I.S. Sounder Circuits

[Benzerari]

Buzz If I have well understood your question, any fire alarm system is designed to support a limited number of sounders... isn't?

In your case your system is designed to support a number of normal sounders per sounder circuit call it 'Xn', with a unit current consumption call it 'In'.

How ever, if replacing the normal sounders by IS (Intrinsically Safe) ones the new unknown number call it 'Xis', these later should have a Unit current consumption of 'Iis'

Where:

Xn: well defined through manufacturer instruction Guide
In:  well defined through manufacturer instruction Guide
Iis:  well defined through manufacturer instruction Guide
Xis: Unknown

Therefore:

Xn ---------------> In
Xis ---------------> Iis         Then: Xis = (Xn * Iis)/In = Number of IS sounder required

Hope this help, and please forgive my ignorance, if I am away from the main subject

Benz, the problem with IS systems is that there is a device known as a barrier that is positioned between the normal sounder circuit and the IS sounders. This barrier restricts the electrical energy available on the IS side of the circuit and therefore will limit the number of sounders that you can connect to it. The problem Galeon has is determining what current his barrier will allow, and how many sounders he can therefore use with it.

Wiz the barrier which I guess is a diode, it is already incorporated within the IS sounder isn't? the current 'Iis' specified by the technical guide of the IS sounder is determined while taking into account the barrier which is part of the IS sounder itself. This is my undestanding and the calculation should still as above...

But. If the diode barrier is not incorporated within the IS sounder itself, then a similar calculation has to be done while taking into account the barrier's ampedance 'Rb' which should be well known.

The following explanation may do the job:

1   -   1/Reqis = Xis * (1/ (Rb+Ris)) --------->    when barrier ampedance not incorporated and when all
                                                                       sounders go off

2   -   1/Req = Xn * (1/ Rs)           ---------->   when using normal sounders and all sounders go off


3   -   Reqis * I²is = Req * I²n          ----------->  as the power at the sounder circuit remains constant (limited)
                                                                       in either cases


After combining these three equations we get:

Xis = ((Rb + Ris) * Xn * I²is ) / Rs * I²n   =  the max. number of IS sounders required


The EOL of the sounder circuit hasn't been counted in equation 1 and 2, just to simplify the calculation, but if added in both, the end result wouldn't be of big difference from the one above.

If adding the EOL resistor the number of sounders required would be:

1   -   1/Reqis = [Xis * (1/ (Rb+Ris))] + 1/Reol --------->    when barrier ampedance not incorporated and when
                                                                                       all  sounders go off

2   -   1/Req = Xn * (1/ Rs)  + 1/Reol          ---------->   when using normal sounders and all sounders go off


3   -   Reqis * I²is = Req * I²n          ----------->  as the power at the sounder circuit remains constant (limited)
                                                                       in either cases

after combining the three equations we get

Xis = [(XnReol + Rs)(Rb + Ris)] I²is / [(Reol + Rb +Ris)] I²n

Hope this help, my timer has timed out now

[Allen Higginson]

Buzz If I have well understood your question, any fire alarm system is designed to support a limited number of sounders... isn't?

In your case your system is designed to support a number of normal sounders per sounder circuit call it 'Xn', with a unit current consumption call it 'In'.

How ever, if replacing the normal sounders by IS (Intrinsically Safe) ones the new unknown number call it 'Xis', these later should have a Unit current consumption of 'Iis'

Where:

Xn: well defined through manufacturer instruction Guide
In:  well defined through manufacturer instruction Guide
Iis:  well defined through manufacturer instruction Guide
Xis: Unknown

Therefore:

Xn ---------------> In
Xis ---------------> Iis         Then: Xis = (Xn * Iis)/In = Number of IS sounder required

Hope this help, and please forgive my ignorance, if I am away from the main subject

Benz, the problem with IS systems is that there is a device known as a barrier that is positioned between the normal sounder circuit and the IS sounders. This barrier restricts the electrical energy available on the IS side of the circuit and therefore will limit the number of sounders that you can connect to it. The problem Galeon has is determining what current his barrier will allow, and how many sounders he can therefore use with it.

Wiz the barrier which I guess is a diode, it is already incorporated within the IS sounder isn't? the current 'Iis' specified by the technical guide of the IS sounder is determined while taking into account the barrier which is part of the IS sounder itself. This is my undestanding and the calculation should still as above...

But. If the diode barrier is not incorporated within the IS sounder itself, then a similar calculation has to be done while taking into account the barrier's ampedance 'Rb' which should be well known.

The following explanation may do the job:

1   -   1/Reqis = Xis * (1/ (Rb+Ris)) --------->    when barrier ampedance not incorporated and when all
                                                                       sounders go off

2   -   1/Req = Xn * (1/ Rs)           ---------->   when using normal sounders and all sounders go off


3   -   Reqis * I²is = Req * I²n          ----------->  as the power at the sounder circuit remains constant (limited)
                                                                       in either cases


After combining these three equations we get:

Xis = ((Rb + Ris) * Xn * I²is ) / Rs * I²n   =  the max. number of IS sounders required


The EOL of the sounder circuit hasn't been counted in equation 1 and 2, just to simplify the calculation, but if added in both, the end result wouldn't be of big difference from the one above.

