5.7 Difference Amplifier

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5.7 Difference Amplifier

Difference (or dif ferential) amplifiers are used in various applications where there is a need to amplify the dif ference between tw o input sig nals. They are first cousins of the instrumentation amplifier, the most useful and popular amplifier, which we will discuss in Section 5.10.

A difference amplifier is a device that amplifies the difference between two inputs but rejects any signals common to the two inputs.

Consider the op amp circuit sho wn in Fig. 5.24. K eep in mind that zero currents enter the op amp terminals. Applying KCL to node a,

______ v¹ − v^a R1 = ______ v^a − v^o R2

$$v_o = \left(\frac{R_2}{R_1} + 1\right) v_a - \frac{R_2}{R_1} v_1 \tag{5.16}$$

Difference amplifier.

The difference amplifier is also known as the subtractor, for reasons to be shown later.

Practice Problem 5.6

Applying KCL to node b,

$$\frac{v_2 - v_b}{R_3} = \frac{v_b - 0}{R_4}$$

or

$$v_b = \frac{R_4}{R_3 + R_4} v_2 \tag{5.17}$$

But v^a = vb. Substituting Eq. (5.17) into Eq. (5.16) yields

$$v_o = \left(\frac{R_2}{R_1} + 1\right) \frac{R_4}{R_3 + R_4} v_2 - \frac{R_2}{R_1} v_1$$

or

$$v_o = \frac{R_2(1 + R_1/R_2)}{R_1(1 + R_3/R_4)} v_2 - \frac{R_2}{R_1} v_1$$

(5.18)

Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that v^o = 0 when v¹ = v2. This property exists when

$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$

\n(5.19)

Thus, when the op amp circuit is a dif ference amplifier, Eq. (5.18) becomes

$$v_o = \frac{R_2}{R_1}(v_2 - v_1)$$

\n(5.20)

If R² = R1 and R³ = R4, the difference amplifier becomes a subtractor, with the output

$$v_o = v_2 - v_1 \tag{5.21}$$

Example 5.7 Design an op amp circuit with inputs v1 and v2 such that v^o = −5v¹ + 3v2.

Solution:

The circuit requires that

$$v_o = 3v_2 - 5v_1 \tag{5.7.1}$$

This circuit can be realized in two ways.

Design 1 If we desire to use only one op amp, we can use the op amp circuit of Fig. 5.24. Comparing Eq. (5.7.1) with Eq. (5.18), we see

$$\frac{R_2}{R_1} = 5 \quad \Rightarrow \quad R_2 = 5R_1 \tag{5.7.2}$$

Also,

$$5\frac{(1+R_1/R_2)}{(1+R_3/R_4)} = 3 \quad \Rightarrow \quad \frac{\frac{6}{5}}{1+R_3/R_4} = \frac{3}{5}$$

or

$$2 = 1 + \frac{R_3}{R_4} \Rightarrow R_3 = R_4 \tag{5.7.3}$$

If we choose R¹ = 10 k Ω and R³ = 20 k Ω, then R² = 50 k Ω and R⁴ = 20 kΩ.

Design 2 If we desire to use more than one op amp, we may cascade an inverting amplifier and a two-input inverting summer, as shown in Fig. 5.25. For the summer,

$$v_o = -v_a - 5v_1 \tag{5.7.4}$$

and for the inverter,

$$v_a = -3v_2 \tag{5.7.5}$$

Combining Eqs. (5.7.4) and (5.7.5) gives

v^o = 3v² − 5v¹

which is the desired result. In Fig. 5.25, we may select R¹ = 10 kΩ and R³ = 20 kΩ or R¹ = R³ = 10 kΩ.

Design a difference amplifier with gain 7.5. Practice Problem 5.7

Answer: Typical: R¹ = R³ = 20 kΩ, R² = R⁴ = 150 kΩ.

An instrumentation amplifier shown in Fig. 5.26 is an amplifier of lowlevel signals used in process control or measurement applications and commercially available in single-package units. Show that

$$v_o = \frac{R_2}{R_1} \left( 1 + \frac{2R_3}{R_4} \right) (v_2 - v_1)$$

Solution:

We recognize that the amplifier A3 in Fig. 5.26 is a difference amplifier. Thus, from Eq. (5.20),

$$v_o = \frac{R_2}{R_1}(v_{o2} - v_{o1})$$

\n(5.8.1)

Since the op amps A1 and A2 draw no current, current i flows through the three resistors as though they were in series. Hence,

$$v_{o1} - v_{o2} = i(R_3 + R_4 + R_3) = i(2R_3 + R_4)$$

(5.8.2)

Figure 5.25 For Example 5.7.

Example 5.8

Figure 5.26 Instrumentation amplifier; for Example 5.8.

But

$$i = \frac{v_a - v_b}{R_4}$$

and v^a = v1, v^b = v2. Therefore,

$$i = \frac{v_1 - v_2}{R_4}$$

(5.8.3)

Inserting Eqs. (5.8.2) and (5.8.3) into Eq. (5.8.1) gives

$$v_o = \frac{R_2}{R_1} \left( 1 + \frac{2R_3}{R_4} \right) (v_2 - v_1)$$

as required. We will discuss the instrumentation amplifier in detail in Section 5.10.

Obtain io Practice Problem 5.8 in the instrumentation amplifier circuit of Fig. 5.27.

Figure 5.27

Instrumentation amplifier; for Practice Prob. 5.8.

Answer: 800 μA.