# I'm trying to apply calculus to electronics.

• Aug 2nd 2010, 07:26 PM
Zanderist
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 $\displaystyle {V}_{EBO}$ ) 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:

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

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

So my only two variables are $\displaystyle {R}_{1}$ and $\displaystyle {R}_{2}$. 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)
• Aug 3rd 2010, 02:04 AM
Ackbeet
I'm confused. Why do you need to take the derivative? If you're looking for $\displaystyle R_{2}$ when $\displaystyle V_{\text{out}}=V_{\text{EBO}},$ then why not simply plug in $\displaystyle V_{\text{out}}=V_{\text{EBO}}$ into your equation and solve for $\displaystyle R_{2}$?
• Aug 3rd 2010, 07:32 AM
yeKciM
put the scheme that u'll try to do :D (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 :D

it's done by basic laws :D (don't need to complicate anything)
• Aug 3rd 2010, 05:34 PM
Zanderist
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.
• Aug 4th 2010, 02:22 AM
Ackbeet
Quote:

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)?