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 5 deleted 1 character in body edited Oct 29 '18 at 20:47 Misunderstood 5,87611 gold badge55 silver badges1919 bronze badges 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 4 added 2 characters in body edited Oct 29 '18 at 20:25 Misunderstood 5,87611 gold badge55 silver badges1919 bronze badges 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 3 added 156 characters in body edited Oct 29 '18 at 20:03 Misunderstood 5,87611 gold badge55 silver badges1919 bronze badges 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 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: Looking on an IV curve: 2 mA ≈ 2.6 V 7 mA ≈ 2.9 V 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. 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. 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. 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. 2 added 1580 characters in body edited Oct 29 '18 at 20:01 Misunderstood 5,87611 gold badge55 silver badges1919 bronze badges 1 answered Oct 29 '18 at 18:56 Misunderstood 5,87611 gold badge55 silver badges1919 bronze badges