Principle of negative feedback

After the previous chapters initially addressed the internal structure of the amplifier and the resulting model, as well as the real effects not described in the ideal model, this subchapter will deal with the circuitry of the operational amplifier.

The following skills should be acquired within the scope of this chapter:

To this end, a problem will be presented for which an operational amplifier is to be used to find a solution. This problem will be used to motivate the use of operational amplifiers step by step.

Part 1: Development of an amplifier circuit for music signals Beispiel Verstaerkung When developing hi-fi devices, amplifiers must be matched to the load (i.e., the speaker). For this purpose, a power amplifier (i.e., a final stage) is required. Such an amplifier, capable of boosting the signal from a mobile phone to drive a speaker, is to be built here. A simplified equivalent circuit of the speaker is represented by an inductance in series with a resistor (see Figure 1). The input voltage level is 1 V, and a 4 \(\mathrm {\Omega }\) speaker is to be driven (4 \(\mathrm {\Omega }\) means that the speaker has an impedance of 4 \(\Omega \) at 1 kHz). Determine the required output voltage of the amplifier so that the speaker consumes 25 W at 1 kHz. How can this be approached if an operational amplifier circuit is used for the solution?

Solution First, the voltage that the amplifier is to output must be calculated. \begin {equation} U_{\textnormal {A}} = \sqrt {P \cdot R} = \sqrt {25~\mathrm {W} \cdot 4~\mathrm {\Omega }} = 10~\textnormal {V} \end {equation} How can an operational amplifier be used to replace the part of the circuit marked with „?“ in order to achieve the required gain of 10 (see Figure 1)?

As we know from previous chapters, operational amplifiers have an inverting and a non-inverting input. If there is a voltage difference between these two inputs, the operational amplifier amplifies this difference many times over until the voltage reaches the maximum output voltage of the operational amplifier (see A in Figure 2). The slew rate indicates how quickly this increase can occur. If the output voltage is not fed back to the input, the output voltage increases until it reaches the level of the supply voltage. This type of amplifier circuit is referred to as a comparator circuit.

This behaviour is undesirable for most applications in which the input voltage signal with low amplitude is to be amplified. Instead, operational amplifiers are usually operated in what is known as negative feedback (B), in which the output is fed back to the inverting input. The switch shown in (B) is initially open and the operational amplifier is not supplied with voltage. At time 0 s, the switch is closed and the supply voltage of the amplifier is switched on, activating the feedback of the output signal \(U_{\text {A}}\) (C). The feedback of the output signal ensures that the voltage at the inverting input is adjusted to the voltage level of the output voltage. The amplifier circuit shown in (C) has a gain of 1 and is referred to as an impedance converter, as the amplifier in this type of circuit can be used to implement a circuit with a high-impedance input due to its high input impedance (this is necessary, for example, in measuring circuits for voltage measurements). However, this circuit is not yet suitable for solving the application problem, as a gain of 5 is required. To achieve this, additional resistors are now incorporated into the amplifier circuit (as shown in (D))¹.

Part 2: Development of an amplifier circuit for music signalsBeispiel Verstaerkung 2 The final step is to determine how the resistors in (D) must be selected in order to achieve a gain of 5. To do this, the mesh equations must first be established. An important assumption for establishing this mesh equation is that the amplifier equalises the two input signals through the feedback of the output signal. Here, we can write „Assumption: \(U_{\text {dif}}~=~0\) V“. This results in \(U_{\text {E-}}~=~1\) V for the inverting input. Since it is further assumed that no current flows into the inputs, \(I_{\text {R1}}~=~I_{\text {R2}}\). Based on these assumptions, the mesh equation is \begin {equation} U_{\textnormal {A}} = U_{\textnormal {R1}} + U_{\textnormal {R2}} = U_{\textnormal {E}} + I_{\textnormal {R2}} \cdot R_{\textnormal {2}} = U_{\textnormal {E}} + \frac {U_{\textnormal {E}}}{R_{\textnormal {1}}} \cdot R_{\textnormal {2}} \end {equation}

This results in the following expression for the input-output behaviour in the time domain. \begin {equation} \frac {U_{\textnormal {A}}}{U_{\textnormal {E}}} = \underbrace {(1+\frac {R_2}{R_1})}_{V} \end {equation}

Solving this equation for \(R_2/R_1\) gives the ratio of resistances for the problem described above as \(R_2/R_1 = 9\). The resistances should not be chosen too large here, as the balancing current between the output and input could then become too small. This can lead to severe noise at the amplifier output. Typical resistance values can usually be found in the example circuits in the data sheet of an operational amplifier. In this case, for example, \(R_2=9~\textnormal {k}\Omega \) and \(R_1=1~\textnormal {k}\Omega \) could be selected.

Similar to the circuit shown above, transfer functions can also be derived for all other operational amplifier circuits. However, this will not be shown here. Further transfer functions for some selected basic operational amplifier circuits can be found in Table 1. This table shows the circuits, the Bode diagrams with phase response (red) and amplitude response (blue), and the transfer functions. Now that the circuitry of amplifiers has been presented, the following chapter will discuss another important property of amplifier circuits that results from the amplifier components used and the external circuitry: the stability of the amplifier circuit.