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Generally process control devices use analog signal. Normally in an industrial automation setup, 0-20 mA (mili amp) or 4-20 mA or 0- 10 Volts are used. These signals are used for both providing control output and to provide feed back signal (say to a PLC (programmable logic controller) or DCS) Example for a control out put is a motorized damper operated by 4 - 20 mA signal. PLC provides 4-20 ma signal. Example of a feed back signal is 4 - 20 mA signal from temperature transmitter, measuring a liquid temperature to PLC as input.
Your total milliamp range is 4 to 20 milliamps. 4 Ma being 0mmH2o and 20 Ma being 2500mmH2o. So 1250mmH2o is 50% which is equivalent to 12Ma signal.
L = +10 µA HIGH/-0.18 mA LOW_ = +40 µA HIGH/-1.6 mA LOWLS = +20 µA HIGH/-0.36 mA LOWS = +50 µA HIGH/-2.0 mA LOWAS = +20 µA HIGH/-0.5 mA LOWALS = +20 µA HIGH/-0.1 mA LOWH = +50 µA HIGH/-2.0 mA LOWF = +20 µA HIGH/-0.6 mA LOWHC = +1 µA HIGH/-1 µA LOW
Yes you can. All the 1300 mA rating means is that adaptor can supply devices up to 1300 mA. The old adaptor's limit was 800 mA. The new adapter has 500 mA more in reserve if it is ever needed.
It might be. Or it might not. You need to specify the transformer's turns ratio or its impedance ratio, and th eload on the secondary.
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Generally process control devices use analog signal. Normally in an industrial automation setup, 0-20 mA (mili amp) or 4-20 mA or 0- 10 Volts are used. These signals are used for both providing control output and to provide feed back signal (say to a PLC (programmable logic controller) or DCS) Example for a control out put is a motorized damper operated by 4 - 20 mA signal. PLC provides 4-20 ma signal. Example of a feed back signal is 4 - 20 mA signal from temperature transmitter, measuring a liquid temperature to PLC as input.
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As of 2012 it is 6.25 percent, due within 10 days.
If you are talking about 4 mA = 0% and 20 mA = 100% then the formula would be a linear equation: y=mx + b y=6.25(x) - 25 y= percentage x=mA ouput So for example put 4 mA in place of x and you get 0% for y and if you use 20 mA in place of x you get 100% for y. You can rewrite the equation if only Percent input is known to find mA output it would be X= (Y + 25)/6.25 this way you input the percent in for (Y) and you have the mA output for X
In a current loop, 4-20 ma is a typical range. We don't use 0 ma as a valid signal, because that is reserved for a broken wire or dead transmitter condition. The receiver is designed to offset and translate the 4-20 as necessary, with the break condition also being reported.
50.20 percent according to a website.
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Previously when transmitters were configured for 0-20mA signal it was very difficult to identify an open circuit at 0mA,hence to supersede this defect it was decided to configure the transmitter as 4-20mA signal,so that if there is open signal it can be easily identified.
4-20ma is a standard range used in process controls to indicate measured values. For instance, a pressure instrument may output 4ma to indicate 0 psi, up to 20 ma to indicate 100 psi, or full scale. Or, to control a variable-speed motor drive, 4 ma may mean to stop the motor, and 20 ma to run at full speed. The reason that 4ma is chosen instead of zero is for fail safe operation. A broken wire always results in 0 ma. Suppose our signal right now is 12 ma, and this causes a ventilation fan to run at 50% speed. Suddenly the signal drops to 0 ma. Now, if our range was 0-20 ma, does this change mean stop the fan? Or does it mean the signal wire just broke? There would be no way to tell. With a 4 ma low range, we would know that 0 ma was a broken wire. Then instead of stopping the fan, we could set it to some default speed, or simply continue at the previous speed until the wire was repaired.