For many installers it’s a case of chuck in whatever cable is in the back of the van and let the commissioning engineers worry about it… So what if they’ve ignored the cabling guidelines on the schematic!
We all know Ohms law V = IR (voltage = current x resistance), but not many are sure about its application when it comes to installing the right cable for powering an electric lock.
First things to consider are as follows:
Without the above information even Mr Georg Simon Ohm himself couldn’t help you!
Use the following calculation…
The cable to power the lock release or magnetic lock should be calculated as follows using the formula for volt drop: 2 x L x R x I, where:
For example, if you were using a magnetic lock where the current consumption of the lock is 0.5 A and the minimum voltage required at the lock is 11.5 V, the distance between the lock and the lock power supply is 30 m and you have used a Cat5 cable the result would be the following:
The DC resistance of Cat5 cable (at 20 degrees Celsius) is approximately 9 ohms per 100 m (per core), so…
2 x 30 x 0.09 x 0.5 = 2.7 V drop
If the output of your power supply were 13.6 V then the voltage at the lock would be 10.9 V and the maglock would not work.
By twisting four conductors together for the 0V and four for the +V it would be possible to lower the cable resistance enough to get the maglock working, but beware…
For a release rated at 2.5 A (which do exist!) and using all eight conductors of a Cat5 cable, for the same distance:
2 x 30 x 0.0225 x 2.5 = 3.375 V drop
…leaving 10.2 V at the maglock with no spare conductors to ‘double up’.
Other factors come into play
If you’re unsure of the cable to use you must find out the maximum current consumption of the release, the minimum supply voltage of the release, the distance between the release and the power supply and maximum output of the power supply.
Factors such as temperature and surge current of the release also come into play, so if the calculation shows the volt drop to be close to the maximum permissible it’s a good idea to allow for a larger cable.
Reducing the length of cable run by moving the PSU closer to the lock can also help, but this may be impractical sometimes as a mains supply will be required local to the door.
However in some cases there may be no alternative: if the current required by the release and the distance between the building riser cupboard, for example, equates to a cable with 3.0 mm cross-section, it’s unlikely that such a cable could physically be connected to the release or access control equipment.
There are many online volt drop calculators that can be used, but as the old saying goes: “Garbage in, garbage out”.
If you cannot feed the calculator with the right information then the results will be inaccurate. It’s as simple as that!
Manji Gami is managing director of Urmet in the UK
*If you have any questions about – or observations on – the content of this article please respond by clicking on the Comments link below or send an e-mail direct to: [email protected]