If adding the EOL resistor the number of sounders required would be:

1   -   1/Reqis = [Xis * (1/ (Rb+Ris))] + 1/Reol --------->    when barrier ampedance not incorporated and when
                                                                                       all  sounders go off

2   -   1/Req = Xn * (1/ Rs)  + 1/Reol          ---------->   when using normal sounders and all sounders go off


3   -   Reqis * I²is = Req * I²n          ----------->  as the power at the sounder circuit remains constant (limited)
                                                                       in either cases

after combining the three equations we get

Xis = [(XnReol + Rs)(Rb + Ris)] I²is / [(Reol + Rb +Ris)] I²n

Hope this help, my timer has timed out now

Benz ,have a look at a data sheet on zener bariers and galvanic isolators and it (may) become clearer.As far as the formulas' go they are double dutch to me - thats what datasheets are for!
The barrier datasheet says the maximum current available on the IS side.If I use the barrier that gives me 70mA and I use the IS sounders rated at 12mA then I can put no more that 5 on this circuit (70/12 = 5.8333>)

[GregC]

Theres no limit to how many Barriers you can have though is there, dont know the layout but if you can have 2 per barrier then you need 6 barriers, if its on an addressable system shouldnt be to hard to get them located outside the protected area and make up the cost of the additional units by saving on labour and cable etc in the protected area.

[Allen Higginson]

Theres no limit to how many Barriers you can have though is there, dont know the layout but if you can have 2 per barrier then you need 6 barriers, if its on an addressable system shouldnt be to hard to get them located outside the protected area and make up the cost of the additional units by saving on labour and cable etc in the protected area.

If it was an adressable loop and it was for detectors or MCP's then it would be easy but no one is doing IS addressable sounders yet (mores the pity!).
However,I would need a 24 vdc power supply to drive the each sounder circuit out in the field if they are being fed via a sounder module.The actual distance involved is minimal from the "safe side" to the IS areas but I had originally allowed only 2 circuits with 5 in area a,4 in area b and 2 in area c.However,I have got it down to 3,3 and 2 now so 3 circuits (and 3 barriers) are now necessary - hope Ive enough in the kitty for therm!

[Benzerari]

Benz, the problem with IS systems is that there is a device known as a barrier that is positioned between the normal sounder circuit and the IS sounders. This barrier restricts the electrical energy available on the IS side of the circuit and therefore will limit the number of sounders that you can connect to it. The problem Galeon has is determining what current his barrier will allow, and how many sounders he can therefore use with it.

Wiz the barrier which I guess is a diode, it is already incorporated within the IS sounder isn't? the current 'Iis' specified by the technical guide of the IS sounder is determined while taking into account the barrier which is part of the IS sounder itself. This is my undestanding and the calculation should still as above...

But. If the diode barrier is not incorporated within the IS sounder itself, then a similar calculation has to be done while taking into account the barrier's ampedance 'Rb' which should be well known.

The following explanation may do the job:

1   -   1/Reqis = Xis * (1/ (Rb+Ris)) --------->    when barrier ampedance not incorporated and when all
                                                                       sounders go off

2   -   1/Req = Xn * (1/ Rs)           ---------->   when using normal sounders and all sounders go off


3   -   Reqis * I²is = Req * I²n          ----------->  as the power at the sounder circuit remains constant (limited)
                                                                       in either cases


After combining these three equations we get:

Xis = ((Rb + Ris) * Xn * I²is ) / Rs * I²n   =  the max. number of IS sounders required


The EOL of the sounder circuit hasn't been counted in equation 1 and 2, just to simplify the calculation, but if added in both, the end result wouldn't be of big difference from the one above.

If adding the EOL resistor the number of sounders required would be:

1   -   1/Reqis = [Xis * (1/ (Rb+Ris))] + 1/Reol --------->    when barrier ampedance not incorporated and when
                                                                                       all  sounders go off

2   -   1/Req = Xn * (1/ Rs)  + 1/Reol          ---------->   when using normal sounders and all sounders go off


3   -   Reqis * I²is = Req * I²n          ----------->  as the power at the sounder circuit remains constant (limited)
                                                                       in either cases

after combining the three equations we get

Xis = [(XnReol + Rs)(Rb + Ris)] I²is / [(Reol + Rb +Ris)] I²n

Hope this help, my timer has timed out now

Benz ,have a look at a data sheet on zener bariers and galvanic isolators and it (may) become clearer.As far as the formulas' go they are double dutch to me - thats what datasheets are for!
The barrier datasheet says the maximum current available on the IS side.If I use the barrier that gives me 70mA and I use the IS sounders rated at 12mA then I can put no more that 5 on this circuit (70/12 = 5.8333>)

1st   -   Probably less than 5, bear in mind the current drawn through the EOL too!
2nd   -  70mA is it the current permitted by the barrier in quiescent or fire condition?
3rd   -   Is the barrier fitted in serial with the IS sounder or fitted in serial with the positive feed just before the
            IS side, in this case the calculation above wouldn't be valid!

