Discussion in 'Math' started by FFtravism, Sep 30, 2009.

1. FFtravism Thread Starter Active Member

Aug 21, 2008
30
0
I am trying to calculate the load of a circuit. I have running LED's and resistors. I am running 48 LEDs in groups of 3 with 252ohm resistors (2x 510ohm parallel). The math was figured at 25mw per LED. I am trying to figure out the size of wire to use. Do I just take the 25mw * 48 = load?
Thanks
Travis

2. mik3 Senior Member

Feb 4, 2008
4,846
63
Calculate the current through each group and then multiply by the number of groups to get the total current. Use the total current to find the appropriate wire.

3. FFtravism Thread Starter Active Member

Aug 21, 2008
30
0
So do I have to account for any resistor calculations?

4. mik3 Senior Member

Feb 4, 2008
4,846
63
Calculate the current through each parallel combination of resistors (252 Ohm) and multiply by the number of branches to get the total current of the circuit. Then according to the total current value you can find the proper wire.

5. FFtravism Thread Starter Active Member

Aug 21, 2008
30
0
So (.25mA + .25mA + .25mA) * 8 = 6mA or .006A correct?

6. FFtravism Thread Starter Active Member

Aug 21, 2008
30
0
hrmm that does not seem right but atleast the circuit is posted.

LED's are 2.2v 25mA resistors are 510ohm each.
Voltage is figured at 12.9v

Travis

7. mik3 Senior Member

Feb 4, 2008
4,846
63
Each Led branch will draw about 25mA. You said you have 48 Leds, thus there are 16 Led branches.

Total current=current through each branch*number of branches

Thus

Itotal=25mA*16=400mA

A 33 AWG wire is enough but it is very thin and it is not practical to use. You can use a lower AWG wire which is thicker and more practical to use like 25 AWG.

8. FFtravism Thread Starter Active Member

Aug 21, 2008
30
0
Ok that makes more sense.
Thanks.