*** M1XEP Current Draw?

Mp333

Active Member
Just wondering what is the actual draw on XEP.
I have got a PD9 with a 12v 6A power supply feeding it. I wanted to feed XEP from PD9. However the current is limited to 250mA.on each output of the PD9. The XEP specs the draw at approx 300mA. I measured the current draw and it comes up at 90mA from start to normal op mode. Does it draw more when M1 goes into Alarm mode.
I have jumped two outputs in parallel on PD9 to supply XEP for now to give max of 500mA.
 
Thanks
 
And by doing such, negated any protection or benefit the PD9 provided via PTC to the connected circuit.....You really need a PD9HC
 
I'd go with what the engineers put on the white paper.
 
Thanks DEL. Not sure why I didn't go with PD9HC??? They are only couple dollars more....
Just curious why it's only drawing the third of whats on spec.....
 
Having a device that draws just 1/3 of its max power is not unusual.
 
Semiconductor chips that are used in most electronic devices can vary significantly in how much power they draw.  Some of the variation is due to the manufacturing process and the characteristics of the particular piece of silicon a chip is made from.  More variation comes from how the chip is operating at any particular moment and how many signals are switching internally and externally.
 
As a result, you can take two chips that were manufactured on different days and set them up doing the exact same thing and see a difference in power draw.  And if they are doing slightly different things (say the data passing through happens to be different), you can see an even bigger difference.
 
Other components, such as resistors and capacitors, can also vary in their values from their spec'd values, which will also change the power draw.
 
The maximum power specs are usually based on the worst case conditions. Your particular XEP might never reach those conditions, but you can't be 100% sure, so you have to allow for it to occur.
 
DELInstallations said:
And by doing such, negated any protection or benefit the PD9 provided via PTC to the connected circuit.....You really need a PD9HC
 
I'd go with what the engineers put on the white paper.
DEL and RAL
 
Why would the PTC not work when connecting a load to two of the 250ma outputs in parallel? It seems to me that the individual outputs would still break at 250ma and that the load would be pretty much equal between the two supplies if they are connected in parallel.
 
Mike.
 
From a strictly theoretical point of view, there is no reason you can't use PTCs in parallel, as long as they are of the same ampacity.  In fact, the installation manual for the PD9 says you can do this.  And Tyco, a manufacturer of PTCs, also says you can  (see page 5).
 
RAL said:
From a strictly theoretical point of view, there is no reason you can't use PTCs in parallel, as long as they are of the same ampacity.  In fact, the installation manual for the PD9 says you can do this.  And Tyco, a manufacturer of PTCs, also says you can  (see page 5).
RAL is correct.  In fact, I believe it even mentions in the PDH manual that you can do this to increase PTC protected current capacities. ;)
 
BraveSirRobbin said:
RAL is correct.  In fact, I believe it even mentions in the PDH manual that you can do this to increase PTC protected current capacities. ;)
 
The way I understand it is that when a load is applied to the two supplies in parallel approximately equal current would flow from each supply having the effect of doubling the circuit breaking point. The PTC heats up in relation to the current flowing through it and each PTC could supply 250ma before reaching it's breaking point.
 
At least that is how it would work with two batteries. With electronic power supplies there may be complications depending on their circuitry.
 
Mike.
 
mikefamig said:
The way I understand it is that when a load is applied to the two supplies in parallel approximately equal current would flow from each supply having the effect of doubling the circuit breaking point. The PTC heats up in relation to the current flowing through it and each PTC could supply 250ma before reaching it's breaking point.
 
At least that is how it would work with two batteries. With electronic power supplies there may be complications depending on their circuitry.
 
Mike.
elkpower.jpg
 
mikefamig said:
The way I understand it is that when a load is applied to the two supplies in parallel approximately equal current would flow from each supply having the effect of doubling the circuit breaking point. The PTC heats up in relation to the current flowing through it and each PTC could supply 250ma before reaching it's breaking point.
 
At least that is how it would work with two batteries. With electronic power supplies there may be complications depending on their circuitry.
 
With the PD9, it is even simpler in that there is only a single supply with two paths through the PTCs.  The important thing in making it work properly is that the circuit board traces and other wiring need to be reasonably matched so that they have equal resistance.  If they aren't, more current will flow through one path than the other and that would prevent it from fully delivering 2x the current. 
 
As the current limit is reached, ideally, the PTCs would trip at exactly the same instant.  But if one trips before the other, the remaining PTC will suddenly see twice the load and trip quickly after that.
 
If you had a situation with two separate power supplies (or batteries), it's more difficult to share the load.  If they are not perfectly matched, they may not supply equal currents.  There are some power supplies that are designed to be used with their outputs coupled together and load balance between them.
 
@ RAL, how the trip occurs varies significantly, as do the performance of PTC's themselves....while they're great to minimize the replacement of fuses, they typically flip over 10-15% of their rated value....part of the reason why fuses still exist, not to mention why they're chosen for sensitive security components/power supplies.
 
Not to mention PTC's can get into a weird half conductance state (read Elk's white paper on their 1640 transformers and Radionics 9000 series panels-we were the ones to find out that one).
 
They're on the dealer side or application notes. Not sure what you guys can see or not; I've been on the dealer side of Elk for almost 10 years.
 
In Fire Alarm circuits a Power Limited circuit (PTC protected) is actually specified over a Class 2 (fuse protected) in NFPA 72 and UL Standards. UL tests the performance of the PTC's across the specified temp range of the device (typically 0 to 49 C) and by the mfg name and part number. So company ABC part number 123 is all the mfg can use unless they test another. Fuses are NOT permitted in many Fire Alarm applications for various safety and performance reasons. Again refer to NFPA 72 and UL864 and UL985 for specifics.
 
Apples to apples. FACP vs. security are two different beasts and design criteria.
 
A FACP is required/mandated (and tested) to a different extent than most peripherals and required to remain operational under many conditions that would render other products non-functional (short, ground fault, overload, etc.)
 
Access control and CCTV typically are specified with fuses for their power sources to protect the host equipment and peripherals where a PTC could cause damage by the over-volt condition prior to it tripping.
 
That said, I'd have to go through docs, but I don't see the PD9 or PD9HC being UL listed for fire service, so they only offer the basic protection the PTC could afford within manufacturing tolerances.
 
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