7.7 † First-Order Op Amp Circuits

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7.7 † First-Order Op Amp Circuits

An op amp circuit containing a storage element will exhibit first-order behavior. Differentiators and integrators treated in Section 6.6 are examples of first-order op amp circuits. Again, for practical reasons, inductors are hardly ever used in op amp circuits; therefore, the op amp circuits we consider here are of the RC type.

As usual, we analyze op amp circuits using nodal analysis. Some times, the Thevenin equivalent circuit is used to reduce the op amp circuit to one that we can easily handle. The following three examples illustrate the concepts. The first one deals with a source-free op amp circuit, while the other two involve step responses. The three examples have been carefully selected to co ver all possible RC types of op amp circuits, depending on the location of the capacitor with respect to the op amp; that is, the capacitor can be located in the input, the output, or the feedback loop.

For the op amp circuit in Fig. 7.55(a), find vo for t > 0, gi ven that Example 7.14 v(0) = 3 V. Let Rf = 80 kΩ, R1 = 20 kΩ, and C = 5 µF.

For Example 7.14.

Solution:

This problem can be solved in two ways:

■ METHOD 1 Consider the circuit in Fig. 7.55(a). Let us derive the appropriate differential equation using nodal analysis. If v1 is the voltage at node 1, at that node, KCL gives

$$\frac{0 - v_1}{R_1} = C \frac{dv}{dt}$$

(7.14.1)

Because nodes 2 and 3 must be at the same potential, the potential at node 2 is zero. Thus, v1 − 0 = v or v1 = v and Eq. (7.14.1) becomes

$$\frac{dv}{dt} + \frac{v}{CR_1} = 0$$

(7.14.2)

This is similar to Eq. (7.4b) so that the solution is obtained the same way as in Section 7.2, i.e.,

$$v(t) = V_0 e^{-t/\tau}, \qquad \tau = R_1 C \tag{7.14.3}$$

where V0 is the initial voltage across the capacitor. But v(0) = 3 = V0 and τ = 20 × 10³ × 5 × 10*⁻⁶ = 0.1.* Hence,

$$v(t) = 3e^{-10t}$$

(7.14.4)

Applying KCL at node 2 gives

$$C\frac{dv}{dt} = \frac{0 - v_o}{R_f}$$

or

$$v_o = -R_f C \frac{dv}{dt} \tag{7.14.5}$$

Now we can find v0 as

$$v_o = -80 \times 10^3 \times 5 \times 10^{-6} (-30e^{-10t}) = 12e^{-10t} \text{ V}, \qquad t > 0$$

■ METHOD 2 Let us apply the shortcut method from Eq. (7.53). We need to find vo( 0⁺), vo(∞), and τ. Since v( 0⁺) = v(0*⁻) = 3 V,* we apply KCL at node 2 in the circuit of Fig. 7.55(b) to obtain

$$\frac{3}{20,000} + \frac{0 - v_o(0^+)}{80,000} = 0$$

or vo( 0⁺) = 12 V*.* Since the circuit is source free, v(∞) = 0 V*.* To find τ, we need the equivalent resistance Req across the capacitor terminals. If we remove the capacitor and replace it by a 1-A current source, we have the circuit shown in Fig. 7.55(c). Applying KVL to the input loop yields

$$20,000(1) - v = 0 \qquad \Rightarrow \qquad v = 20 \text{ kV}$$

Then

$$R_{\text{eq}} = \frac{v}{1} = 20 \text{ k}\Omega$$

and τ = ReqC = 0*.1.* Thus,

$$v_o(t) = v_o(\infty) + [v_o(0) - v_o(\infty)]e^{-t/\tau}$$

= 0 + (12 - 0)e^{-10t} = 12e^{-10t} V, \qquad t > 0

as before.

For the op amp circuit in Fig. 7.56, find vo for t > 0 if v(0) = 4 V*.* Assume that Rf = 50 kΩ*, R1 = 10 kΩ,* and C = 10 µF*.* Practice Problem 7.14

Answer: −4 e ⁻2*^t* V*, t >* 0*.*

Figure 7.56 For Practice Prob. 7.14.

Example 7.15 Determine v(t) and vo(t) in the circuit of Fig. 7.57.

