# Thread: I'm trying to apply calculus to electronics.

1. ## I'm trying to apply calculus to electronics.

What I'm basically trying to do is have transistors turn fully on when a certain resistance is achieved on a dial (potentiometer). Each transistor's middle pin (base pin) has a fixed resistor in series with it, this is to create a voltage drop. As the transistors turn on they allow charge to flow to an led.

The transistors turn on when 5 volts ( this limit is what it's design parameters are, this limit will be referred to as ${V}_{EBO}$ ) is applied to that base pin.

Now for the calculus question:

The formula I am using is the one from the Boards Manual for voltage drop.

It is as follows (I've added in the limit, which is not part of the actual formula:

$\frac{({V}_{in} * {R}_{2})}{({R}_{1}+{R}_{2})}= {V}_{out} \lim_{{V}_{out}\rightarrow {V}_{EBO}}$

The only thing that's fixed for that formula in this circuit I am designing is ${V}_{in} (6 volts)$

So my only two variables are ${R}_{1}$ and ${R}_{2}$. To find the Derivative, could I just do the quotient rule or do I have to make use of Implicit differentiation? Basically which part of calculus should I be looking into to solve this.

(Just a note, R1 in this case is the dial resistor(potentiometer), R1 in it's self has it's own limit, 0 to 10K (10000 ohms). What I'm searching for is different values for R2 which is an actual fixed resistor when voltage will equal 5volts. I hope you all understand what I'm looking for. I know that what I'm looking for will require calculus, which makes me happy that I can apply it)

2. I'm confused. Why do you need to take the derivative? If you're looking for $R_{2}$ when $V_{\text{out}}=V_{\text{EBO}},$ then why not simply plug in $V_{\text{out}}=V_{\text{EBO}}$ into your equation and solve for $R_{2}$?

3. put the scheme that u'll try to do (u don't need to derivate anything)

when u have voltage divider ( your two resistors) they are connected in such way that when manipulating with them, you decide how much u wont to put voltage on transistors "base-emitter" connection so that your transistor will no longer be "open switch" and become "closed switch" so your led should glow

it's done by basic laws (don't need to complicate anything)

4. I actually got some results when I just did it with Algebra last night, but I did spend the rest of the night trying to figure out why both LEDs came on.

But still, if I wanted to find the derivative, would I use implicit differentiation.

5. But still, if I wanted to find the derivative, would I use implicit differentiation?
That depends on what is a function of what, and what your equation looks like. What's your independent variable(s), and what's your dependent variable(s)?