5.3 Ideal Op Amp

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5.3 Ideal Op Amp

To facilitate the understanding of op amp circuits, we will assume ideal op amps. An op amp is ideal if it has the following characteristics:

• 1. Infinite open-loop gain, A ≃ ∞.
• 2. Infinite input resistance, Ri ≃ ∞.
• 3. Zero output resistance, Ro ≃ 0.

An ideal op amp is an amplifier with infinite open-loop gain, infinite input resistance, and zero output resistance.

Although assuming an ideal op amp provides only an approximate analysis, most modern amplifiers have such large gains and input im pedances that the approximate analysis is a good one. Unless stated otherwise, we will assume from now on that every op amp is ideal.

For circuit analysis, the ideal op amp is illustrated in Fig. 5.8, which is derived from the nonideal model in Fig. 5.4. Two important properties of the ideal op amp are:

1. The currents into both input terminals are zero:

$$i_1 = 0, \quad i_2 = 0$$

(5.5)

This is due to infinite input resistance. An infinite resistance between the input terminals implies that an open circuit e xists there and current cannot enter the op amp. But the output current is not necessarily zero according to Eq. (5.1).

2. The voltage across the input terminals is equal to zero; i.e.,

$$v_d = v_2 - v_1 = 0 \tag{5.6}$$

or

$$v_1 = v_2 \tag{5.7}$$

The two characteristics can be exploited by noting that for voltage calculations the input port behaves as a short circuit, while for current calculations the input port behaves as an open circuit.

Thus, an ideal op amp has zero current into its tw o input terminals and the v oltage between the tw o input terminals is equal to zero. Equations (5.5) and (5.7) are e xtremely important and should be regarded as the key handles to analyzing op amp circuits.

Rework Practice Prob. 5.1 using the ideal op amp model. Example 5.2

Solution:

We may replace the op amp in Fig. 5.7 by its equivalent model in Fig. 5.9 as we did in Example 5.1. But we do not really need to do this. We just need to keep Eqs. (5.5) and (5.7) in mind as we analyze the circuit in Fig. 5.7. Thus, the Fig. 5.7 circuit is presented as in Fig. 5.9. Notice that

$$v_2 = v_s \tag{5.2.1}$$

Since i¹ = 0, the 40 - and 5 -kΩ resistors are in series; the same current flows through them. v1 is the voltage across the 5 -kΩ resistor. Hence, using the voltage division principle,

$$v_1 = \frac{5}{5+40} v_o = \frac{v_o}{9}$$

(5.2.2)

According to Eq. (5.7),

$$v_2 = v_1 \tag{5.2.3}$$

Substituting Eqs. (5.2.1) and (5.2.2) into Eq. (5.2.3) yields the closed loop gain,

$$v_s = \frac{v_o}{9} \quad \Rightarrow \quad \frac{v_o}{v_s} = 9 \tag{5.2.4}$$

which is very close to the value of 9.00041 obtained with the nonideal model in Practice Prob. 5.1. This shows that negligibly small error re sults from assuming ideal op amp characteristics.

At node O,

$$i_o = \frac{v_o}{40 + 5} + \frac{v_o}{20} \text{ mA}$$

(5.2.5)

From Eq. (5.2.4), when v^s = 1 V, v^o = 9 V. Substituting for v^o = 9 V in Eq. (5.2.5) produces

$$i_o = 0.2 + 0.45 = 0.65 \text{ mA}$$

This, again, is close to the value of 0.657 mA obtained in Practice Prob. 5.1 with the nonideal model.

Repeat Example 5.1 using the ideal op amp model. Practice Problem 5.2

Answer: −2, 200 μA.