# Differences

This shows you the differences between two versions of the page.

 electronics [2018/10/09 19:29]yehowshua electronics [2018/10/09 19:29] (current)yehowshua 2018/10/09 19:29 yehowshua 2018/10/09 19:29 yehowshua 2018/10/09 19:23 yehowshua created 2018/10/09 19:29 yehowshua 2018/10/09 19:29 yehowshua 2018/10/09 19:23 yehowshua created Line 1: Line 1: - I believe I have achieved the correct derivation. + *[[How OpAmps Really Work]] - + - 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 + - + - $$\texttt{EQ 8.} A \cdot u(t)e^{-(\frac {A\cdot R_g}{R_g+R_f})t}$$ +
• /var/www/html/container/dokuwiki-2018-04-22a/data/pages/electronics.txt