Will this give us three individual voltages to the microcontroller (analogue pins in this case) - or will the state of one resistor affect the voltages across the other resistors? (I reckon they'll all be individual since they appear to be in parallel: but I figured I'd double check before throwing anything together)
Many thanks!
Yes this give three individual voltages.
Don't worry, it works
Pelle
Thanks very much xD
I wasnt sure if I had to connect them each individually to +ve / -ve xD
Many thanks!
Caveat: The resistance between Vcc and Ground will be reduced by having parallel resistances to ground, so you could end up with current-draw issues if you use too many, or they're low resistance pots. If you're using >10k each, not much to worry about -- but you should be aware, and know how to calculate it, especially if you're using batteries.
In this particular circuit there are 3 in total: 2x 1k and 1x 10k
Can you just confirm it's 1/Rt = 1/R1 + 1/R2 + ... 1/Rn for n resistors? (and then just use V=IR for current/resistance as required?)
The 10k resistor will be tied to a resistor before going to the probe (it's part of a mini-oscilloscope) - might it be an idea to shove a few resistors (say, 470 ohm) in front of all the resistors (I'm a learned-n00b at electronics xD: I'm still learning good an bad practice - so thanks in advance for the feedback!)
cprobertson1:
In this particular circuit there are 3 in total: 2x 1k and 1x 10kCan you just confirm it's 1/Rt = 1/R1 + 1/R2 + ... 1/Rn for n resistors? (and then just use V=IR for current/resistance as required?)
The 10k resistor will be tied to a resistor before going to the probe (it's part of a mini-oscilloscope) - might it be an idea to shove a few resistors (say, 470 ohm) in front of all the resistors (I'm a learned-n00b at electronics xD: I'm still learning good an bad practice - so thanks in advance for the feedback!)
Vcc= 5 volt
Analog input current ?0
every 1k? pot needs 5 mA (U/R=I)
10k? pot needs 0,5 mA
Gives at total of 10,5 mA (3 pot)
Yes 1/R= 1/R1+ ....and so on are correct but you do not need to use it for this calculation.
Why a 470? resistor?????
In series with the 1k? pot it gives an output signal from 0 to 3.40 volts as maximum!
A little bit better with the 10k?, up to 4.78 volts
Perhaps the 470 should be connected to the pot tap, still why?
Pelle
I just have a huge bundle of 470's lying around 
(around 100 of them xD: I recycle my old electroncis and they seemed to be (very) popular on this old boiler controller board xD)
Let them lying around, do not connect them to the potentiometers.
Pelle
To clarify about why you would or wouldn't want to add resistors to your potentiometer:
You will usually need to select the pot's resistance based on a few things -- a balance of current draw vs. noise (higher resistance causes more noise in a circuit, but low resistance draws excess current), to suit some formula (e.g., using it as a rheostat in a feedback loop to control gain), and so on.
When you're just setting up a voltage divider (having the wiper output a voltage somewhere between that of the two terminals), the resistance of the pot isn't so important -- except of course the associated current draw aspect. You can place a resistor at either end terminal if you want to "bias" the range -- e.g., if you want to ensure there's always a minimum value above the lower terminal, or a maximum value below the higher terminal. It's kind of like putting a physical stop somewhere before the end of travel on the wiper. If you put the same value of resistor at both ends of the pot, it's like using a pot with a higher resistance -- you're not changing the balance any, just decreasing current flow. (Useful if you don't have the right value pot, but you can fake it with resistors on-hand.)
Using a resistor in series with the wiper is rarely useful, but I'm sure there are times when it would be necessary. However, placing one between the wiper and one of the end terminals is handy to change the output curve from linear to some exponential shape. (Commonly used to fake a logarithmic output curve for audio applications.)
Of course, there's nothing inherently wrong with sticking resistors any old place, but you should know why you're doing this rather than just doing it to make it feel more correct. 
Finally, huge nod of agreement on the circuit sim. I use that thing all the time.
Let them lying around
Lol xD Okaydokeys
Thanks a bundle SirNickity! That's helped a lot! xD
My reasoning for having in place was that I was worried about shorting the pin on the arduino if the potentiometer was set to minimum resistance (do they actually go to 0 ohms? I should probably have checked that first... I don't have the datasheet handy - will look it up later (I'm just in from work!))
Thanks to everybody by the way for the assistance And sorry for being a n00b. Well... I know everybody has to start somewhere - but I always feel guilty about taking up other people's time! So thank you all!
The resistance between either end terminal and the wiper changes, and can be anywhere from 0 ohms (ish) to the maximum resistance of the pot. That's normal, and is usually what you want. There's no harm connecting an input pin directly to either Vcc or Ground, so there's likewise no harm in having little to no resistance through the pot. An input pin has very very high resistance, as it doesn't need to consume current, so it's safe.
Remember (because this is a mistake I made early in my days), when you turn the pot, you don't change the resistance between the two end terminals. That's always 1K (or 10K, or whatever..) You only change the resistance between each end to the wiper. The wiper forms a voltage divider with continuously variable resistances -- 0R on one side, 1K on the other, or 500R on each end, or any combination thereof. Still 1K from end to end, regardless.
cprobertson1:
Let them lying around
Lol xD Okaydokeys
Thanks a bundle SirNickity! That's helped a lot! xD
My reasoning for having in place was that I was worried about shorting the pin on the arduino if the potentiometer was set to minimum resistance (do they actually go to 0 ohms? I should probably have checked that first... I don't have the datasheet handy - will look it up later (I'm just in from work!))
Thanks to everybody by the way for the assistance
Hmmm
Short the pin to ground?
The "pin" must be an INPUT not an output
Pelle
You can connect input pins directly to 5v Orr ground. They are high impedance and hardly draw any current at all.
My reasoning for having in place was that I was worried about shorting the pin on the arduino if the potentiometer was set to minimum resistance (do they actually go to 0 ohms? I should probably have checked that first...
Make sure that the Arduino pins are set to inputs, not outputs.
Aha! Of course!
For some reason I was treating them as if they were outputs! Auch - sorry - that was a truly daft moment Don't know why I was thinking there was current coming out of them... that's weird =/
Thanks you all very much for clearing that up 