8/16/2007 8:09 PM | |
Posts: 30 Rating:
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I have a temperature transmitter that has a 1 - 5 VDC output that corresponds to 0 - 135 deg F. This is wired to 0 - 10VDC analog input. I am having trouble determining the gain and offset to convert this to engineering units. I ultimately want to display the temperature in the message text. I have the message text working fine but the scaling is wrong. Any help would be appreciated! |
8/17/2007 4:21 PM | |
Joined: 1/5/2007 Last visit: 4/7/2023 Posts: 1690 Rating:
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Hello, use an amplifier immediatebetween an analoge input (ai) and an output (ax). For the ampilfier is ax = ai * GAIN + Offset For you parameter-translation ai 1= 1 V (= 100 LOGO-internal units) --> Ax1 = 0 F and ai2 = 5 V (= 500 LOGO-internal units) --> Ax2 = 135 F
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8/17/2007 4:49 PM | |
Posts: 30 Rating:
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Thanks for the response Betel! |
8/18/2007 3:15 AM | |
Posts: 191 Rating:
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Hi!! there is a different approach that could work.. in any analogical function you can select 0-10V input if 5V is 135ºF then 10V would be 270ºF just select an imput from 0 to 270ºF and this will work!! hope this helps!! ROSA |
8/19/2007 10:54 PM | |
Joined: 1/5/2007 Last visit: 4/7/2023 Posts: 1690 Rating:
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Hello, Ax = Ai * GAIN + OFFSET
Ai,maximum * Ax,minimum and
The analog functions of the LOGO can at entrance values Ai and at output values Ax (= "current values Ax") only work with integer values, so that a roundoff is carried out where appropriate (cf. LOGO handbook). In this case the integervalue x becomes from x,0 to x,4 and from x,5 to x,9 the result is the integer VALUE x+1!!! At the example 5/4=1,25: Ai = 5; GAIN = 0,25; OFFSET = 0 shows a value of Ax = 1! In order to be able to announce also decimal values now, in the corresponding functions exclusively for the representation in message texts the number of decimal places can be defined (n = 1 to 5). From for example 12345 as current value of an analog amplifier is indicated then therefore with 2 decimal places 123,45. Every decimal place is in this case then a multiplication of the actual value by the factor 1/10 = 0,1 equally significant, something with a concrete "circuit" is to be considered (for example through preceding multiplication by n * 10 by means of n of further in series pre-connected amplifier functions). Simultaneously the absolute range of values of the advertisement keeps on being located so: 32767 --> 327,67 !!! Unfortunately neither the "number of the decimal places" nor the parameter "Gain" (the corresponding analog functions) through a reference to a value of another function can be defined! Considering this state of affairs makes itself now with values of Ai not too big also to the rounding errors reduce or exclude. At the example with Ai = 5 is let, through regulation of 2 decimal places (in the 2. amplifier for the advertisement) and the use of 2 (or more, because -10,00 < GAIN < 10,00)amplifier functions switched in series with values adjusted on each other for GAIN, to avoid the rounding error (= through multiplication of Ai by 100): GAIN_amplifier1 = 10, GAIN_amplifier2 = 2,5 (Instead of 0,25 and a further pre-connected amplifier with GAIN=10). All further decimal places are always 0 at GAIN = 0,25 and integer Ai [On the other hand at a GAIN of 0,40 an advertisement with 1 decimal place is sufficient (for example 7 * 0,4 = 2,1), all further decimal places are always 0 at GAIN = 0,40 and integer Ai (in particular also the 2nd decimal place that can be renounced therefore in the advertisement and the calculation!).] During the preselection for the sensor type attention is to be paid to following:
The settings "0..10V" and "no sensor" show also identical output values Ax with handicap of identical values for Gain and offset at identical inputs Ai. At "0..10V" the specifications of the "measuring range" are helpful, when a sensor is supposed to be evaluated 0 to 10 V. In this case the signal voltages become 0 to 10 V "translates into the range of values 0 to 1000, if Gain = 1 and offset = 0 (cf. for this purpose handbook)! If using now 2 decimal places (for the advertisement), these values 0 to 1000 displayed as 0,00 to 10,00 are indicated then (what corresponds to the input signals in V). Ergo means every decimal place a division of the bloom output value Ax/10 (only for the advertisement, otherwise knows the LOGO only integers!) Best regards AttachmentCalculation_of_GAIN_and_OFFSET_for_analoge_amplifier_V2.zip (1698 Downloads) |
Last edited by: Betel at: 5/26/2011 11:13 PMadd attachment with EXCEL-files ==> Meine TAG-Listen: "deut." |
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8/21/2007 4:04 AM | |
Posts: 191 Rating:
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Wow!!!! sorry.. i mised the 1-5V i usualy divide the range by 4 ( 135 / 4 = 33.75 ) and generate a table.. -33.75º > 0V 0º > 1V 33.75 > 2V .. .. 135.00º > 5V .. .. 303.75º > 10V then i try using the -33.75 and the 303.75 as max and min parameters.. the closest i could get was -337 and 3033 with 1 decimal shift in the message text ( which is basicaly what everyone else got ) have a nice week!! ROSA |
8/21/2007 8:51 PM | |
Joined: 1/5/2007 Last visit: 4/7/2023 Posts: 1690 Rating:
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Hello Rosa, ido not understand your comment and neither, that you missing 1 V to 5 V. Ok, the equations are placed according to the specifications in the handbook (4.3.6) and in the 1st paragraph of my executions. Therefore they require the values Ai in LOGO units, that is in the range of -32767 to 32767 and in the special case, when the entrance values of analog hardware entrance come, 0 to 1000. From volts no speech is there. According to handbook input voltages of 0,00 V over... 5,12 V to... 10,00 V from the unit "analog input" are translated into integer values of 0 over... 512 to... 1000 (Format xx,nn) and these are set into the equation Ax = Ai * GAIN + OFFSET for use!. Accordingly must therefore all values of Ai converted from volts (that is Ai multiplied by 100) into LOGO-Units before they can be used to the calculation by OFFSET and GAIN into the indicated equations (example to Controls_Guy matches). If one considers this, when a value or a range (linear) are supposed to be transformed in a wanted value or range of values, the indicated equations are to be used universally. Further considerations and tables are not necessary! You need only one (for e point) or two (fur e range) pairs of korresponding values. Best regards Betel |
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4/7/2008 6:34 PM | |
Posts: 8 Rating:
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Hi I need some help with scaling. Your explanation isnt really clear. Im using a sensor which gives an output voltage which is proportional to the angle of tilt of the device.I want to output angle of tilt on the logo screen. The voltage varies between 0.3 and 4.5 V . The question ! I want 0.3 V to represent 0 degrees and 4.5 V to represent 90 degrees. I managed something for 0-10V output because I was going to amplify the output of the device but changed my mind as I was advised that this might cause noise which I want to minimise to prevent eroneous readings. I am using an analog input block to an analog amplifier block to an analog threshold trigger block ( I dont want to go beyond 90 degrees) a messsage text block and terminate with a q output. I tried uploading the file but message said it is not allowed |
4/7/2008 9:34 PM | |
Posts: 191 Rating:
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Please post your program.. ROSA |
4/7/2008 9:42 PM | |
Joined: 1/5/2007 Last visit: 4/7/2023 Posts: 1690 Rating:
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Hello fantas, look at the example into the attachment, I think that is what you need. The EXCEL-file contains the values you tell us and the result to use in a amplifiver... remember: to upload a fil it must be zipped or remame to *.zip... Best regards Betel AttachmentExample_degree.zip (2103 Downloads) |
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