4 - 20 mA current loop, short circuit isolator

Question

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 tried to make fast isolator that opens sensor line, before intruder panel, detect a short circuit across the sensor zone terminals.

System block schematic:

Brief about system:

Working voltage for zone 24 volt DC.

3.9k end of line resistor for monitoring line break. Line resistance go higher than 3.9k reports fault. Connected in parallel to zone terminals working in 24 volt DC.

1k resistance or less for fire/intruder detection resistance. Connected in parallel 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 terminals.

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

For the above requirement, below short circuit isolator with mosfet and external power supply designed, and it's not functioning as expected.

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 positive terminal, thus a open circuit in line is reported instead of a over current situation.

I used N-channel Mosfet (IRF Z44N), connected (gate and drain) across intruder alarm zone terminals in parallel, 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.

1. What's 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?

Answer 1

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:

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

2. 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.

Answer 2

If you just want to generate signals that can be monitored by a 4 - 20 mA analog input then try this.

^{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.