Consider the following circuit. The switch closes at \$t=0\$. The voltage source is \$V_0e^{-t/t_0}\$. We assume to op is ideal, so there's no voltage difference between the input terminals and no current through the input.

Approach 1

For \$t > 0\$ we have immediately that \$V=\frac{R}{R+R}V_0e^{-t/t_0}=\frac{1}{2}V_0e^{-t/t_0}\$.

Let \$i_L\$ be the current through the inductor from \$V_{out}\$ to \$V\$. We see that \$i_L = V/R = \frac{V_0}{2R}e^{-t/t_0}\$. We have \$V_{out}-V=L\frac{di_L}{dt}=-\frac{V_0L}{2Rt_0}e^{-t/t_0}\$, and we find \$V_{out}=V-\frac{V_0L}{2Rt_0}e^{-t/t_0}=\frac{V_0}{2}\left(1-\frac{L}{t_0R}\right)e^{-t/t_0}.\$

Approach 2

Let's analyse the circuit in the Laplace domain. We thus let the voltage source be \$V_0e^{-t/t_0}H(t)\$ in the time domain, where \$H(t)\$ is the step function. The Laplace transform of this is \$V_0\frac{1}{s+1/t_0}\$.

In the same way as before, we find \$V(s)=\frac{V_0}{2}\frac{1}{s+1/t_0}\$. We also have \$V_{out}-V=\frac{sL}{sL+R}V_{out}\$, whence \$V_{out}=\frac{sL+R}{R}V=\frac{V_0L}{2R}\frac{s+R/L}{s+1/t_0}=\frac{V_0L}{2R}\left(1+\frac{R/L-1/t_0}{s+1/t_0} \right).\$ Taking the inverse gives \$V_{out} = \frac{V_0L}{2R}\delta(t)+\frac{V_0}{2}\left(1-\frac{L}{t_0R}\right)e^{-t/t_0}.\$


Apparently, we pick up a delta-distribution with the second approach. Obviously, this cannot occur in a real circuit, so I assume the problem somehow lies in our assumptions about ideality of the OP?

Is it then in general true that we cannot analyse transient behaviour of ideal OP circuits in the Laplace domain? (Or rather, if we do so, we have to neglect all non-physical terms?).


simulate this circuit – Schematic created using CircuitLab


2 Answers 2


Obviously, this cannot occur in a real circuit

You have several problems.

You are using derivatives on something that is discontinuous. That usually tends not to work. Also, the opamp's output voltage will go to infinity at t=0. I will now explain...

The switch closes at \$t=0\$. The voltage source is \$V_0e^{-t/t_0}\$.

So, after the two rather useless resistors which divide this by 2, when \$ t=0^- \$ which is a tiny instant before t=0, we have \$ V= 0 \$. Then at \$ t=0 \$ the switch closes and we have \$V=V_0/2\$, thus V is discontinuous.

Our nice perfect opamp must adjust its output voltage to keep its two inputs at the same potential. So, at t=0 the output of the opamp will go up. However, there is a perfect inductor in the feedback loop. And the current in an inductor cannot change instantly. And the opamp's negative input voltage V is \$ R i_L \$. And at t=0, \$ i_L =0 \$.

Thus, the voltage at the two inputs of the opamp is not equal. IN+ is at Vo/2 and IN- is at 0V. So, the perfect opamp model stops working, but you got other problems. For the ideal opamp, there is only one choice... its output voltage instantly jumps to +infinity.

Now, at t=0 voltage across the inductor which is \$ V_{inductor} = L\frac{di_L}{dt}\$ is thus +infinity.

If we break enough math, spill the pieces over the floor, then roll over it forward and reverse with a tank, which you did, we can thus integrate that and conclude that \$ \int V_{inductor} = L i_L\$ and therefore right after t=0, give or take, more or less, \$ i_L = 1/L \int +\infty = V_o/2R\$ ...

Tada! Done.

