Basic 20mA Current Loop

[wuchy143]

Hi All,

I'm trying to understand a 20mA current loop so I can use it in a design I'm working on. What I have gathered so far:

1. The micro sends 0-5V serial data over to the opto coupler. This then gets transmitted over the opto which then switches the output transistor on and off. Right so far?

2. This then gets fed over to the receive side of the 20mA current loop. This is where it falls apart for me. What are the input resistors doing?(10k, 2.7k, 220) Also, why -12V and +12V how does that come into play? What's the 9639 doing?

I know that the transistor must be biased so that when it is on there is 20mA of current allowed to flow(constant current source?)

If anyone can shed some light on this it would be much appreciated thanks!

 

[dr pepper]

I dont think any of the schematic has to do with current loop - edit: this is incorrect see kiss's thread entry.
The 9639 is a serial data receiver.
It looks like the circuit is showing serial datacomms interface from the micro to another device (maybe a current loop controller).
The + and -12v will be required for the line receiver to do its job, the 10k and 2k7 resistors look like they are biasing the line receiver so it can discriminate a logic 1 from 0, and the 220r is a pull down to reduce noise sensitivity.
Whats the schematic for?

 

[KeepItSimpleStupid]

Nowdays when you say 20 mA current loop, someone might think 4-20 mA or 0-20 mA for transmission of analog signals. Not the case here.

This really looks like current loop to RS232. The older standards for RS232 were -24 to -3 and +3 to +24 volts. This uses closer to +12 and -12. Rs232 does go as low as -5 and +5 and then there is RS232TTL which is usually inverted.

I don;t know what the 9639 is, but lets just assume it's an OP Amp for now. It's really acting as a comparitor. A reference is set with a voltage divider which sets the voltage at 2.55V. The 10K and 2.7K. The 220 resistor is where the X mA will be impressed on. It should be (5-0,6)/1220. So, when the voltage is >2.55 V, the OP amp switches state. I think the diode is adding some reverse polarity protection and the transistor inversion. The secondary current is (5-0.6)/1220 or at least 4 mA. Something seems fishy with my calculations. So, if the power supply was bigger (+12 instead of +5 on the OPTO) and/or the 1K resistor smaller, it would make sense.

It's been a while and I forget the quiescent state and the definition of mark and space, but switching the inputs tot he OP amp, just reverses the output. Mark and Space never concerned me because I always used a breakout box.

There are standard RS232 converter chips that probably could be used.

 

[KeepItSimpleStupid]

dr pepper:

Current Loop is an OLD standard used in the mid 70's which was when I was familiar with it. The old ASR-33 teletypes used it at 110 baud. http://wd4eui.com/Pictures/ASR_33_Teletype.jpg and so did the DEC Decwriter II. There was an optional CL to RS232 board available for it.

Also see: http://en.wikipedia.org/wiki/Current_loop

and even: **broken link removed** and http://www.radio-electronics.com/in...ent-loop-serial-interface/basics-tutorial.php

 

[dr pepper]

Yes I just looked it up.
I was thinking industrial controls where instrumentation used current loops, also from the 70's.