5 deleted 1 character in body
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I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev

it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? – Anton Stafeyev

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED.
This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give useus 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev

it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? – Anton Stafeyev

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED.
This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give use 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev

it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? – Anton Stafeyev

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED.
This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give us 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

4 added 2 characters in body
source | link

I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev

it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? – Anton Stafeyev

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED.   
This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give use 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev

it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? – Anton Stafeyev

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED.  This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give use 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev

it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? – Anton Stafeyev

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED. 
This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give use 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

3 added 156 characters in body
source | link

I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev  

it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? – Anton Stafeyev

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED. This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give use 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev  

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED. This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give use 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

I assume you used the same 325 Ω in both cases.

5V with a 325 Ω resistor and Vf = ≈ 7 mA
3.3V with a 325 Ω resistor and Vf = ≈ 2 mA


plugging your numbers in to a resistor calculator:

enter image description here

enter image description here
Source: LED Series Resistor Calculator



Looking on an IV curve:

2 mA ≈ 2.6 V
7 mA ≈ 2.9 V

enter image description here
Source: OSRAM blue LED


is there any mathematicval formula to calculate voltage drops at different curerrent ? – Anton Stafeyev

it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? – Anton Stafeyev

This is easier to show with a high power LED.

Let's say we want to make a flashlight with 1000 lumen output.
We select this 900 lumen LED. This luminous intensity is measured at 400 mA and 85° C.
This LED's maximum current is 750 mA.

enter image description here


We have to up the current from 400 ma to get 1000 lumens.

1000/900 = 111%

So we go to the Relative Luminous Intensity graph.
Draw a line across at 111%
Draw a line down from the point where the 111% line intersects with the 85° curve.
We see that 475 mA should give use 1000 lumens.

enter image description here


We go to the IV curve and draw a line from 475 mA up to the 85˜ curve.
The draw a line from were they intersect to the forward voltage.

The forward voltage for this LED at 475 mA is 17.75V.

enter image description here


Let's say we are using a supply voltage of 24V.
We go to the resistor calculator and enter 24V supply, 475 mA, and 17.75V for the forward voltage.

So for 1000 lumens we need a 13.3Ω, 5 Watt resistor.

enter image description here

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