useful to quantify the performance of our amplifier for purely Differential Mode signals and so

we evaluate the Differential Mode Gain as by removing the Common Mode component, thus

v1(t)=VDM/2 and v2(t)= - VDM/2. From equation (5) we find:

12'1'2'1'2'1212)()(2)()(RRtVRRRRRRRRtVtvDMDMo if R1’= R1 and R2’= R2