2
\$\begingroup\$

Hayt's (8th edition) Engineering Curcuit Analysis (Chapter 9, p360) contains an illustration of Operational Amplifier solving \$\frac{d^2v}{dt^2} = -9v\$. It is hard for me to understand how it was constructed. To get better understanding, I created simpler diagram to solve first order ODE: \$\frac{dv}{dt} = -v\$

Here is the diagram: ODE and Op. Am.

Here is my derivation

\$-c\frac{dV_{C1}}{dt} = \frac{V_{R1}}{R1}\$

\$V_{C1}= V_{R1}+V_{R2}+V_{R3}\$

\$\frac{V_{R1}}{R1} =\frac{V_{R2}}{R2} = \frac{V_{R3}}{R3}\$

\$\frac{V_{C1}}{R1R2R3} =\frac{V_{R1}}{R1R2R3} +\frac{V_{R2}}{R1R2R3}+\frac{V_{R3}}{R1R2R3}\$

\$\frac{V_{C1}}{R1R2R3} = \frac{V_{R1}}{R1}(\frac{1}{R2R3} + \frac{1}{R1R3} + \frac{1}{R1R2})\$

\$V_{C1} = V_{R1}(1 + \frac{R2}{R1}+ \frac{R3}{R1})\$

Let \$\alpha = (1 + \frac{R2}{R1}+ \frac{R3}{R1})\$

\$-c\frac{dV_{C1}}{dt} = \frac{V_{C1}}{\alpha}\$

Picking proper values of resistance we get the desired result.

Is this reasonably accurate?

\$\endgroup\$
1

1 Answer 1

2
\$\begingroup\$

Just analyzing here your proposed circuit considering ideal components. Also, in my drawing below, I've reversed the inputs \$+\$ and \$-\$ of op. amp. \$\mathrm{U1}\$ (I dont' know your intention was even that) in order to put in the form of a classic integrator.

1. Firstly:

When writing

$$\frac{v_{R1}}{R1}=\frac{v_{R2}}{R2}=\frac{v_{R3}}{R3}$$

you are assuming that

$$i_{R1}=i_{R2}=i_{R3}$$

which is incorrect. Note that \$\mathrm{U2}\$ also provides a current in its output, in order to keep the virtual ground. Here follows an explanation considering the op. amp working in linear mode:

$$i_{R2}=i_{R3} = \frac{v_{out}}{R_3}$$

$$i_{R1}=-\frac{R_2}{R_1R_3}v_{out}$$

which, obviously are different from your statement.

2. Secondly:

The complete circuit presented in the question reference (book from Hayt et al) has two integrators and yours, only one. In that original form it is possible to make an oscillator, with an associated second order differential equation (and \$\mathrm{V_{BT1}=6 \space V}\$):

$$ \frac{dv_{out}^2}{dt^2}= -9v_{out} $$

which solution is a sinusoidal signal:

$$v_{out} = 2\sin(3t)$$

Because you suppressed an integrator, you cannot get an oscillator like that. Just to remember, there are waveform generator circuits, including those based on just one op. amp., such as the Wien-bridge oscillator (sinusoidal) or even those exploring the non-linear mode of operation, based on configurations using Schimitt-Triggers (triangular-square). In fact, your configuration will introduce a form of positive feedback, which will end up (in practice) saturating the operational amplifier. Follow the calculations of the currents and voltages involved, in the figure below (with \$\mathrm{BT1}\$ being removed from circuit in \$t=0\$, just to provide the initial condition or bootstrap):

enter image description here

So, it's possible to get:

$$\frac{dv_{C_1}}{dt}=-\frac{R_2}{R_1R_3C_1}v_{out} $$

Or, since \$v_{out} = -v_{C_1}\$:

$$ \boxed{ \frac{dv_{C_1}}{dt}=\frac{R_2}{R_1R_3C_1}v_{C_1} \qquad [1] } $$

With solution wrt:

$$\frac{R_2}{R_1R_3C_1}\int_{0}^{t}d\tau=\int_{\text{VBT1}}^{v_{C1}}\frac{d\xi}{\xi}$$

$$\frac{R_2}{R_1R_3C_1}t=\ln \left(\frac{v_{C_1}}{\mathrm{V_{BT1}}} \right )$$

$$ \boxed{ v_{C_1(t)}=\mathrm{V_{BT1}}e^{\frac{R_2}{R_1R_3C_1}t} \qquad[2] } $$

Thus, the circuit you've proposed is unstable with \$v_{C1}(t)\$ growing without limits in both cases, for \$\mathrm{V_{BT1}} < 0\$ and \$\mathrm{V_{BT1}} > 0\$. The temporal responses (from \$[2]\$) and phase portraits (from\$[1]\$), for \$t \ge 0\$, are presented in the sketch below:

enter image description here

Finally, an example of simulation with \$ \mathrm{V_{BT1} = v_{C_1(0)}= 2 \space V} \$:

enter image description here

Where \$ \mathrm{v_{C_1(1)} \simeq 5.4 \space V}\$, according expression \$[2]\$, using those component values.

\$\endgroup\$
8
  • \$\begingroup\$ Thank you very much for your answer! I am looking at your formula : \$ i_{R1} = -\frac{R_2}{R_3}v_{out} \$. Unit on the left are amp and on the right are volts. Am I reading it correctly? \$\endgroup\$ Apr 15 at 16:28
  • \$\begingroup\$ It's the voltage on output of \$U2\$. \$\endgroup\$ Apr 15 at 17:11
  • \$\begingroup\$ So you think your formula is correct? \$\endgroup\$ Apr 15 at 17:14
  • \$\begingroup\$ That's the standard formula for the output voltage in an inverting, op. amp. based, amplifier given the input voltage and the gain \$-\frac{R_2}{R_3}\$. \$\endgroup\$ Apr 15 at 17:22
  • 1
    \$\begingroup\$ You are correct. After your first comment, I was looking on schematics, but now I see the missing of \$R_1\$ on denominator of \$i_{R1}\$ in the first part of the text the text. Already correct it. Thanks. \$\endgroup\$ Apr 15 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.