Multiple PIR's on same fsl zone

[mel]

Hi,

 

I have a Castle care-tech euro mini and need to add another PIR to a FSL zone which already has a PIR. i.e. I need to wire two PIR's in series. Unfortunately the manual does not give any details how to do this and I have not found any helpful information on the net either.

 

I would be most grateful if someone can point me in the right direction to achieve this.

 

Thanks in advance.

 

Mel.

[Secware_Tech6]

I have attached a drawing showing an example

[mel]

Thanks for the diagram. I knew I had to get the alarm switches, wired in series, in parallel with 4k7. I just couldn't visualise it.

 

From the diagram, it looks like I will need to use three wires to achieve this.

 

Assuming I placed the 4k7 in the last device and using a spare core form the cabling to connect the other end of the resistor to one of the alarm contacts on the first device. What would happen if that return wire was severed for what ever reason? Ordinarily, I would expect the system to sound a tamper or alarm. However, because all the alarm switches are closed, the system does not detect either of those conditions. Would that be correct?

[Secware_Tech6]

As you have seen there is no secure way to wire fsl using this method.

Its advised that you have one device per zone.

 

You can use a 3 wire method and keep the 2k2 at the end of the line. This will give some protection.

[mel]

I have rearranged the circuit slightly. I think this is more secure, although it does mean using four wires.

[Secware_Tech6]

yes that will work but you wont have short circuit protection at the first device.

[mel]

Yes. You're absolutely correct. I didn't spot that.

 

If my understanding is correct, please correct me if I'm wrong, on some panels the 'COM' terminal are the same as 0v terminal. Are you able to confirm if that is the case with the mini euro please?

 

Assuming for the moment that that is the case, if the circuit was rearrange as shown, I think it satisfies all conditions.

[Secware_Tech6]

YOu are right in most cases.

ANd this is the case with the euro. Most modern panels have a common circuit neg as well as power neg.

 

James

[mel]

I had emailed castle care-tech several days ago regarding this issue. They replied with the following description:-

 

"The first N on the first PIR to panel Zone, wire the 4K7 resistor across the N/C and just a link between C and the first T. The second T goes to N on the second PIR and so on. The final PIR has the 2K2 resistor in place of the link and the second T to the panel Common."

 

I drew a diagram to that description which castle confirmed as being correct. Please see below. To my amateur eyes, it doesn't look right. Anyone walking in front of PIR2 or PIR3 will cause a tamper condition, surely?

[Secware_Tech6]

yes the circuit would go open circuit causing a tamper.

 

The easy thing to remeber is that

 

2k2 = closed

6k9 = open (alarm)

>7k = Tamper

 

That is over simplified as there are high res and other monitoring in between, but will allow you to work it out.

ANother way is to use double pole and series this up (using 6 cores per pir)

Then use relay coils for each pole (ie tamper and alarm) but use a pos feed on the one circuit and a neg on the other to give short circiot protection.

 

James

TEch

[mel]

James, I would be most grateful if you could post a diagram for me.

 

Regards,

 

Mel.

[Secware_Tech6]

Attached is a quick drawing showing 2 relays

[mel]

Hi James,

 

Thanks for the diagram.

I don't understand what the +ve and -ve terminals are. You refer to those terminal as 'feed'. Could you explain what they are please.

 

Thanks.

 

Mel.

[mel]

After some more analysis of the circuit I posted, it looks like it is still susceptible to having the alarm lead shorted to 0v without tamper detection, thereby disabling the alarm contacts of the consecutive devices in the chain.

 

So, I have come up with another idea which seem to detect tamper in all cases. Please let me know if I have missed anything. See attached diagram.

 

How it works:-

Panel programmed for 4k7 EOL and 4k7 shunt.

 

When in normal condition, the panel must see a 4k7 between A and 0v.

I chose 2k7 for R1. Therefore, R2 in parallel with Rs1+Rs2 must = 2k.

