Constant current HELP

Hello all! Great to see such a forum, with such great information on it.

I am trying to power my new 3w red leds, which are 2.4-2.8 VF. I have looked everywhere for a suitable power adapter, and only came across one and its 5V @ 700Ma, which is only enough to power 2, works great and I wish it was 10V. I have found several other ones laying around but they are all 1000ma.

I have a question on making a constant current of 700Ma circuit, which would be fully customizable. ( For 5-10 LEDS )
[MODNOTE]link deleted[/MODNOTE] I followed some circuit guides but I cannot find certain things like an N channel logic level NFET at the store. I can guarantee they have it though under a different name.

Can anyone give me some idea's on perhaps a solution for me to power 5-10 of these. Thanks!

An old laptop power supply, plus a power MOSFET, could perhaps be used to run 7 or 8 in series ?

I can guarantee they have it though under a different name

Click to expand...

MOSFET ?

I think your missing the point entirely, but it comes up quite a bit.

Fist, Take the house: It may come with a 200 Amp 240/120 service. It's OK if I plug in a 5 W night light.

Same deal, a 5 V 700 mA supply, you can use anything bigger. so a 5V 200 Amp supply will work.

There is a a couple of laws in Electronics that state a voltage source can be replaced with a current source and a series resistor and a voltage source can be replaced with an ideal current source and a parallel resistor. An IDEAL current source has an output Z of infinity and an ideal voltage source has an output Z of zero.

That being said, the resistor needed for n LEDS in series is < R = (Vs-n*Vf)/If; so for 2 LEDSs and 5 V; R < (5-2*2.8)/0.7 amps in your case.
The Wattage of that resistor should be > (0.7^2)*R Watts.

Your right, it can only support 2 LEDs, but it doesn't matter what current the power supply can supply as long as it's bigger than 700 mA.

Now suppose,, you want to support 3 LED's. There could be one strong of 2 as above and one string of 1 in parallel.
The string of 1 requires an R of (5-2.8)/0.7

This adds another parallel path or current, so you will need a transformer that can supply at least 1400 mA.

4 LED's would require a resistor change, but no transformer change.

Get the idea?

Alright, thanks for clearing up some basics.

I am however, still not understanding why I would need a resistor if I am matching the Volts, with my devices. I want to have 4 in series, I would need ANY 12v power supply 700ma+, with no resistors?

Using just this V-REGULATOR ADJ. 1.2V-37V 1.5A TO-220, and a 1w resistor, with a power supply of 37V - 1.5A. I would be able to power 13-14 of these 2.4-2.8 Vf @ 700Ma in series ?

Is this correct ? ?

Im real excited to get started especially since I found this at a garage sale for 5$, its a switchable power supply looks like for a printer.

It says 32v = 925ma LPS
16v = 625ma

So when I chop the wires I would just cap off the 16v end and not use it ? ?

LED's is not like a radio. the LED's need a specific CURRENT to operate. All devices in a SERIES circuit use the same amount of current.
See: https://www.ibiblio.org/kuphaldt/electricCircuits/DC/DC_5.html

As the voltage from the battery passes through each LED, it gobbles some voltage. (I'll use 2 V)
So, if we had a 12 V battery and put in 1 LED, at then end of the LED I's have 10 volts left. If I put 2, I'd have 12-2(2) or 8 V left.

If I don;t gobble up that 8V, my LEDs will burn up.

So, if they needed 100 mA for each LED, then I would need a resistor that will gobble up the last 8 volts, so I would need a resistor of 8 V/100 E-3 or 800 ohms. Now since resistors blow up when their power rating is exceeded, we have to check.
P=V*I or (8V)(100e-3 Amps) or 0.08W, so my resistor has to have a wattage rating > 0.08 W.

Your power supply still isn't enough (n*2.8)<32 and you need a resistor that gobbles up the rest of the voltage (32-whats left)/0.7Amps

Now what you can do, is operate the LED's at 625 mA and use both power supplies.
Same deal (a*2.8)<16 and ( b*2.8) < 32 and a+b =13 or 14
Now select two resistors (16-what's left2)/0.625 and 32-what's left)/0.625

Im still not understanding why cant I just use the 16v ~ 625ma, and add up enough LEDS to equal the 16V. My lights are rated at 700, im sure 625 would still light them.

That's why I picked 625. It would mean the same brightness for both strings.

Well, you have to meet Thevinin and Norton. These guys said that an IDEAL voltage source has a series reistance of zero and and IDEAL current source has a PARALLEL resistance of infinity.

They also said that you can replace a real voltage source (an ideal source and a series resistance) with a current source in parallel with a finite resistor and vice versa.

In the model for an LED, there is a parameter called rds(on) which is the ON resistance and it's actually a function of current. It's fairly small.

Your EXTERNAL resistance must be much greater than the internal sum of all of the rds's so the external resistor dominates.

The LED must be current FED. Just trust us.

Okay thanks for explaining some advanced topics, however right now all I have to work with is this 5.5V ~ 700Ma, and the printer PSU 32v = (925ma LPS
16v = 625ma 40watts max)

I also have 1/4 watt resistors on hand, they say " 100k 1/4w 5% CF ".

Right now my plan for the PSU is to use the 32V ~ 925Ma and chop that down to 700Ma with these resistors I have.

Is this going to work ?

No,

The 32 and 16 V supplies may be unregulated which means the voltage will change appreciably with load.

32/2.8 is 11.x, so let's pick 10

32-28 = 4 Volts

R=4 V/0.625

That's 6.4 ohms. Check the power rating I*I*R or (0.625)(0.625)*6.4

So, your resistor has to be > 2.5 Watts; so use a 3 to 5 Watt Resistor

That's it.

Aside:
Each time you take 2 resistors in parallel, you half the value and double the wattage. 100k||100K = 50K and
100K||100K||100K||100K = 25K

So, you would have a 25K, 1 W resistor. It will take lot of resistors to get to 6.2 or something close.

LEDs have a very sharp voltage to current transfer function at their voltage turn on point. This means that they will draw almost no current below that turn on point, draw a reasonable operating current at the turn on point, then draw enough current to fail quickly above the turn on point.

In practical terms, you need to provide enough voltage to be above the highest possible turn on point, but limit the current so that you don't burn the LEDs up with too much current. The simplest way to limit the current is with resistance. A more precise way is to use an active current limit circuit. The most power efficient way is to use a dedicated switching DC to DC converter.

The 100k resistors you have have way to much resistance to use in this case. See kiss' posts above for how to select the resistor.

Electriceye said:

Im real excited to get started especially since I found this at a garage sale for 5$, its a switchable power supply looks like for a printer.

It says 32v = 925ma LPS
16v = 625ma

So when I chop the wires I would just cap off the 16v end and not use it ? ?

Click to expand...

That supply will work fine, but YOU need to limit the current. Unless a power supply is specifically designed as a limited current source, you must assume that it is a voltage source. The output current rating is the maximum that you should ever draw from it, but it is not internally limited to that level. It will go ahead and deliver voltage into a load drawing more current than the rating. But it won't do so for long. It may not fail for days, weeks, or even months, (depending on the overload factor) but it will fail sooner than it would if running within it's spec'd output current.

Running an LED at less than it's spec'd current is OK. It will just reduce the brightness a little. The plus side is that it will run cooler, and last longer.