Backyard Aquaponics
A place to discuss aquaponics

"Exploded" is the default position of my brain.

RE: the half way point (or any other point) I know that's what happens because I've played with the MM, but I cant see how.

I come from a position of not understanding how this stuff works in any way. As a result I cant tell you which bit I don't understand.

It says something about electronics that this should be the case. Or it says something about my brain. Either way, I cant see how it doesn't matter if you use a 10k pot or a 100 k pot.

I know it doesn't matter, and I rely on it not mattering when I made that part of my fish feeder, but that doesn't help me understand why

Does the chip know what the maximum and minimum voltage is? And does it use it to reference the 50%? If so how does a voltage divider work without a chip and without any reference to the maximum voltage?

No the chip only knows the max and the min, in this case (I'm assuming) +5v and 0v.

Pots are basically wire-wound resistors with a "tap off" point.
So as you move the shaft, it alters the "tap off" point of your 2 series resistors.
So, with a 100k pot in the center position, you have 2 x 50k resistors in series, which will always give you 2.5v regardless of the overall value of the pot.
Move the shaft 75% to the right and you now have:
75K and 25k in series, giving you 75% of the supply voltage.

Hope this helps and doesn't further confuse things.
Try to see a pot as 2 variable resistors connected between the supply, not just 1

The ADC gets its reference from the chips power supply. So it divides its ADC range 0-255 up to fit between 0v and the 5v supply. So 0v will be ADC 0 and 5v will be ADC 255 (or whatever the max of the ADC is)

If you connect the chip up to 4.9v then the ADC range will be between 0v and 4.9v, anything over 4.9 will still read the max (255) on the ADC (too high will kill it obviously)

Obviously you cant connect the chip up to 12v because that will let the smoke out.



Yes, it doesn't "know" where 50% is, but when the input voltage on the ADC is at 50% it just gives you the ADC number of 255/2 which is 127.



A voltage divider doesn't need anything attached to it, its just 2 resistors connected together.
If you use 2 equal resistors then the midpoint of the resisters will ALWAYS be half of the voltage that is applied across both. To say another way, if you connect 2 equal resistors (any values, just equal) between 0v and Xv then the midpoint of the resistors will always be Xv/2.
X could be a billion volts, you will just get half a billion at the midpoint.
The midpoint voltage is equal to a RATIO of the values of the resistors.

There might be other reasons to connect two series resistors between 0v and 5v, its only called a voltage divider if you decide to use the voltage at the midpoint of the 2 resistors.

We are not going to give up on you Bullwinkle Read our posts a couple of times... all of a sudden... DING!

Changing the value of the two pots changes the current flow, not the voltage.
There will be more current flowing through the 10k pot than the 100k pot.

Remember ohms law? I = V / R
So at 5v, a 10k pot gives (5 / 10,000) 0.5mA
a 100k pot gives (5 / 100,000) 0.05mA

Simple

So the ADC does a select case type of thing by dividing the the high voltage by 255 and assigns a unit value to each step. In the case of a 5 volt power supply, that would be a "step" every .02 volts or so.

The pot acts as a voltage divider by effectively creating a pair of resistors, one above and one below the output pin to the chip. This gives us everything from 0 to 5 volts and the ADC functionality of the chip assigns a value to that voltage ranging from between 0 and 255, where 2.5 volts (pot set at 50%) would return 127 as a digital value.

My confusion stems from the fact that if the two resistors (one above and one below the point at which I take output) are acting like a tap and restricting the flow, I can see how the second "tap" diverts more flow to the output pin the higher it's value, but I still struggle with why the same flow goes through the first "tap" regardless of that "tap" being half a 1k pot or half of a 10k pot. This seems to me, that a .5k resister and a 5k resister are both doing the same amount of resisting.

I can use it's functionality, but find the behaviour of the first "tap" counter intuitive. I prefer my understanding to be intuitive

I suspect my problem is some massive failing in my understand of resistors, and I suspect that stems from my lacking formal education, and lacking mathematical ability.

The "resistance" is to current...
ALL of the voltage "flows" through the resistor, you have +5v at one side, 0v at the other.
Varying the value of R just varies the current flowing from 5v to 0v.

By putting two same value R in series, you "drop" half the voltage across each.
By varying R (within the pot), you "drop" a varying percentage of the voltage across each.
By varying the overall R value of the pot, you still "drop" those same percentages of voltage, but at a different current.

Hey Bullwinkle I just thought of a great (I hope) analogy.
First let me just say that Voltage is the "force" or "pushing force" (its actually called EMF or Electromotive force) and current is the amount of "flow" (its actually numbers of electrons per second)

The analogy is water flowing through a pipe.
First of all pushing force or potential difference is referenced to the ground (meaning it is 0 at ground level)
The flow of the water measured in L/hr or L/s is the Current

1. Think of a water tank on a tower, the height of the tower is equal to 5 units. Think of this as a "pushing force" of 5 (pressure maybe?) There is also a tank on the ground (if you want)
2. At the ground level the pushing force is 0.
3. Now there is a pipe that goes from the top tank into the bottom tank.
4. This pipe has 2 sections, joined in the middle, Lets say there is a T in the middle, where we can get some water out, but just a tiny little tap, not much water flow.

Explanation (I hope)
The height (voltage) of the tank is 5
The Height (voltage) of the middle is 2.5 - this doesn't change..yet.
Each section is a resistor and must stay equal in size(trust me).
So the flow of the water through the pipe is the Current.
If you make the pipe bigger you get more flow or Current, smaller, less flow

Now if you are able to slide the T up and down you are changing the height and there for the Potential difference or voltage. It doesnt matter how big the pipe is going to the ground, you are only changing the Potential DIFFERENCE from the T to the ground.

Voltage is the Potential DIFFERENCE between 2 points. The potential difference between something at 5v and something at 8v is 3v, the SAME potential DIFFERENCE between 0v and -3v.

Current is the flow of electrons. So between 2 points if you increase the Voltage, you increase the flow.
OR if you decrease the Resistance you also increase the current / flow.

Hmmm, hope that helps. Think that is a good explanation guys ?

Hmmm it does, doesn't it.

Yeah ok.

I think I understand, and I think I've been concentrating on the water flowing, rather than the pressure difference.

and I need one of those sliding T junctions in my aquaponics system

yeah I realised in one of your posts you didnt get the voltage vs current concept.
I hope its all becoming clear now

Float switch 200v 1A horizontal waterproof

There are a few different ones at Altronics. FYI





http://www.altronics.com.au/index.asp?a ... &id=S1160A

It is, but it's still a bit counter intuitive.

It's never a good idea to learn things incorrectly. Once you get those wrong ideas stuck in your head, it's really hard to shake them loose.

I'm on a borrowed windows system at the moment, but on my linux box that blew up, I was running an electronics simulator called Ktechlab, which made all this stuff easy to understand, because you could just drop components in and see the volts and amps all the different points. I cant wait to get it back.

Definitely worth a look for anyone running linux or willing to run it in a window.

Ktechlabis free and open source.

Its a circuit design and simulator IDE that simulates all the time, so as you add a component, its live as soon as it hits the circuit.

It seems to be actively under development, and has chip programming and simulating functionality.

As a solution to my near fish death experience?

I cant remember who is was but someone really needs that. A user here in the tropics. The device looks all plastic and wouldnt corrode.

Yeah it looks pretty good. I'm assuming it uses a reed switch, which is a small magnetic switch that switches when in comes close to a magnet, which would be in the floating part.