FireNet Community
FIRE SERVICE AND GENERAL FIRE SAFETY TOPICS => Technical Advice => Topic started by: Allen Higginson on September 06, 2008, 01:02:35 AM
-
Quick question - how many sounders can you have on an IS sounder circuit?
The answer may caue damage to my head through impact with the desk!
-
Quick question - how many sounders can you have on an IS sounder circuit?
The answer may caue damage to my head through impact with the desk!
What you mean by 'IS sounder circuit'? for what so ever, it is the (proper calculation) that can tel you how many sounders in any given sounder circuit you need. taking into account the max. current consumption of the whole load.
-
Quick question - how many sounders can you have on an IS sounder circuit?
The answer may caue damage to my head through impact with the desk!
The purposes of the zener barrier between a normal sounder circuit and an Intrinsically Safe sounder is to restricts the electrical energy available on the IS side to prevent 'sparks' that could ignite flammable vapour. Therefore there must be some restriction that will affect the amount of current that can be drawn. I do not have my Fulleon password available here at home to download the technical info to confirm any answer about their IS equipment and I'm on holiday from Monday. Maybe someone else can look it up on the Fulleon website.
-
Buzz ,
35 ma on a sounder , allow 50 ma with a beacon , Apollo sounder controller cut the link for I.S unit fused as 1 amp .
-
Buzz ,
35 ma on a sounder , allow 50 ma with a beacon , Apollo sounder controller cut the link for I.S unit fused as 1 amp .
Hi mate - the Fulleon IS sounder draws 14mA.
The guys in our office have told me that through a galv barrier you can only put on 2 sounders max per circuit - due to the fact I have designed 11 split over two circuits you see my problem!!!
-
Quick question - how many sounders can you have on an IS sounder circuit?
The answer may caue damage to my head through impact with the desk!
What you mean by 'IS sounder circuit'? for what so ever, it is the (proper calculation) that can tel you how many sounders in any given sounder circuit you need. taking into account the max. current consumption of the whole load.
IS = Intrinsically Safe as per Wiz!
-
Buzz ,
If its any help a company called e2s warning signals have an alarm driver for IS sounders rated @ 80 ma load so with your 14ma Fulleons you may get away with 5 .
-
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
-
Buzz ,
If its any help a company called e2s warning signals have an alarm driver for IS sounders rated @ 80 ma load so with your 14ma Fulleons you may get away with 5 .
Hi mate - been on their site the other day but can't find anything other than sounders and beacons.
-
Alright Buzz ,
Detection supplies tel 01425 658239 Catalogue Page 23 Ref MT5022 web link is the catalogue , good hunting.
www.detectionsupplies.co.uk
-
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.
-
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.
Its actually my problem but Galeon is welcome to it!!
-
Buzz ,
No I must insist that you keep your I.S Barrier , I will stick with me eol.(previous thread).
-
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
Hey Benz I reckon that explanation warrants a prize of some kind. Dont know what it means but I think it at least deserves free membership of the FBS and personal attention from Matron. Perhaps even the award of FIFBS.
Best wishes
Kurnal GradIFBS
-
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
Hey Benz I reckon that explanation warrants a prize of some kind. Dont know what it means but I think it at least deserves free membership of the FBS and personal attention from Matron. Perhaps even the award of FIFBS.
Best wishes
Kurnal GradIFBS
Kurnal, what I meant by this simple explanation is that the number of sounders is unversaly proportional to the current drawn by each sounder type, it is just a lineair variation, the less current drawn by the sounder the more sounders you can fit, but the max number of sounders is limited any way, for what so ever type of sounders...
The power at the sounder circuit using a normal sounder is limited (Req * I²). If 'Req' gets down due to some sounder types, the current 'I' raises up to keep up (Req * I²) remains constant and so more sounders can be used and vis versa... if 'Req' gets up when using IS sounder i.e. the current 'I' drops down and so less sounders are permited by sounder circuit.. the calculation in the next post explains how to find out the max number of sounders required for what so ever. :)
If I am wrong please feel free to correct me :)
-
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 :)
-
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>)
-
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.
-
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!
-
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… :)
-
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.
-
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...
-
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.
-
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'. !