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-   -   Are 100k Ohm pots necessary? (http://www.chiefdelphi.com/forums/showthread.php?t=32935)

kirov 19-01-2005 23:39

Re: Are 100k Ohm pots necessary?
 
hi, value of pots depends on current, check the voltage and you can also calculate the pots value by yourself(sorry I dont recall the formula for calculating). but if you use pots which are in lesser value than the recommended, then the function of particular thing may be suffured. bye :cool:

Max Lobovsky 16-02-2005 22:15

Re: Are 100k Ohm pots necessary?
 
I'm sorry, am I an idiot or is it not possible for RC to use anything from 250 - 100k ohm potentiometers without something else in the circuit? For example, how can the RC possibly tell if its connected to a 10k ohm in the mid position, or a 5k ohm at the max position? Will you just be limited to some small section of the resolution if you use poteniometers less than 100k ohm?

Dave Flowerday 16-02-2005 22:25

Re: Are 100k Ohm pots necessary?
 
Quote:

Originally Posted by Max Lobovsky
For example, how can the RC possibly tell if its connected to a 10k ohm in the mid position, or a 5k ohm at the max position?

A potentiometer is really just a variable voltage divider. Using Ohm's Law (voltage = current multiplied by resistance or V=IR), then at 5 volts there is (5V / 100K Ohms) or 50 uA of current flowing across the pot. Now, if you set that pot at its mid-point, you have 50K on either side of the resistor. Again, using Ohm's Law, you have V = (50uA * 50K Ohms) = 2.5 volts of a drop across that half of the potentiometer.

If you repeat this exercise for a 10K pot, you'll find that if it's set at its mid-point (5K Ohms on either side), you still have 2.5 volts dropping on either side. So in either case (10K or 100K), with the wiper set at mid-point, the RC sees 2.5 volts. The only difference is that at 10K more current is flowing. The reason they limit you to no less than a 250 Ohm pot is because any pot with a smaller resistance will draw more current than the RC is capable of providing.

Hopefully at least some of that made sense ;)

Max Lobovsky 16-02-2005 22:32

Re: Are 100k Ohm pots necessary?
 
Ahhhh. I understand Ohm's law and such, I just didn't realize that potentiometers usually had the a 3rd lead that has the full resistance between it and the other non variable lead. Well that is very convenient... And it did make sense (I just needed to read the part about "volts dropping on either side"), thanks.

ConKbot of Doom 18-02-2005 00:14

Re: Are 100k Ohm pots necessary?
 
Here is what I'm wondering, if the OI uses an internal resistor for the "bottom half" of the divider leg, if you were to use a joystick with 10k pots, would the current drawn by the pull down resistor make the output of the joystick that far off? I know it would add some non-linearity. But if there is a joystick you really want, couldn't you fix that in code? I know my saitek cyborg (my own personal one, were not trying to use it...) has the pots wired up as dividers, not rheostats like the white sticks we've had for the past two years.

Of course if you are really really bent on using a particular joystick, just program a PIC to use the ADC to sample the joystick axis value and output the value to a 100k digi-pot. :D

gburlison 18-02-2005 00:36

Re: Are 100k Ohm pots necessary?
 
If you read the documentation for the OI it tells you to only wire +5V and the wiper for a potentiometer connected to the OI. We mistakenly wired the third leg of our 100K potentiometer to ground and had some strange results. After reading about the pullup circuitry, our strange results make sense.

Kyle Fenton 18-02-2005 08:37

Re: Are 100k Ohm pots necessary?
 
Rhode Warrior 8 uses 10k pots, and we haven't had any problems with them. The numbers on the programming output always vary a little, but we built in a tolerance factor to correct for that.

Al Skierkiewicz 18-02-2005 09:16

Re: Are 100k Ohm pots necessary?
 
Quote:

Originally Posted by ConKbot of Doom
Here is what I'm wondering, if the OI uses an internal resistor for the "bottom half" of the divider leg, if you were to use a joystick with 10k pots, would the current drawn by the pull down resistor make the output of the joystick that far off? I know it would add some non-linearity. But if there is a joystick you really want, couldn't you fix that in code? I know my saitek cyborg (my own personal one, were not trying to use it...) has the pots wired up as dividers, not rheostats like the white sticks we've had for the past two years.

Of course if you are really really bent on using a particular joystick, just program a PIC to use the ADC to sample the joystick axis value and output the value to a 100k digi-pot. :D

As Dave Flowerday pointed out earlier, the OI is a different beastie than the RC. Joystick ports are designed to take into account the need to produce a solid 127 if the joystick is disconnected from a port so that a robot doesn't drive around by itself. (The actual spec is less than a 0.05 volt input) The 100K on the OI is different from the discussion on analog inputs on the RC. The 1k OI analog input resistance is not a "pull down" but a current limiter. Since a 10k pot willl produce more current, it will be converted differently than the current through a 100k pot. The result is control non-linearity which can be overcome through software, driver practice and speed contoller calibration.
While we are discussing the OI, the current rules are specific about the devices you can and cannot connect to the OI input ports. A PIC controller that interprets an odd joystick and passes it along to the OI must use the joystick power, which is current limited.

Stephen Kowski 18-02-2005 09:30

Re: Are 100k Ohm pots necessary?
 
Quote:

Originally Posted by BrianBSL
Sorry about that. The part # is RV4NJ104C-ND. It is made by Precision Electronics Components or something like that.


Does the same company make multi-turn pots also?


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