It’s been quite a while since I’ve taken Calculus, so asking an AI chatbot might be a good starting point if you don’t know what references to use. Computers are pretty decent at mathematics, certain tools might help explain a mathematical principle in terms that you can better understand. I utilized Microsoft’s Co-Pilot to help me figure this portion out. It explained that a good approach is to use two related equations with something called a La Grangian multiplier. Don’t worry if you have never studied this method or haven’t taken Calc in years. The goal here is to simplify the math as best as I can.
In my first post, we learn that we should not exceed 55mA on the discharge pin of the 555 timer. That gives us the following equation:
This is going to be called our “constraint equation” (or g(x)). If you solve CT from the period equation, you get the following:
You can now consider “Ra + 2Rb” as a “term” because it is inside parenthesis. We are no longer trying to find arbitrary values for Ra and Rb. We need a specific combination that allows Ct to become as large as possible while obeying the 55mA current limit. This will be our “S” equation (in the La Grange setup it is the same as f(x)):
The format for La Grange is L = f(x) + λ*g(x). Here is what the full equation would look like with our values:
Take the partial derivatives with respect to all variables. If you have never come across this, all this means is that you take the derivative with respect to that particular variable. Any other term by itself is treated as a constant (any number). A derivative of any number is 0. An excellent tool to use for this is Wolfram Alpha. It’s easier to do this one step at a time.
It turns out when you differentiate with respect to lambda, you will just get the constraint equation again. In our case, it is irrelevant because now we have Ra and Rb by themselves. With La Grange, you just take these terms, set them equal to zero, and then solve for lambda.
Since lambda is two results, you can set them equal to each other and solve for either Ra or Rb. I decided to solve for Ra.
Instead of including square roots, it’s easier to write the approximate form:
We can take this value and substitute it back in our constraint equation:
Solving for Rb is a lot more simple now. That ends up being about 113 Ω’s. If you know Rb, you can use the constraint equation again and solve for Ra. Ra would approximately be 196 Ω’s. Looking back to the solved equation for Ct above, we can now substitute the values in (remember, we know what T is: 1/60 or 0.0167):
This is where the 57.105uF comes from. This ensures knowing what value we should not exceed to overdrive the Q14 transistor at 60Hz. In other words, I cut it really close with 50µF, 250 ohms for Ra, and 110 ohms minimum for Rb. Let me be clear on this, do not take the 57.105µF as a recommendation! Aim for a much lower capacitance. My last comment on the calculus, the resistor values obtained from this analysis should not be interpreted as a universal minimum. They are the values associated with the La Grange process of “optimization.” Different capacitor values will generally require different resistor combinations and several combinations can still manage to be below 55mA total.