|
|
OHMS LAW AND CURRENT LOOPS Process instrumentation commonly uses a 4-20 mA DC (4 to 20 milliampere or 0.004 to 0.020 amperes) signal to provide a 0-100% process or control value. A unique feature of a current signal is that the circuitry of instrument drives the signal at the desired level between 4 and 20 mA and will keep that signal constant, even with changes of resistance and voltage in the signal loop. This makes the signal very desirable for the instrument loops because the accuracy is not affected by changes in connection resistance, line resistance or the addition of other loads in the circuit. However, there is a limit to how much load can be connected without causing error to the signal. Many question arise concerning the wiring of these circuits and the effect of multiple devices. Ohms Law is the primary rule that controls all wiring arrangements and connection limitation. Ohms Law defines the rules of electrical voltage, current and resistance. It states:
or: voltage equals current times resistance (ohms) or: volts = amperes x ohms For example: 24 Volts = 20 mA (0.020) x 1200 ohms
This equation can also be rearranged for obtaining the other two values: I = E/R current equals voltage divided by resistance R = E/I resistance equals voltage divided by current
Let's look at a common current loop using a power supply, transmitter and controller (See fig. 1). Often a 24 Volt DC power supply is used and the controller input in often 1 to 5 Volts DC so we will use them in this example. The transmitter output is usually 4-20 mA, meaning it will allow 4 milliamps to pass through it when it's input is zero and 20 milliamps when it's input is 100% and proportional current signals in between. The sum of all voltage drops around the loop must not exceed the voltage of the power supply (24 v or 20 mA x 1200 ohms). If the loop resistance is negligible and the transmitter is at 100% then it's passing 20 mA through the loop and causes a 5 volt drop (20 mA x 250 ohms) at the controller. Because the power supply provides 24 volts, the transmitter must make up for the difference and drop 19 volts (24v-5v). At the low end of its' range, when the transmitter is at 0 %, it passes 4 mA causing a 1v drop (4 mA x 250 ohms) at the controller. In this case the transmitter will drop 23v. Let's consider a transmitter that requires a minumum of 13v to operate. With a 24v power supply, that leaves 11v left to drive other instruments in the loop. Therefore, up to 550 ohms load is available for the other instruments (11v divided by 20 mA). 24v - 13v = 11v Power supply minus transmitter equals balance 11v / 20 mA = 550 ohms 11 volt balance divided by current equals 550 ohms load left for other instruments and loop resistance If we add another instrument to our example loop above and it also had a 1-5v input (250 ohms) our loop would still work fine because the loads would still be within range. The transmitter would drop 14v and could still operate. The transmitter actually adjusts its voltage drop an maintains the signal at 20 mA. An important feature is that at 100%, the current remains 20 mA and the signal at the controller remains 5v in spite of the change in loop resistance. 24v - 5v - 5v = 14v (good) However we could not add a third instrument with 1-5v input because there would not be enough voltage left to operate our transmitter with a 13v minimum rating. 24v -5v -5v -5v = 9v (not good) In evaluating current loops we use the 20 mA signal because that is the maximum load condition. If the loop resistance and voltage drops are within operating range at 20 mA the loop will operate at the lower current signal levels.
fig. 1
This, obviously, is an simplified explanation. We hope it has been of some use to you.
If you have any questions or comments please email us |
