Operational Amplifier Applications with Capacitors

Operational Amplifier Applications with Capacitors

In this unit we consider the most common applications of operational amplifiers where the circuit uses a capacitor as the key component. Included are the opamp integrator and a basic opamp square-wave oscillator. Both forms are investigated in projects.

12.1 Operational Amplifier Integrator

The noninverting or inverting resistance-feedback amplifier can be converted into an integrator with the addition of a feedback capacitor, C_f, as shown in Fig. 12.1. Specifically, the feedback resistor is replaced with C_f, except that the resistor R_f must remain for dc stabilization purposes.

Figure 12.1. Opamp integrator. The resistor R_f in shunt with C_f is for opamp stabilization purposes.

The operation of the integrator can be explained as follows. For the initial discussion, assume that R_f is infinite. Upon application of a constant voltage V_s, a current is induced in R_y, which is

Equation 12.1

The approximation relies on the fact that 656i816g V 0, which can be expected. The resulting output voltage is

Equation 12.2

To a good approximation, v_o(t) is proportional to the integral of the input. Performing the integration, the result is

Equation 12.3

If V_s is the magnitude of a square-wave, the output will be a triangular wave. In the integrator project we explore the integrator by using it to generate a triangular wave. The project design calls for a peak of the triangular wave equal to a few volts, that is, a significant portion of the power-supply voltage. The time of integration for the plus or minus segment of the square-wave input is T/2, where T is the period and T = 1/f, that is, the inverse of the square-wave frequency. The peak of the triangular wave then is based on

Equation 12.4

The 2 V_opeak is due to the fact that the integration is from minus V_opeak to plus V_opeak. The voltage drop across R_y has been neglected in the output.

Suppose we choose C_f as the design parameter where V_s, T, V_opeak, and R_y (from Unit 11) are given. The capacitor is then selected according to

Equation 12.5

For example, for V_s = 1 mV, V_opeak = 5 V, R_y = 220 W and T = 1 mS (f = 1 kHz), C_f = 23 nF.

In the integrator project, a sine-wave source is used. The integrated output for this case is

Equation 12.6

The sine term can be neglected and the resulting formulation can be used to predict the peak of the cosine for a given C_f and peak V_s.

A more general solution for the output voltage for the noninverting integrator includes the current through the resistor R_f. If we neglect the voltage across R_y, ( ) becomes, for this case (constant input or square-wave input)

Equation 12.7

The solution is

Equation 12.8

Replacing the exponential with the first three terms of an infinite series gives

Equation 12.9

This demonstrates that for the circuit to function properly as an itegrator, the choice of R_f should satisfy R_fC_f >> T/2. In the above example, C_f = 23 nF. With this capacitor and R_f = 1 MW, for example, R_fC_f = 23 ms, compared with T/2 = 0.5 ms. The magnitude of R_f must also satisfy the requirement of sustaining the operational amplfier in the active mode. The solution for the periodic square-wave input, giving an approximate trianglar-wave output, unlike ( ), includes the fact that the transient does not start at zero volts. This case is considered in Exercise12.mcd, Part 2. The output voltage function equivalent to ( ) for the inverting-mode integrator is

Equation 12.10

This would apply to the circuit of Fig. 12.1 with the plus input grounded and the source, V_s, attached at the grounded end of R_y.

12.2 Operational Amplifier Oscillator

The opamp can serve as a square-wave oscillator by adding positive feedback with R₁ and R₃ and replacing R_y with a capacitor C_o, as shown in Fig. 12.2. The frequency of the operation can be computed on the basis of the following derivation. With positive feedback, the circuit is unstable and the output voltage is either V_op or -V_om, that is, the positive or negative limit. The limits are close to V_DD or V_SS. These voltage values are applied to the input through the positive feedback network such that the plus input terminal is either at

Equation 12.11

or

Equation 12.12

Figure 12.2. Opamp oscillator. The frequency is determined by the value of resistors and the capacitor.

where V_om is positive.

However, the negative-feedback circuit consisting of R_f and C_o will cause the capacitor to charge up toward the voltage at the output. For example, suppose that for an interval of time, v_o(t) = V_op V_DD. Also assume that R₃ = R₁ such that the plus input is at V_op/2. When the capacitor charges up to and just beyond V_op/2, the opamp input V changes sign and the output will switch over to v_o(t) = -V_om. At this point, the capacitor begins charging toward the opposite polarity, and so forth.

For a given output, the time-dependent capacitor voltage is a simple RC transient. For example, for v_o(t) = V_op,

Equation 12.13

with a steady-state value of V_op. As an oscillator, the capacitor voltage will never reach the steady-state value; the transient terminates at the value given by (

The output is switching back and forth from V_op to -V_om. Therefore, the initial value of the capacitor voltage in ( ) is

Equation 12.14

Using this to obtain a complete solution for ( ) results in

Equation 12.15

We define t = T_plus as the positive time segment of the square-wave period. As noted, this corresponds to the time where the output switches to -V_om; that is,

Equation 12.16

Using ( ) in the left-hand side of ( ) gives an equation for T_plus as follows:

Equation 12.17

Then, solving for T_plus results in

Equation 12.18

Following the reasoning that led to ( ), the falling capacitor-voltage transient is

Equation 12.19

The total time interval (negative segment of the square-wave period) of the falling transient, t = T_minus, corresponds to the capacitor voltage, decreasing to

Equation 12.20

Using ( ) in ( ) gives the equation for T_minus, which is

Equation 12.21

This leads to

Equation 12.22

A special case is for R₁ = R₃, where ( ) and ( ) become

Equation 12.23

and

Equation 12.24

The oscillator period is T = T_plus + T_minus or, for this special case,

Equation 12.25

A good estimate comes from the approximation V_op V_om. The result is

Equation 12.26

Note that although T_minus and T_plus can be significantly different [from ( ) and ( )], the sum is well represented by ( ). Note also that the waveform is strictly a square-wave only if V_op = V_om.

LabVIEW uses ( ) to calculate the actual value of the capacitor used in the project on the oscillator. In the project, period T is a measured value. Shown in Fig. 12.3 is a project result of the measurement of the output voltage and capacitor voltage for the oscillator. Power-supply voltages were set at plus and minus eight volts and R₁ = R₃.

Figure 12.3. Oscillator v_o(t) and v_co(t) versus t (ms). Note that the switching takes place where the capacitor voltage reaches about one-half of the square-wave peak magnitude. This is for R₁ = R₃.

12.3 Summary of Equations

12.4 Exercises and Projects