🚀 go-pugleaf

Nucharin W. Jansen wrote:
> Typical Graph between V(f) and I(f) of each Ultra LEDs are not the same too.
> that made me very hard to calculate R drop and I think It can't
> calculate R on paper.
> I have to try adjust the POT in real world one by one with every strings.
> And don't swap LEDs between strings any more.


>>Yes, I mean in series.
>>But two strings can not use the same R values.
>>So sad it very hards to calculate the current and R.
>>the forward voltage drop does not identical in each LED.
>>and it changes when current change too.

>>ex:

>>strings 1 :   ( 12 volt - ( 3.453 + 3.6324 + 3.833 ) ) / 40 ohm = 27 mA

>>Now I want 20 mA ,  How to calculate the resistor ?
>>Since forward voltage change when the current change too.



> No, If you answer R = 40 * 0.027 / 0.020 = 54 ohm 
> It is totally wrong. You can't get 20mA from R = 54 ohm.
> Because now at R = 54 ohm, the V(f) of 3 LEDs are change.

> headache.. We can't calculate R drop at the exact 20mA in the paper.

Well, it doesn't have to be _exactly_ 20 mA.  If it is within + or - 20 
percent, then it will be good enough.  If you reduce the current by 20 
percent, you will probably not notice any difference in the brightness 
of the LEDs.

If you are concerned about matching the brightness of all LEDs so they 
look the same, then having the same current thru all of them will not 
give you equal brightness.  One LED might be much brighter at the same 
current.  (Maybe I should say that the same current will not guarantee 
equal brightness.  This is because you could have LEDs from the same 
batch or lot that just happen to have been picked to be closely matched 
in brightness.  But that is probably not likely if you just bought LEDs 
from a store, individually packaged, for example.)

If you must have current that is more accurate than this, then you can 
use fewer LEDs in series, which gives more voltage across a higher value 
of series resistor, which will give you more accurate current.  Or you 
can use a current source.  You can use a current diode, or you can 
connect the gate to the source of a JFET, and use it as a current 
source.  But JFET currents are so widely variable that you might have to 
buy a dozen to get a few that match the current that you need.  Or else 
you could connect two 10 mA JFETs in parallel to get 20 mA, or three 6.7 
mA JFETs to give 20 mA, etc.  But I bet you will hear a lot of others 
here say that you shouldn't do that.

Here is a circuit that is a simple current regulator.  It is not very 
accurate because the voltage across the B-E junction diode varies with 
temperature.  Use a 33 ohm resistor and then connect a 300 ohm resistor 
in parallel to give about 30 ohms, or a 220 ohm in parallel to give 
about 28.7 ohms, etc.  You should be able to get the current up to 20 mA 
easily.  If you must ground the LEDs you can invert the circuit by using 
PNP transistors.  2N4401 works well for NPNs, or else 2N4403 for PNPs.
The circuit requires about 2V across it to keep it working.  I think if 
you use this curcuit, it will do what you want.


     LED Current Regulator Using Two Transistors
           (must view with courier font)


                      o +V
                      |
                      |
                     ---  1 or
                     \ /  more
                     ---  LEDs
                      |    in
           +----------+   series
           |          |
           3          |
           3          |
           k         /
           |       |/   NPN GP
           +-------|    2N4401
           |       |\
           |        E\
           |          |
            \         |
             \|       |
     NPN GP   |-470---+
     2N3904  /|       |  Sets LED
            /E        3  current = 0.6V / R
           |          3  33 ohms gives about 18 mA.
           |          |
           +----------+---o GND

Since the 33k current also goes thru the LED, the lower I limit is when 
the emitter resistor approaches infinity, when the only current thru the 
LED is what is going thru the 33k.  So the 33k can be connected to the 
positive supply instead of thru the LED.  This 33k value may have to be 
changed if the circuit is regulating low voltages or high currents.

-----------------------------------------------------------------


I modified the circuit at the following URL to use just a single LED (it 
is *strongly* recommended that you do not connect LEDs in parallel as 
they did in this project).  I used a regular 2N4401 transistor instead 
of the power FET.

"A linear Constant Current LED Lamp Dimmer" 
http://radiolocation.tripod.com/LEDdimmer/LEDlampDimmer.html


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