Now, your problem is you didn't notice this, so your you got the wrong result in "Approach 1". No blame, we all make mistakes, and the ideal opamp model really invites mistakes too, since any deviation from ideal operating point creates impossible conditions.

I didn't check the Laplace Transform calculations, but let's examine the circuit again. Let's throw away the switch and both resistors, and use the "V" node as input. We now have a bog-standard non-inverting amplifier. Its gain is:

\$ G = 1+ Ls/R \$

Now, as is obvious from this equation, G goes to infinity as frequency goes up. This kinda works like a differentiator. But it is given a discontinuous signal as an input. Thus it derivates a non-derivable input. Hence your Laplace results have a delta.

At infinitely high frequency, the inductor's impedance is infinite, thus it is an open circuit, and we can remove it from the schematic: on discontinuous inputs, the opamp no longer has feedback, so the ideal opamp model cannot apply.

In real life, opamps are not infinitely fast, therefore the phase lag caused by the inductor in the feedback loop will turn the opamp into an oscillator.

So this circuit is a trap ;)

  • \$\begingroup\$ 1. Why the two reisistors are useless? 2. Where did we divide by zero as you said above? 3. At high frequency , the inductor is open and we don't have the feedback and so the can't apply the property V+ = V- here. So how do you calculate Vout for that circuit? The method should work for any frequency not just low frequency when the feedback is sitll available. 4. If possible, can you talk more about how oscillation occur in real life opamp? \$\endgroup\$
    – emnha
    Commented Sep 3, 2017 at 8:58
  • 1
    \$\begingroup\$ 1. because they introduce a trivial problem ("divide by two") whch detracts from the big picture (circuit won't work). 2. was a reference to internet meme, now removed it. 3. At high frequency, as you say feedback disappears, so the gain of the circuit becomes infinite, so the output will clip. Thus the circuit is non-linear, and we can't use linear tools (like Laplace transform) to study it, the result will be unrealistic. \$\endgroup\$
    – bobflux
    Commented Sep 3, 2017 at 9:13
  • 1
    \$\begingroup\$ Note that we could consider the opamp as a comparator instead, but I believe that was not the intent. 4- Read up on en.wikipedia.org/wiki/Nyquist_stability_criterion and the simpler Bode-Nyquist criterion, control theory. The inductor introduces a pole, and while the ideal opamp has no poles (it has perfect frequency response), a real opamp will have some poles in its open loop transfer function (because it is not infinitely fast). Using the simpler criteria on a bode plot of open loop transfer function, phase shift introduced by the extra poles will cause oscillation. \$\endgroup\$
    – bobflux
    Commented Sep 3, 2017 at 9:27
  • 1
    \$\begingroup\$ (hard to explain without explaining all of control theory! if you're in eng. school, you'll have a class about this, otherwise you can find material on the internet, it's rather necessary when using opamps if you want them to be stable...) \$\endgroup\$
    – bobflux
    Commented Sep 3, 2017 at 9:28
  • \$\begingroup\$ Thanks for the responses. So with the circuit above, at low frequency, the circuit still in linear domain and we can still use linear theory as Laplace transform to study about it as approach 2 above. But at high frequency, the opamp behaves as a comparator, so we have to treat it in a different way. Is there an exact threshold frequency where we can classify "low frequency" and "high frequency" here? \$\endgroup\$
    – emnha
    Commented Sep 3, 2017 at 10:36

Obviously, this cannot occur in a real circuit, so I assume the problem somehow lies in our assumptions about ideality of the OP?

Either that or the ideality of the inductor.

A real op-amp would not output a voltage higher than its power supply voltage.

A real inductor will have an equivalent parasitic parallel capacitance that limits the actual voltage needed to drive a current through it (for a short time) to equalize the op-amp's inputs.

Is it then in general true that we cannot analyse transient behaviour of ideal OP circuits in the Laplace domain?

In general you can't make a true ideal differentiator with real circuit components.


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