I did the trick aout transistor and 1-3 bridge on my two cores as seen here:
http://www.ucapps.de/midibox_lc/midibox_ng_switching_psu.pdf
how can I calculate the resistor to put in serial with the led between the ground and the 5V ???
any ideas or explanations could be fine and I’d increase my knowledge too 
all the best,
protofuse
LED’s are current devices.. if you pass too much current, they die.
So we add a resistor to limit the current that goes through them.
The HIGHER the resistor value (in Ohms), the LOWER the current will be through the LED.
To figure it out, we use “ohms law” (google that for more).
+5 Volts----------////-----------|>|------------GND
So we need to know how many volts will be dropped across the resistor.
We start with the total voltage, usually 5 volts for MIDIBox stuff.
then we subtract the “forward voltage” of the LED. That is how many volts the LED uses when it’s lit.
What we are left with is the voltage that we need to drop across the resistor.
So, 5 volts - 2.2 volts (LED forward voltage) = 2.8 volts across the LED.
Ohms law says that if we know the voltage across that resistor (2.8) and we
know the current through that resistor (20ma = 0.02 Amps) then we can calculate the
resistance as R=V/I. V=2.8, I=.02, so 2.8/.02 = 140 Ohms. We round UP to the next higher standard value, and we get 150 Ohms.
Resistors come in many power ratings, usually expressed in watts. This tells us how much heat will be created at this resistor. To calculate the wattage required by this circuit, we use Watts = Volts*Amps. Since we are dissipating (burning off) 2.8 volts at .02 amps, Out resistor will be dissipating 0.056 watts. We usually use 1/4 watt or 1/8 watt resistors, these will be fine in this app.
If this doesn’t help, ask questions!
LyleHaze
no problem with all of that (current, voltage, ohm’s law etc) , but…
why do I have to link +5v to Gnd before my whole circuit?
I mean, if I have to do that, I guess the specifications of my psu are involved …?! right ?
I’m not sure I understand the question.
It’s about completing a circuit. Ground is the usual return point for most circuits.
It is possible to power the LEDs between +12 Volt and +5 volts, you would calculate using the difference between the two, a 7 volt difference.
Some people hook up their computer fans that way to slow them down..
In all cases, you need to complete the circuit or there will be no current.. and if there’s no current,
the LEDs will not light.
LyleHaze
hum…
of course.
I didn’t succeed in asking, sorry.
ok to close the circuit, of course.
BUT, on this schematics, there is a big short circuit, for my opinion … but indeed, I have to understand something that seem to be hard to explain.
the circuit will be close by putting the 5V to the 5V input on the core, and the gnd of the core to the gnd of the psu. right?
so, why should I put closer to the psu a line between 5v and gnd ??? maybe, it could be ONLY to light a pretty led … I thought this line had a specific function…
Do you men the heavy red line on the CORE in that drawing?
ok to close the circuit, of course.
BUT, on this schematics, there is a big short circuit, for my opinion … but indeed, I have to understand something that seem to be hard to explain.
the circuit will be close by putting the 5V to the 5V input on the core, and the gnd of the core to the gnd of the psu. right?
If you have +5V supply, you don’t need any voltage stabilizer circuit. Because shorted circuit In with Out.
In these stabilizers voltage drops~2,5volts, and they working with higher voltage ~+7,5volts to~+35volts.
Regards, Janis
I mean the little circuit between GND and +5V … Under the number (2) on the drawing.
When reading a schematic, crossed wires do NOT connect unless there is a dot there.
I hope that helps.
LyleHaze
…
Noone sees the line between the fuse and the ground …?
This line contains a led/diode and a resistor…
This is about that piece of wire i have doubt
…
Noone sees the line between the fuse and the ground …?
This line contains a led/diode and a resistor…
This is about that piece of wire i have doubt
Read ( 2 ) under drawing .
…i read before posting,but didnt understand the whole concept …
In my first post i asked about psu spec
how could i calculate the right value for my psu ?
PC-PSUs usually turn off, if the there is no load on their output.
No load means no current. To create a load, the LEDs and resistors are added.
There will be some current through those, so your PSU won’t turn off.
How much current it needs to stay on, actually depends on the PSU, but isn’t specified anywhere.
So you have to try. You won’t fry the PSU with wrong values, because it would either just turn off, or burn the LED…
ok I understood.
it NEEDS a dummy load.
thanks a lot.
I didn’t know that… 
PC-PSUs usually turn off, if the there is no load on their output.
No load means no current. To create a load, the LEDs and resistors are added.
There will be some current through those, so your PSU won’t turn off.
How much current it needs to stay on, actually depends on the PSU, but isn’t specified anywhere.
So you have to try. You won’t fry the PSU with wrong values, because it would either just turn off, or burn the LED…
I exhume an old post.
on the psu datasheet, it is written :
for 5V, mini current is 0.5A
5/0.5 = 10 ohm.
if I put a little led + a 100ohm resistor, I guess it could do the trick
I mean, putting a higher value resistor doesn’t seem to be a problem..
Am I right?
on the psu datasheet, it is written :
for 5V, mini current is 0.5A
5/0.5 = 10 ohm.
if I put a little led + a 100ohm resistor, I guess it could do the trick
I mean, putting a higher value resistor doesn’t seem to be a problem..
depends on what “trick” you are trying to accomplish..
If you’re trying to make the minimum current, then an LED and a 100 ohm resistor won’t be enough.
Getting back to ohms law:
assuming 5 volts, - 2 for the LED, with the 100 ohm resistor you specified..
you’ll get about 30 ma of current..
30 ma is more than most LEDs are rated for, so it will be bright, but will probably not last too long.
Your power supply “minimum” is 0.5A, or 500ma. so you’re not drawing the minimum it needs either.
So the power supply will probably just shut off, saving your LED.
Are you trying to light an LED, or are you trying to set up a dummy load for your switching power supply?
We have given you instructions to do each, but you seem to be somewhere in between.
Have Fun,
LyleHaze
Hmm. maybe you want to use an LED circuit AND make the minimum for your switching power supply..
Assuming 20 ma per LED, and a 500 ma minimum for your power supply, you’ll need at least 25 leds all lit up at once, and all connected in parallell, to make 0.5 amps
i talked about dummy load only.
my controller is the protodeck (you can check it on my website or on google)
it is a big controller that probably doesnt requirent a dummy load
but…i’d prefer to avoid unwanted switch off ..
Hmm. maybe you want to use an LED circuit AND make the minimum for your switching power supply..
Assuming 20 ma per LED, and a 500 ma minimum for your power supply, you’ll need at least 25 leds all lit up at once, and all connected in parallell, to make 0.5 amps
yes, still the same thing: make a minimum for my switching psu… only that.
My protodeck controller involves arou 96 RGB leds, 2 cores, 2 DOUT, 3 AIN, 3 DIN …
all is here : http://www.julienbayle.net/diy/protodeck/blog/
So probably I don’t need a minimum current load at the output of my PSU.
the datasheet pdf is here: http://www.farnell.com/datasheets/28497.pdf and mine is the ECM40.
for 5V, it is written 0,5 A minimum.
perhaps, only with all the midibox modules it involves, it doesn’t require any dummy load.
I’ll try without anything. it could be the first step