# Current Divider

Discussion in 'General Electronics Chat' started by SnakeByte, Apr 21, 2008.

1. ### SnakeByte Thread Starter New Member

Apr 21, 2008
1
0
Have 24v 1.5amp power supply.
Would like to take off .75 amps for power to a plc, then use the rest of the power for other device's , (dividing up the remaining power)

ie=p---1.5x24=36watts supply.

using a parallel branch circuit with resistors seems a bit harsh, wouldn't the power resistors have to be able to handle 20-40 watts?

2. ### SgtWookie Expert

Jul 17, 2007
22,201
1,807
You'd be better off using a buck regulator.

They are quite efficient.

3. ### Externet AAC Fanatic!

Nov 29, 2005
1,339
169
Snake : You cannot decide how much current from the capability of the power supply is splitted to circuits here or there.
The resistance of the circuit uses the current it needs, if the supply is capable of giving it.

In other words, the sum of currents of all circuits powered by one single supply should be within the total current capability of the supply and forget about adding resistors in parallel. The powered circuits are the loading resistors.

As in an automobile, you do not split/decide the ~400 Amperes capability of the battery to run the wipers, or the starter, or the radio. They each take what they need.

Miguel

4. ### mik3 Senior Member

Feb 4, 2008
4,846
70
Externet is right. You cant say that the PLC draws 0.75 amps all the time because is contains transistors, logic circuits which switch on and off and the current drawn by the PLC changes all the time.

you can estimate the maximum current drawn by the PLC (say 0.5 amps) and then you will have 1 amp more to use for other circuits. You cant go over 1 amp for the other circuits, if the PLC draws a max current of 0.5 amp because the power supply is not capable to supply over 1.5 amps.