The fundamental problem I have with the various explanation of OpAmps is that they claim OpAmps need feedback. Naturally, I was not very satisfied with this explanation. Why can't we use an amplified without feedback? Well it turns out, that most transistors have inherent current variations that vary wildly with temperature. The natural linear region of transistors is also insufficiently small for many applications. After about an hour of struggling, I was finally able to show that the traditional feedback op amp with resistors Rf and Rg can stabilize as system that has a tendency to amplify, but not necessarily linearly. The implications extend beyond electronics and reach into the realm of mechanical systems and other systems I'm sure.

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I believe I have achieved the correct derivation.

We consider some arbitrary amplification factor A which can take on any value atm. Let us us Vo(s) to represent Vout in the Laplace domain, and Vin(s) for Vin. Note that to find the response of our system to say perhaps a unit step - we could substitute in 1/s for Vin(s) in the Laplace domain. OK. Here we go.

We know be definition that, $$\texttt{EQ 1.} V_o(s) = A(V_{IN}-V_A) $$

Let’s see if we can get in terms of to make EQ 1. amenable to analysis. Applying the voltage divider rule

$$\texttt{EQ 2.} V_A(s) = \frac {A\cdot R_g} {R_g + R_f} \cdot V_o(s) $$

Substitute and expand: $$ \texttt{EQ 3.} V_o(s) = A \cdot V_{IN}(s) - \frac {A \cdot R_g} {R_g + R_f}V_o(s) $$

Collect Terms $$\texttt{EQ 4.} V_o(s) = A \cdot V_{IN}(s) - \frac {A \cdot R_g} {R_g + R_f}V_o(s). $$

Some Algebra $$\texttt{EQ 5.} V_o(s) = \frac {A \cdot V_{IN}(s)} {1 + \frac {A \cdot R_g}{R_g + R_f}} $$

If we assume A is infinite, and this is critical, because in reality, this is not true - so lets apply a limit allowing A to approach infinity. $$ \texttt{EQ 6.} \lim_{A \to \infty} V_o(s) = \lim_{A \to \infty} \frac {A \cdot V_{IN}(s)} {1 + \frac {A \cdot R_g}{R_g + R_f}} $$

Since A is the driving term, all else fades away except for constants multiplying A, which in this case is Vin in the numerator, and the voltage divider fraction in the denominator. $$\texttt{EQ 7.} \lim_{A \to \infty} V_o(s) = \frac {R_g + R_f}{R_g}\cdot V_{in}(s) = (1 + \frac{R_f}{R_g})\cdot V_{IN}(s) $$

Which is the proper equation we wanted to start with. So what so special about this equation? We could have easily derived it in the time domain for the same result and not used the Laplace domain. The problem is that we chose to force A to infinity. If we don’t force A to infinity and plug in a unit step for Vin we find that our Vout becomes

$$\texttt{EQ 8.} A \cdot u(t)e^{-(\frac {A\cdot R_g}{R_g+R_f})t} $$

If we apply the Laplace frequency shift property. This indicates that if A is not infinite, that the op amp will forever stabilize, approaching, at an exponentially decaying rate, A*u(t). Making A infinite kills the exponential term (because of the negative sign in the exponent) giving the illusion of some instantaneous change in Vout from Vin that confuse TimNJ. The problem of course is that in practice, values are never infinite.

In addition, we didn’t have to use the Laplace domain to solve this system - we could have used algebra - but in my experience, the Laplace domain naturally generates sensible answers for systems involving feedback with much less effort than does the time domain.

Lastly, a bit more analysis reveals that an op amp with a non-infinite A will respond differently to inputs of different frequencies and will actually shift the frequency of the input. This is why I suppose data sheets often specify op-amps’s frequency response ranges.

Of course I could be wrong, so please correct me.

  • how_opamps_really_work.txt
  • Last modified: 2018/10/09 20:17
  • by yehowshua