The Innovation First RC Reference Guide tells how connect a potentiometer to an Analog Input. He suggests the use of 250Ω – 100KΩ potentiometers.
I have a lot of 22R potentiometers and I would like to know if you can use them?
if not, why?
thaks
Gustavo
brVex
Don’t use them.
Your potentiometers have a resistance of 22 ohms. That will place a relatively high load on the sensor power supply. At 5 volts, it will be drawing more than 200 milliamps and constantly dissipating more than a watt of power. It will probably “work”, but it’s outside the recommended range of values by an order of magnitude.
Gustavo -
The answer is depends on if you intend to use the IFI system or the 2009 system.
I do not know the IFI answer, so I will not attempt to fabricate one.
For 2009, the minimum resistance of the potentiometers is dictated by current draw and heat dissipation. You may draw up to 750mA for one analog sensor* (3A total), which would limit you to about 7 ohms. However, this would dissipate about 4 Watts, which is probably not what you want.
At 22 Ohms, you will dissipate roughly a watt. Make sure that your potentiometer can handle it.
The maximum resistance is dictated by the input impedance of the NI9201 Analog Input Module. If your potentiometer’s resistance is too high, the small current the 9201 draws to measure the voltage will impact the reading. Since I do not know the NI answer, I will not attempt to fabricate one.
Please note that my post only represents the capabilities of the system and do not reflect the robot rules. Robot rules may place additional limitations on the system.
At the risk of looking like an EE neophyte, can you explain how amperage draw, the resistance (ohms) and the energy dissipated are related? There must be a formula for that.
Thanks
Sure, I used these to remember:
V = C R (even if it blinks 12:00 forever)
P = V C
V(olts) C(urrent) R(esistance) P(ower)
To do your calculation, I took:
Power = Voltage^2 / resistance
To double check, you can use the google calculator - It does units! Just google “(5V)^2/22 ohms”
While you may have done this to simplify things for non EEs, I find your choice of symbols for current very unique. Most, if not all, EEs use the symbol “I” for current. I thought you were talking about capacitance there for a moment. :rolleyes:
Wait, there’s more!!
Being a EE, I too found the symbols a little odd, but I understand their usage for the purpose of remembering easily.
In fact, the way I remember is:
P=IE ====> Power=Current *Voltage
That’s right, “I” is for current AND “E” is for Voltage.
In actuality, “E” is for “Electro Motive Force”.
Honestly, I don’t quite remember exactly why “I” is used for current. Maybe one of our other EE’s can fill us in?
(Forgetting is what happens when you don’t use what you know. Or, maybe it is what happens when you get old. Darn, I forget ) 
So, if the potentiometer is of 5W I do not have problems?
What about the analog input value range?
Sorry for my english!
Gustavo
And to break it down into words:
Voltage (in volts) = Current (in amps) * Resistance (in ohms)
Power (in watts) = Voltage (in volts) * Current (in amps)
This is for a DC system. AC is similar, but you need to be careful what voltage you use.
I used C(urrent) instead of I(mperage?) to make it easier for non EEs. I find VCR and PVC easy to remember.
But yes, I use “I” in real life.
Gustavo,
To answer your question about whether the 5W potentiometer is okay -
Yes it’s okay to use but I would recommend that you find something like a 220Ohm or higher potentiometer with 1/4W power rating. It’s generally wasteful to use > 1W for a sensor and it will probably get warm. But it will work.
There is an upside to using a 22 Ohm 5W pot, though - it’s probably a very durable part and unlikely to be damaged during competition. There’s something to be said for ruggedness.
current needed = 5V over a 22 Ohm resistor = 5/22 Amps = 0.23 Amps (which is less than the 0.75 Amp rating for each pin of the Analog Breakout module)
power dissipated by the 22 Ohm resistor = 5V * 0.23 Amps = 1.14W (which is less than the 5W rating for the part)
The 2009 Analog Breakout module can supply enough power and the NI 9201 Analog Input module will provide 10 bits of resolution between 0 and 5V (ie the full range of the potentiometer).
Be very careful to connect the wiper of the pot to the analog input. If you connect 5V power to the wiper, you’ll have a variable current draw from the power supply which will short circuit the supply at one end of the potentiometer’s range.
Russ
Thanks for the basic EE training…
Sorry about hijacking your thread Gustavo:o
