Signal Calculation

It is a linear function. y=mx +b

the function reduces to...

(mA - 4) / 0.16 = %

For your question...

(4.5 - 4) / 0.16 = 3.125%

Hi, could you let me know where 0.16 come from?

Rgds,

Hien

Y = mX+b

Where:
Y = Y axis, 4 = 20 ma
m = slope of the line, delta Y / delta X
X = X axis, Full range say, 0 - 100%
b = the Y axis offset, 4 ma

The slope = m = delta Y / delta X = (20 - 4) / 100 = 0.16

Re-arranging Y = mX+b to find % value:

(Y - b) / m = X

Now lets say that the current being measured is 12ma - mid range of the 4 - 20 miliamp range

Substituting values:

(Y - b) / m = X:
(12 - 4) / 0.16 = 50

AussieBod, could you let me know what is the name of this equation and its application range? We learned it in school? High school or what else?

I dont know if I should post such a silly question here cause' this forum is just for pro. engineer. I am sorry for this.

Rgds,

I thought it was just the equation of a straight line, but its been a while so I did a check, it is called the slope intercept form of the equation. Try this

http://zonalandeducation.com/mmts/functionInstitute/linearFunctions/lsif.html

http://www.mathsisfun.com/equation_of_line.html

the max signal is 20ma, minus the 4 ma offset= 16

the max percent is 100

16/100=.16, the value for each %

It might not necessarily be linear. It might be a sine wave, square wave, etc.

the whole point for having a 4 to 20 ma is to have some current indicating the low end of the transducer and 20ma the high end. Zero ma is not recommended because this does not differentiate if you have an open. However, a 4 to 20 ma output is proportional in a linear fashion to its input and the relationship is indeed a linear relationship with an offset of 4ma by the equation y=mx+b.

Hope this helps.

One thing to be careful when applying any formula is to know the limits of its application. You asked for a conversion to percent signal and that's what this equation gets you. But in a typical use it gets you only halfway to where you want to be and that is "what are the engineering units?". Most of the time it is simply multiplying your percent signal (i.e. 0.03125) by the span of the instrument. If that is a pressure transmitter spanned for 0 - 100 psig for example, your pressure is 3.125 psig. But if you are using an orifice plate to measure a flow then the flow rate in engineering units is proportional to the square root of the percent. For a flow transmitter spanned 0-100 gpm this would be the square root of 0.03125 or 0.17678 * 100 giving you about 17.7 gpm.

Hien,

Here is how to calculate when the signal is at 4.5mA or whatever

% = (100%/16mA) x (mA - 4mA)

for example

for 12mA % = (100%/16mA) x (12mA - 4mA) = 50 %

for 4.5mA % = (100%/16mA) x (4.5mA - 4mA) = 3.125%

This work for linear system

Regards

JP

Hi all,

Tks for all of your support,

Rgds,

Hien

Without the linear equation, pure logic solves it too. Total range is 16 units. At 4.5ma, you are now 0.5 units into the range. What percent of 16 is 0.5? Same answer as the equation, 0.5/16 = 0.03125 or 3.125%. Similarly 12ma is 8 units into the range, so 8/16 = 0.5 (50%). (this really is the linear equation, but sometimes a bit easier to remember for people who have trouble remembering equations)