Lock release cable calculations
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?
L is the length of cable in Meters
R is the resistance per meter of the cable in Ohms
I is the current of the lock in Amps
For example if you were using a magnetic lock where the current consumption of the lock is 0.5A and the minimum voltage required at the lock is 11.5V, the distance between the lock and the lock power supply is 30M 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 100M (per core) so –
2 x 30 x 0.09 x 0.5 = 2.7V drop.
If the output of your power supply were 13.6 volts then the voltage at the lock would be 10.9V 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 2.5A (which do exist!) and using all 8 conductors of a CAT5 cable, for the same distance –
2 x 30 x 0.0225 x 2.5 = 3.375V drop
Leaving 10.2V at the maglock with no spare conductors to ‘double up’.
If you are 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 is a good idea to allow for a larger cable.
