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How to read current instead of voltage?
#1
Posted 06 December 2011 - 05:13 PM
#2
Posted 06 December 2011 - 05:23 PM
#3
Posted 06 December 2011 - 05:31 PM
Voltage is easier to measure, you can simply add a resistor with known value to the sensor output and measure the voltage drop across it using Netduino's analog input. The current is then calculated using the Ohm's law, I = U/R (R is known). You need to choose the resistor value so the voltage does not exceed Netduino's analog input limit (3.3V) and has appropriate power rating (it has to dissipate U*I power as heat).Do I need to convert this into volts first, somehow?
#4
Posted 06 December 2011 - 06:19 PM
#5
Posted 07 December 2011 - 08:30 AM
Both the above indications are correct on to perform a measure, but bear in mind that the upper limit is +3.3V, not +5V.
So, the actual resistor value should be:
R = 3.3 / 0.02 = 165
Use a couple of 330 Ohms resistor in parallel, et voilà!
Cheers
Thanks Mario, but excuse my ignorance here - what is the unit of measurement for the 165? If it's ohms, then why would I need a 'couple of 330 Ohms resistors' if all I need is a resistor for 165? Or will the resistors bring it down to 165? Maybe I'm reading it incorrectly
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Thanks again for your help
#6
Posted 07 December 2011 - 08:48 AM
Yes, it is ohms. You can simply use one 165Ω resistor, if you can get it - Mario has suggested two 330Ω resistors (in parallel), because 330 is much common value than 165 (330 is in the E6 standard EIA decade values, while 165 is in E96 - but these should be rather easy to source nowadays).Thanks Mario, but excuse my ignorance here - what is the unit of measurement for the 165? If it's ohms, then why would I need a 'couple of 330 Ohms resistors' if all I need is a resistor for 165?
#7
Posted 07 December 2011 - 09:22 AM
Yes, it is ohms. You can simply use one 165Ω resistor, if you can get it - Mario has suggested two 330Ω resistors (in parallel), because 330 is much common value than 165 (330 is in the E6 standard EIA decade values, while 165 is in E96 - but these should be rather easy to source nowadays).
Thanks. So is the aim of putting two 330Ω in parallel to cut their resistance in half? How does that work?
#8
Posted 07 December 2011 - 09:33 AM
Thanks. So is the aim of putting two 330Ω in parallel to cut their resistance in half? How does that work?
Oh, well...the quick formula for just two resistors is:
R = (R1*R2) / (R1+R2)
More generalized, for any number of resistors in parallel, is:
R = 1 / (1/R1 + 1/R2 + ... + 1/Rn)
BTW, from the last one, you can easily get the first formula.
The resistor stands as the brake, thus you may associate the resistance to the friction between two materials.
So, the higher is the resistance, the higher will be the friction.
I'd like to imagine as the friction that the water (i.e. current) has inside a pipe. If you have just one pipe having a friction of "330", two pipes lead a better water flow...
Hope that clarify.
Cheers
#9
Posted 07 December 2011 - 09:47 AM
#10
Posted 07 December 2011 - 09:59 AM
Yes, it is ohms. You can simply use one 165Ω resistor, if you can get it - Mario has suggested two 330Ω resistors (in parallel), because 330 is much common value than 165 (330 is in the E6 standard EIA decade values, while 165 is in E96 - but these should be rather easy to source nowadays).
CW2,
I think reactive wants to know how two 330 ohm resistors in parallel would magically get him to the value of 165 ohms.
To answer that questions you have to understand the difference between resistors in series and parallel. In series, resistors are additive - that is the total resistance is equal to the sum of all the resistors in the chain. So for resistor values of 10, 47, and 100 the total resistance would be 157 ohms (10 + 47 + 100).
In parallel, it divides. with the formula being R = 1 / ((1/R1) + (1/R2) + (1/R3)...). So for values of 3, 9 and 18 the resistance would be 2 ohms -
1 / (1/3 + 1/9 + 1/18) = 1 / (6/18 + 2/18 + 1/18) = 1 / (9/18) = 1 / (1/2) = 2 ohms
In the case of two 330 ohm resistors the calculations would be:
1 / (1/330 + 1/330) = 1 / (2/330) = 1 / (1/165) = 165 ohms
-dan
#11
Posted 07 December 2011 - 12:09 PM
CW2,
I think reactive wants to know how two 330 ohm resistors in parallel would magically get him to the value of 165 ohms.
To answer that questions you have to understand the difference between resistors in series and parallel. In series, resistors are additive - that is the total resistance is equal to the sum of all the resistors in the chain. So for resistor values of 10, 47, and 100 the total resistance would be 157 ohms (10 + 47 + 100).
In parallel, it divides. with the formula being R = 1 / ((1/R1) + (1/R2) + (1/R3)...). So for values of 3, 9 and 18 the resistance would be 2 ohms -
1 / (1/3 + 1/9 + 1/18) = 1 / (6/18 + 2/18 + 1/18) = 1 / (9/18) = 1 / (1/2) = 2 ohms
In the case of two 330 ohm resistors the calculations would be:
1 / (1/330 + 1/330) = 1 / (2/330) = 1 / (1/165) = 165 ohms
-dan
Nailed it! Much appreciated
#12
Posted 26 December 2011 - 09:32 PM
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#13
Posted 26 December 2011 - 10:54 PM
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#14
Posted 27 December 2011 - 06:13 AM
#15
Posted 27 December 2011 - 03:20 PM
Not sure about what you man for "grounding loop". Is the probe cable very long? On what are you worried about?
Thanks Mario.
I have been reading different information on-line about multiple 4-20mA sensors being run from the same power supply. It seems that multiple sensors with the same power source will effect each other through what is known as a "ground loop". The universal solution (again from what I've been reading) is to use a "ground loop isolation module". These are small boards that are connected in-line with the sensor and removes the ground loop effect. The problem is that they are about $175 each for two channels. If I had say.... 48 sensors this would get costly.
Terry
#16
Posted 27 December 2011 - 04:44 PM
I think what they mean by ground loop is that if the sensors are far apart or not grounded close to each other, there may be a voltage differential that affects the measurement due to the fact that ground is not actually 0 Ohms.
If you have the 165Ohm resistors close to your netduinos, i.e. all the sensors are grounded in the same spot, that should take care of it. This is also referred to as "star ground" and is very important in audio amplifier applications, since there you actually "hear" any ground loop problems as the 50/60Hz hum in your speakers.
Here's some additional detail if you are interested: Star Grounding
HTH,
Mike
#17
Posted 27 December 2011 - 04:56 PM
#18
Posted 27 December 2011 - 07:16 PM
#19
Posted 27 December 2011 - 07:48 PM
#20
Posted 27 December 2011 - 07:53 PM
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