Solution:

This problem can be solved in two ways, just like the previous example. However, we will apply only the second method. Since what we are looking for is the step response, we can apply Eq. (7.53) and write

$$v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}, \qquad t > 0 \tag{7.15.1}$$

vo +

where we need only find the time constant τ, the initial value v(0), and the final value v(∞). Notice that this applies strictly to the capacitor voltage due a step input. Since no current enters the input terminals of the op amp, the elements on the feedback loop of the op amp constitute an RC circuit, with

$$\tau = RC = 50 \times 10^3 \times 10^{-6} = 0.05 \tag{7.15.2}$$

For t < 0*,* the switch is open and there is no voltage across the capacitor. Hence, v(0) = 0*.* For t > 0*,* we obtain the voltage at node 1 by voltage division as

$$v_1 = \frac{20}{20 + 10}3 = 2 \text{ V}$$

(7.15.3)

Since there is no storage element in the input loop, v1 remains constant for all t. At steady state, the capacitor acts like an open circuit so that the op amp circuit is a noninverting amplifier. Thus,

$$v_o(\infty) = \left(1 + \frac{50}{20}\right)v_1 = 3.5 \times 2 = 7 \text{ V}$$

(7.15.4)

But

$$v_1 - v_o = v \tag{7.15.5}$$

so that

$$v(\infty) = 2 - 7 = -5$$

V

Substituting τ, v(0), and v(∞) into Eq. (7.15.1) gives

$$v(t) = -5 + [0 - (-5)]e^{-20t} = 5(e^{-20t} - 1) \text{ V}, \qquad t > 0 \tag{7.15.6}$$

From Eqs. (7.15.3), (7.15.5), and (7.15.6), we obtain

$$v_o(t) = v_1(t) - v(t) = 7 - 5e^{-20t} \text{ V}, \qquad t > 0 \tag{7.15.7}$$

Find v(t) and vo(t) in the op amp circuit of Fig. 7.58.

Answer: (Note, the v oltage across the capacitor and the output v oltage must be both equal to zero, for t < 0, since the input w as zero for all t < 0.) 40(1 − e⁻10*^t* ) u(t) mV, 40( e⁻10*^t* − 1) u(t) mV.

Practice Problem 7.15

vo +

1 μF

100 kΩ

‒

Figure 7.58 For Practice Prob. 7.15.

Figure 7.57 For Example 7.15.

Example 7.16 Find the step response vo(t) for t > 0 in the op amp circuit of Fig. 7.59. Let vi = 2u(t) V, R1 = 20 kΩ, Rf = 50 kΩ, R2 = R3 = 10 kΩ, C = 2 µF.

Solution:

Notice that the capacitor in Example 7.14 is located in the input loop, while the capacitor in Example 7.15 is located in the feedback loop. In this example, the capacitor is located in the output of the op amp. Again, we can solve this problem directly using nodal analysis. However, using the Thevenin equivalent circuit may simplify the problem.

We temporarily remove the capacitor and find the Thevenin equivalent at its terminals. To obtain VTh, consider the circuit in Fig. 7.60(a). Since the circuit is an inverting amplifier,

$$V_{ab} = -\frac{R_f}{R_1} v_i$$

By voltage division,

Figure 7.60

Obtaining VTh and RTh across the capacitor in Fig. 7.59.

To obtain RTh, consider the circuit in Fig. 7.60(b), where Ro is the output resistance of the op amp. Since we are assuming an ideal op amp, Ro = 0, and

$$R_{\text{Th}} = R_2 \parallel R_3 = \frac{R_2 R_3}{R_2 + R_3}$$

Substituting the given numerical values,

$$V_{\text{Th}} = -\frac{R_3}{R_2 + R_3} \frac{R_f}{R_1} v_i = -\frac{10}{20} \frac{50}{20} 2u(t) = -2.5u(t)$$ $$R_{\text{Th}} = \frac{R_2 R_3}{R_2 + R_3} = 5k\Omega$$ $$v_o(t) = -2.5(1 - e^{-t/\tau})u(t)$$

where τ = RThC = 5 × 103 × 2 × 10⁻⁶ = 0.01. Thus, the step response for t > 0 is

$$v_o(t) = 2.5(e^{-100t} - 1)u(t)$$

V

Figure 7.61 Thevenin equivalent circuit of the circuit in Fig. 7.59.

Answer: 27(1 − e⁻50*^t* )u(t) V.