 

Also, when any one of the alarm contacts open, R1+R2 must = 9.4k

therefore R2 must = 6.7k. Preferred value of 6.8k.

2.7k+6.8K = 9.5k, which should be within panel tolerance (remember also that resistor tolerance would also effect the final value).

 

To calculate the value of resistors in parallel we use 1/(1/R1+1/R2+...1/Rn)

 

Remember that R2 in parallel with Rs1+Rs2 needs to be 2k. Lets call Rs1+Rs2, Rst.

 

So we now have,

1/(1/6.8+1/Rst) = 2

 

If we rearrange that, we have

1/Rst = 0.5 - 1/6.8

1/Rst = 0.353

therefore Rst = 2.83 i.e Rs1+Rs2 = 2.83. therefore Rs1=Rs2= 1.415k. Preferred value of 1.5k.

 

Now, if one attempts to short the alarm lead to 0v, this would cause the circuit to change its over all resistance. In effect, it will become R1 + (R2 in Parrnell with Rs1),

1/(1/6.8+1/1.5) = 1.23k

2.7+1.23 = 3.9K. This is less than 4.7k and should cause a tamper condition.

 

Mel.

[mel]

I think this can still be simplified. See attached diagram. With this configuration, one can add as many devices as one requires (within system limitations) in series providing full FSL.

 

If I change the resistor values to better suit the situation, the values become R1 = 1.2k, R2 = 8.2k and Rs = 6.2k

 

When alarm contact become open circuit, we have 1.2+8.2 = 9.4 (alarm condition).

Short circuit tampering on alarm wire to 0v, circuit resistance becomes 1.2K - tamper.

 

Can anyone spot a problem with this?

[Secware_Tech6]

The problem yu will have with parralel wiring is that the resistance will vary wildly as individual deviced open

 

YOu are correct in your theory of parralel calcs

 

but when both devices are openyou will have an approx 4k7 res. But the panel will see this as a problem as the resistance is too high to be closed yest too low for open.

James

[mel]

The problem yu will have with parralel wiring is that the resistance will vary wildly as individual deviced open

 

YOu are correct in your theory of parralel calcs

 

but when both devices are openyou will have an approx 4k7 res. But the panel will see this as a problem as the resistance is too high to be closed yest too low for open.

James

 

Hi James,

 

If you're referring to eolInSeries5, you're absolutely correct. I have scrapped that idea. Too complex.

 

However, in eolInSeries6, if any or all the alarm contacts open, there is only one path. And that path is R1+R2. So, if I chose R1=1.2K and R2=8.2k, the total resistance in that path will be 9.4k.

Since my panel is programmed for 4k7 and 4k7, this will give me 9.4k for alarm condition. Exact resistance match!

 

Whilst the alarm contacts are closed, total resistance need to be 4.7k (EOL resistor of 4.7k).

That's, R1+(R2 in parallel with Rs)

R2=8.2k, Rs = 6.2k, parallel value is 3.53k

Therefore, total resistance when alarm contact are closed is 1.2k+3.53k = 4.73k. This is exactly what we need for normal condition.

 

As far as I can see, one can use as many devices in series with this configuration without altering resistor values.

 

This also provides short circuit protection of the alarm lead. Any attempt to short circuit to 0v will effectively make the total resistance of circuit the value of R1, 1.2K, which hopefully is low enough to trigger a tamper condition.

 

There are other resistor combinations that can be used to chive the same results. One other combination I worked out is R1=1.8k, R2=7.5k, Rs=4.7.

In this case R1+R2 = 9.3k (hopefully with in tolerance of panel) when alarm contact open.

In normal condition, R1+(R2||Rs); R2||Rs = 2.89k, therefore total resistance is 2.89k+1.8k = 4.69k. (hopefully this is also with in panel tolerance)

 

Please do correct me if I have missed anything. I appreciate your response.

 

Mel.