Many times an existing wire run is undersized for the current draw of the load device.  This could either be due to an unexpected change in the device being powered, a miscalculation at the planning stage of the job, or a retrofit situation where the existing wire size is not able to be changed.  This undersized wire results in a large voltage drop, leading to improper or erratic operation of the load device.  Even if the voltage at the device is at the low end of the acceptable range, the voltage will quickly drop to unacceptable levels when on battery power.

What is Voltage Drop?

Wire has a specific resistance per foot of length.  The smaller the wire gauge, the higher the resistance (given equivalent wire types).  As the length of the wire increases, the total resistance between the power source and the load device increases.  When current is now drawn through this wire resistance, some of the voltage is "burned off" in the wiring as voltage drop, as defined by Ohm's Law (V=IxR).

Per Ohm's Law, two main factors affect the amount of voltage dropped within the wire run:  the wire resistance, and the current drawn through the wire.  This is why a lower-current device can get away with a smaller wire gauge. 

This is also one of the main reasons the lifesafety industry has, and continues to, switch from 12V to 24V.  A given device will use a certain number of Watts.  If that device is designed to use a 24V input rather than 12V, the current required will be halved (per Ohm's Law I=P/E), which in turn will halve the voltage drop.

How can the B100 help?                         

If powering a 12V device, a B100 can be used to give an adjustable output voltage greater than the 12.5V nominal setting of an FPO power supply.  The FPO will need to be set for a 24V output, and the B100 placed into the adjustable range by moving JP3 to position 2.  The output can then be set by adjusting VR2 to a level giving an acceptable voltage at the load device.  Since this voltage is run off of a 24V supply with a 24V battery set, this voltage will remain constant until the battery set drains to well below 20V.

Please note that if the device being powered has varying current levels during normal operation, the voltage at the device will change with this current, possibly with damaging results.  As an example:

A B100 is set to a level of 16.5V to overcome the voltage drop through 500 feet of 18AWG wire powering a 12V edge device and a 12V maglock at a door.  The draw of the lock is 400mA and the edge device is 100mA, giving a total draw of .5A when the lock is powered.  When the lock is powered, the voltage drop will be 3.24V, leaving 13.26V at the door.  However, when the lock is released, and the current draw drops to 100mA, the voltage drop will decrease to 0.65V, giving 15.85V at the door, which is likely too high for the 12V edge device. 

In the above example you could decrease the B100 voltage to compensate, but you must take the whole operating current range into account.  If the operating current range is too large, there may not be an acceptable voltage to cover all load conditions.

For more information on the B100, see the B100 manual and Application Note AN-07.  Also, be sure to download our FlexCalculator Suite for quick voltage drop calculations.  And remember, our This e-mail address is being protected from spambots. You need JavaScript enabled to view it department is always here to help.