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I am trying to design a short circuit isolator, for 4 - 20 ma current loop based intruder alarm system.The below system,identify short circuit in sensing line and sensor activation(closing 1 k in line) as same signal.I tryed to make, fast isolator, that opens sensor line, before intruder panel, detect a short circuit across the sensor zone terminals.

system block schematic.

enter image description here

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Brief about system.

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Working voltage for zone 24 volt dc.

3.9k end of line resistor for minotoring line break.Line Resistance go higher than 3.9k reports fault.connected in parllel to zone terminals working in 24 volt dc.

1k resistance or less for fire/intruder detection resistance.Cinnected in parllel to zone, working in 24 volt dc.

System has its own fire and fault relay operates, in case 1k resistance appears across its zone or fire reported, if line resistance goes higher than 3.9k across tetminals.

Currently, any short circuit across , intruder alarm zone, turns system to alarm mode, instead of a fault.

For the above requirement, below shortcircuit isolator with mosfet and external power supply designed, and its not functioning as expected

enter image description here

Design assumption:

Normally mosfet will be in on and powering optoisolator, any short circuit across terminal turns voltage across gate to source voltage zero.Turning off the optocoupler transistor, by isolating sensor and system postive terminal, thus a open circuit in line is reported instead of a over current situation.

I used, n chanel Mosfet(Irf z44n), connected(Gate and drain) across intruder alarm zone terminals in parllel, to sense, short circuit and its feed back activating a series optocoupler working in 12 volt.

After connecting, isolator circuit is not working as expected, once a short circuit simulated across, current loop terminal, opto turns off, and isolating line, to limit short circuit current.

Experts,Please advise

1)whats wrong in my approach ? 2)How, can i correct the design error, in my circuit ? 3)Other simple solutions for, fast short circuit isolator for 4 - 20 ma current loop?

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    \$\begingroup\$ A circuit diagram says 1000 words as they say. Please add a schematic, including part numbers for sensors. \$\endgroup\$
    – Ron Beyer
    May 18 '20 at 20:12
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    \$\begingroup\$ You can add a schematic using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar and "Save and Insert" on the editor an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. \$\endgroup\$
    – Transistor
    May 18 '20 at 20:50
  • \$\begingroup\$ @Transistor please advise on the above schematic. \$\endgroup\$
    – Cathode
    May 21 '20 at 3:23
  • \$\begingroup\$ @RonBeyer please see the upadate and advise. \$\endgroup\$
    – Cathode
    May 21 '20 at 3:25
  • \$\begingroup\$ I don't know what your circuit is supposed to do. It's just a battery, some resistors and a bunch of switches. There is no indicator light or computer interface. What is the output supposed to be? If "Short Circuit" is switched on when the opto-LED is on then the opto-transistor will be destroyed. \$\endgroup\$
    – Transistor
    May 21 '20 at 16:31
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To address both parts of your question:

  1. What's wrong with the current circuit: So remember that a MOSFET is not actually a symmetric device. It can't turn-off current going in both directions, only in one direction. Your schematic has current flowing from source to drain on the MOSFET, but that is the direction of the body-diode in a MOSFET as well. Not all circuit symbols show it, which can lead to this sort of simple mistake, but if you look it at the circuit symbol from the datasheet of the very MOSFET part number you used, you can see the more complete circuit symbol that shows the diode:

NFET with Body Diode Shown

So even after you turn-off the MOSFET in the circuit you have, it will still conduct current through that body-diode.

  1. Practical note: This kind of circuit for fault-isolation is generally in the class of ICs known as E-Fuses, and can be bought in a convenient form factor from the likes of TI or STM or Analog Devices. You can even have it re-attempt to turn on in short pulses that won't take down the system but allow you to tell when or if the short has been removed from the system.
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  • \$\begingroup\$ Main challenge, is isolator should be fast enough, that any short circit, wont be reported back to control panel by breaking the supply to sensor from, before panel senses as a overcurrent situation. \$\endgroup\$
    – Cathode
    May 21 '20 at 3:35
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If you just want to generate signals that can be monitored by a 4 - 20 mA analog input then try this.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A circuit that will give readings in the range 4 - 20 mA for a variety of circuit status indications.

  • Line open: \$ i = 0 \ \text {mA} \$.
  • End of line only: \$ i = \frac {24}{3k9 + 1k + 250} = 4.7 \ \text {mA} \$.
  • EOL + STANDBY: \$ i = \frac {24}{(3k9 || 3k)+ 1k + 250} = 8.1 \ \text {mA} \$.
  • EOL + STANDBY + ACTIVE: \$ i = \frac {24}{(3k9 || 3k || 1k)+ 1k + 250} = 12.7 \ \text {mA} \$.
  • SHORT CIRCUIT: \$ i = \frac {24}{1k + 250} = 19.2 \ \text {mA} \$.

R4 will convert these currents to 0 to 5 V for an ADC.

You can work out other combinations for missing END OF LINE resistore, etc.

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  • \$\begingroup\$ The reason, why am doing is, its obselete old system. Only reference is observing system response from current. \$\endgroup\$
    – Cathode
    May 21 '20 at 17:00
  • \$\begingroup\$ if we increase end of line value morethan 3.9 k or standby line current decreases from 5 ma it wil lcreate a trouble in line.only a tolerance of .5 ma allowed.so i choose mosfet its impedence so high and input current draw from intruder alarm system will be negligible and providing aditional power source to avoid loading in the orginal system. \$\endgroup\$
    – Cathode
    May 21 '20 at 17:05
  • \$\begingroup\$ OK. Sorry, I don't understand your question then. Your circuits don't show any interface with another system monitoring the current. I don't think I can help any more until you provide a better diagram of the existing system. \$\endgroup\$
    – Transistor
    May 21 '20 at 17:08
  • \$\begingroup\$ okay i wil post a detailed diagram with control panel and block schema of entire panel and field device. \$\endgroup\$
    – Cathode
    May 21 '20 at 17:11
  • \$\begingroup\$ please, have look on the link .firealarmproblem.blogspot.com/2013/10/… \$\endgroup\$
    – Cathode
    May 21 '20 at 17:27

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