5.8 Cascaded Op Amp Circuits

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5.8 Cascaded Op Amp Circuits

As we know, op amp circuits are modules or building blocks for designing complex circuits. It is often necessary in practical applications to connect op amp circuits in cascade (i.e., head to tail) to achie ve a large overall gain. In general, tw o circuits are cascaded when the y are con nected in tandem, one behind another in a single file.

A cascade connection is a head-to-tail arrangement of two or more op amp circuits such that the output of one is the input of the next.

When op amp circuits are cascaded, each circuit in the string is called a stage; the original input signal is increased by the g ain of the individual stage. Op amp circuits have the advantage that they can be cascaded without changing their input -output relationships. This is due to the fact that each (ideal) op amp circuit has infinite input resistance and zero output resistance. Figure 5.28 displays a block diagram represen tation of three op amp circuits in cascade. Since the output of one stage is the input to the next stage, the overall gain of the cascade connection is the product of the gains of the individual op amp circuits, or

$$A = A_1 A_2 A_3 \tag{5.22}$$

Although the cascade connection does not afect the op amp input-output relationships, care must be e xercised in the design of an actual op amp circuit to ensure that the load due to the ne xt stage in the cascade does not saturate the op amp.

Figure 5.28

A three-stage cascaded connection.

Find vo and io in the circuit in Fig. 5.29. Example 5.9

Solution:

This circuit consists of two noninverting amplifiers cascaded. At the output of the first op amp,

$$v_a = \left(1 + \frac{12}{3}\right)(20) = 100 \text{ mV}$$

At the output of the second op amp,

$$v_o = \left(1 + \frac{10}{4}\right) v_a = (1 + 2.5)100 = 350$$

mV

The required current io is the current through the 10-kΩ resistor.

$$i_o = \frac{v_o - v_b}{10}$$

mA

Figure 5.29 For Example 5.9.

But

$$v_b = v_a = 100$$

mV. Hence,

mV. Hence,

$$i_o = \frac{(350 - 100) \times 10^{-3}}{10 \times 10^3} = 25 \,\mu\text{A}$$

For Practice Prob. 5.9.

Example 5.10
--------------

If v¹ = 1 V and v² = 2 V, find vo in the op amp circuit of Fig. 5.31.

For Example 5.10.

Solution:

• 1. Define. The problem is clearly defined.
• 2. Present. With an input of v1 of 1 V and of v2 of 2 V, determine the output voltage of the circuit shown in Figure 5.31. The op amp circuit is actually composed of three circuits. The first circuit acts as an amplifier of gain −3(−6 kΩ∕2 kΩ) for v1 and the second functions as an amplifier of gain −2(−8 kΩ∕4 kΩ) for v2. The last circuit serves as a summer of two different gains for the output of the other two circuits.
• 3. Alternative. There are different ways of working with this circuit. Because it involves ideal op amps, then a purely mathematical

approach will work quite easily. A second approach would be to use PSpice as a confirmation of the math.

4. Attempt. Let the output of the first op amp circuit be designated as v11 and the output of the second op amp circuit be designated as v22. Then we get

$$v_{11} = -3v_1 = -3 \times 1 = -3 \text{ V},$$ $$v_{22} = -2v_2 = -2 \times 2 = -4 \text{ V}$$

In the third circuit we have

$$v_o = -(10 \text{ k}\Omega/5 \text{ k}\Omega)v_{11} + [-(10 \text{ k}\Omega/15 \text{ k}\Omega)v_{22}]$$

= -2(-3) - (2/3)(-4)
= 6 + 2.667 = 8.667 V

5. Evaluate. To properly evaluate our solution, we need to identify a reasonable check. Here we can easily use PSpice to provide that check.

Now we can simulate this in PSpice. The results are shown in Fig. 5.32.

We obtain the same results using two entirely different techniques (the first is to treat the op amp circuits as just gains and a summer and the second is to use circuit analysis with PSpice). This is a very good method of assuring that we have the correct answer.

6. Satisfactory? We are satisfied we have obtained the asked for results. We can now present our work as a solution to the problem.

Practice Problem 5.10

If v¹ = 5 V and v² = 5 V, find vo in the op amp circuit of Fig. 5.33.

For Practice Prob. 5.10.

Answer: 35 V.