Solution:

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between the ground and the in verting terminal. We are interested in the output voltage and the voltage gain. Application of KCL at the inverting terminal gives

$$i_1 = i_2 \Rightarrow \frac{0 - v_1}{R_1} = \frac{v_1 - v_o}{R_f}$$

(5.10)

But v¹ = v² = vi. Equation (5.10) becomes

$$\frac{-v_i}{R_1} = \frac{v_i - v_o}{R_f}$$

or

$$v_o = \left(1 + \frac{R_f}{R_1}\right) v_i$$

(5.11)

The voltage gain is A^v = vo∕vⁱ = 1 + Rf∕R1, which does not have a negative sign. Thus, the output has the same polarity as the input.

A noninverting amplifier is an op amp circuit designed to provide a positive voltage gain.

Again we notice that the gain depends only on the external resistors.

Notice that if feedback resistor Rf = 0 (short circuit) or R¹ = ∞ (open circuit) or both, the gain becomes 1. Under these conditions (Rf = 0 and R¹ = ∞), the circuit in Fig. 5.16 becomes that shown in Fig. 5.17, which is called a voltage follower (or unity gain amplifier) because the output follows the input. Thus, for a voltage follower

$$v_o = v_i \tag{5.12}$$

Such a circuit has a ve ry high input impedance and is therefore use ful as an intermediate-stage (or buffer) amplifier to isolate one circuit from another, as portrayed in Fig. 5.18. The voltage follower minimizes interaction between the tw o stages and eliminates interstage loading.

Example 5.3 For the op amp circuit in Fig. 5.19, calculate the output voltage vo.

Solution:

We may solve this in two ways: using superposition and using nodal analysis.

■ METHOD 1 Using superposition, we let

Figure 5.17 The voltage follower.

Figure 5.18 A voltage follower used to isolate two cascaded stages of a circuit.

where vo1 is due to the 6-V voltage source, and vo2 is due to the 4-V input. To get vo1, we set the 4-V source equal to zero. Under this condition, the circuit becomes an inverter. Hence Eq. (5.9) gives

$$v_{o1} = -\frac{10}{4}(6) = -15$$

V

To get vo2, we set the 6 -V source equal to zero. The circuit becomes a noninverting amplifier so that Eq. (5.11) applies.

$$v_{o2} = \left(1 + \frac{10}{4}\right)4 = 14 \text{ V}$$

Thus,

$$v_o = v_{o1} + v_{o2} = -15 + 14 = -1$$

V

■ METHOD 2 Applying KCL at node a,

$$\frac{6-v_a}{4} = \frac{v_a - v_o}{10}$$

But va= vb=4, and so

$$\frac{6-4}{4} = \frac{4-v_o}{10} \Rightarrow 5 = 4-v_o$$

or vo= −1 V, as before.

Calculate vo in the circuit of Fig. 5.20. Practice Problem 5.5

Answer: 21 V.