| Electronics > Projects, Designs, and Technical Stuff |
| 4-20mA current loop |
| << < (6/7) > >> |
| Hiemal:
Glad my suggestion worked dagobah! I've made the same mistake before too, trying to use an op amp that isn't specified for single supply on just that. The LM358 is a really nice general purpose op amp, cheap and readily available. Biggest downsides are that it's a bit on the older side, and it's not fully rail to rail. The TL072 is also a really good op amp, but it's primary usage is for audio applications I believe, since it has a really low noise spec that comes in handy for that purpose. |
| soldar:
--- Quote from: rs20 on May 07, 2019, 10:45:50 pm --- I think you've missed the point. He has some hypothetical sensor that outputs 0-5V, as represented by V2, and his circuit (everything else in the box to the right), turns that into a 4-20mA sensor (all passively powered by the 4-20mA current loop, although how V2 is powered is left as an open question). Aside from the fact that V2 should be 1-5V, the circuit does that correctly AFAICT. --- End quote --- Yes, I am not clear on what he is trying to do. So he is trying to convert 0-5 V to 4-20 mA? Oh, OK then. I think I get it now. |
| Zero999:
--- Quote from: OM222O on May 07, 2019, 07:24:13 am ---you can't just use ohms law for a complex IC :-DD they will not magically create their own "voltage drop" and just work. The "viable design" which you speak of, is making sure at least the basics are right :-// even if once circuit works under specific conditions, it's still a terrible practice to just assume the exception is the rule. Even dave showed in one of his videos that you can power an IC from it's IO pins using the protection diodes ... obviously it is not a good design, but it works under certain conditions :-/O Please prove me wrong by setting up a circuit that powers op amps or some other ICs using a current sink / source and is stable with all the operating conditions of the ICs (i.e: different temperatures and different input / output conditions). --- End quote --- No, a current sink is not used to deliver power. A constant voltage is used to power the 4mA to 20mA device and the current drawn by it is used to transmit data. Here's an example of a 4mA to 20mA sensor being monitored by an ADC. A 250Ohm resistor is used to give a range of 1V to 5V. In reality it will be better to use 240R to allow a bit of headroom for component tolerances. In modern 4mA to 20mA systems, the constant voltage powers the op-amp and the transducer connected to it. The op-amp adjusts the current drawn by the entire circuit so it matches an input voltage. I can see you're confused by the OP's circuit. Here's a more basic implementation. The current taken from V1 varies from 4mA to 20mA, as V2, the transducer's voltage, is swept between 1V and 5V. The op-amp adjusts its output voltage, thus Q1's base voltage to ensure V(R1) = V(R3). V(R1) = V2*R3/R2 Therefore: I(V1) = (V2*R3/R2)/R1 Any current through the op-amp's negative rail is taken into account. The circuit will work, as long as the current taken from V1 and passed to R1 is below the minimum desired output current or 4mA, in this case. In reality, the voltage drop across R1 will probably need to be much lower and some additional amplification and level shifting will be required to scale the voltage given by the transducer, V, to something giving a sensible voltage drop across R1. |
| Zero999:
--- Quote from: soldar on May 08, 2019, 07:12:51 am --- --- Quote from: rs20 on May 07, 2019, 10:45:50 pm --- I think you've missed the point. He has some hypothetical sensor that outputs 0-5V, as represented by V2, and his circuit (everything else in the box to the right), turns that into a 4-20mA sensor (all passively powered by the 4-20mA current loop, although how V2 is powered is left as an open question). Aside from the fact that V2 should be 1-5V, the circuit does that correctly AFAICT. --- End quote --- Yes, I am not clear on what he is trying to do. So he is trying to convert 0-5 V to 4-20 mA? Oh, OK then. I think I get it now. --- End quote --- But his schematic says 0 to 4V. In any case, this isn't that difficult to solve. The circuit I posted previously will do, with some modifications. The op-amp's non-inverting input can be thought of as a virtual ground. If another voltage is added, to another series resistor, the voltage across R1 will simply equal the sum of V3 and V2. With a 1V reference, and another 100k resistor: R4. We now have 4mA to 25mA, when V2 is swept from 0V to 5V. Where R2 and R4 meet form a summing junction. If we want 0V to 4V input, then that's it: mission accomplished, it'll only go up to 20mA. For 0V to 5V input, change the value of the R3. Of course this isn't ideal. We might want to used a higher reference voltage and a lower voltage across R1, but that's just a matter of changing the resistor values. Here's one with standard precision resistor values and a 5V reference, which only drops 2.5V across R1. In real life V3 would be a voltage reference IC. |
| soldar:
--- Quote from: Zero999 on May 08, 2019, 11:55:49 am --- In any case, this isn't that difficult to solve. --- End quote --- Yes, I agree, it is probably easy to solve once you know what it is we are trying to solve but, like in many other threads, I spend more time trying to figure out what is being asked than in solving it. I have worked with plenty of 4-20 mA transducers and controllers so I am well familiar with them. If I am going to analyze anything I am looking for something that looks like the diagram on post #5 where you have the remote sensor and the controller separated by the two wire loop. If I see a diagram where both controller and transducer are not clearly separated and I have to start scratching my head and asking questions about what is what ... it gets frustrating for me. |
| Navigation |
| Message Index |
| Next page |
| Previous page |