I have done the calculation while considering the barrier in serial with each IS sounder, in which now I can guess is not the case!

The solution of an external power supply and sounder controller would be the ideal way to come out of the bottleneck…  

[Benzerari]

If the barrier is fitted in serial with the positive feed  side then:

When using normal sounders:

1/Req = Xn (1/Rn)  + 1/Reol  ------------------->   1/Req  =  (XnReol  +  Rn) / Reol  +  Rn


                                                                         Req  =  (Reol  *  Rn)  /  XnReol + Rn)


When using IS sounders:

Reqis  =  Rb  +  Req  =  Rb  +  [(Reol  *  Ris)  /  XisReol + Ris)]


The power in the sounder circuit is limited thus:

Reqis  *  I²is  =  Req  *  I²n          Then  :            


 [Rb  +  [(Reol  *  Ris)  /  XisReol + Ris)]] * I²is  =  (Reol  *  Rn)  /  XnReol + Rn)  *  I²n



After combining the three equations we find the following :

Xis  =  Ris[ [1 / [((ReolRnI²n)/(XnReol+Rn))-Rb]] -  1]

I now it’s even complicated and horrible formula representation in here , but that’s I think the max. number of IS sounders permitted.

[Benzerari]

Buzz here is a simplified calculation of the number of IS sounders permitted:

Ib :  max current allowed by the barrier which is fitted in serial within the positive feed, just before hazardous area.
Iis :  max current drawn through IS sounder
Xis :  the number of IS sounders permitted
Ieol : the current drawn through the EOL 'Reol'


Ib = Ieol + Xis * Iis ----------->  Ib - Ieol = Xis * Tis

                     Then:         Xis = (Ib - Ieol)/Iis

While: Ieol = 24Vdc/Reol; while the circuit is in fire condition.

I am still not sure about 70mA, is it in quiscent or fire conditions? bear in mind the current drawn in the sounder circuit in fire condition is the most important...

[Allen Higginson]

Buzz here is a simplified calculation of the number of IS sounders permitted:

Ib :  max current allowed by the barrier which is fitted in serial within the positive feed, just before hazardous area.
Iis :  max current drawn through IS sounder
Xis :  the number of IS sounders permitted
Ieol : the current drawn through the EOL 'Reol'


Ib = Ieol + Xis * Iis ----------->  Ib - Ieol = Xis * Tis

                     Then:         Xis = (Ib - Ieol)/Iis

While: Ieol = 24Vdc/Reol; while the circuit is in fire condition.

I am still not sure about 70mA, is it in quiscent or fire conditions? bear in mind the current drawn in the sounder circuit in fire condition is the most important...

Hi Benz - judging by your combined quotes you are very proficient in addressable systems.However,this scenario is a conventional system with barriers protecting the detection circuits and the sounder circuits.The currnt limits via barriers is not that important in that these devices are usually active individually upon detection/activation.In the case of sounders there is no quiescent current on the circuit as they only draw current once active and this is what I have to take into consideration - the high powered barrier will only permit 60mA load on the load side and as a maximum I can only put 4 14mA IS sounders on it.
Check here for the sounder barrier http://www.mtl-inst.com/products/isolators/5000/datasheets/MTL5022.pdf  & here http://www.mtl-inst.com/products/isolators/5000/datasheets/MTL5061.pdf  for the one for the detection side.

[Benzerari]

Buzz here is a simplified calculation of the number of IS sounders permitted:

Ib :  max current allowed by the barrier which is fitted in serial within the positive feed, just before hazardous area.
Iis :  max current drawn through IS sounder
Xis :  the number of IS sounders permitted
Ieol : the current drawn through the EOL 'Reol'


Ib = Ieol + Xis * Iis ----------->  Ib - Ieol = Xis * Tis

                     Then:         Xis = (Ib - Ieol)/Iis

While: Ieol = 24Vdc/Reol; while the circuit is in fire condition.

I am still not sure about 70mA, is it in quiscent or fire conditions? bear in mind the current drawn in the sounder circuit in fire condition is the most important...

Hi Benz - judging by your combined quotes you are very proficient in addressable systems.However,this scenario is a conventional system with barriers protecting the detection circuits and the sounder circuits.The currnt limits via barriers is not that important in that these devices are usually active individually upon detection/activation.In the case of sounders there is no quiescent current on the circuit as they only draw current once active and this is what I have to take into consideration - the high powered barrier will only permit 60mA load on the load side and as a maximum I can only put 4 14mA IS sounders on it.
Check here for the sounder barrier http://www.mtl-inst.com/products/isolators/5000/datasheets/MTL5022.pdf  & here http://www.mtl-inst.com/products/isolators/5000/datasheets/MTL5061.pdf  for the one for the detection side.

According to this senario, your calculation of the number of IS sounders permitted just by deviding:

60mA / 14mA = 4.288....,

But you are not taking into account, the current drawn throught the EOL, the number should be I think :

(60 - (24Vdc/Reol)) / 14 = max. number of IS sounders, if Reol = 4K7 then, the number will be:


(60mA - (24Vdc / 4700 Ohms)) / 14mA = 3.92, you may be allowed to use only '3'.  !