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I have an LED with no specs for it. I decided to measure the voltage drop across the LED, so I connected it to 5 V power supply and 325 ohm resistor. I measured the resistor with multi-meter, then I measured the current and I had 6.38 mA. I then calculated voltage drop across the resistor which was 2.07 V (\$IR=V\$) and then I calculated the voltage drop across LED which is \$5-2.07 = 2.93\mathrm{V}\$. So I wrote down my voltage drop across the LED.

The day after I was using the same LED on 3.3V circuit. I decided to measure the LED's voltage drop again and it turned out to be 2.633 V and according to Kirchhoff's law it would affect my current because I am connecting this LED before the resistor.

Can somebody explain to me what happens? Why is it that the same LED has different characteristics with different voltages?

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    \$\begingroup\$ Please show circuit diagrams. Please break up the text to show the two test cases. \$\endgroup\$ – winny Oct 29 '18 at 18:08
  • \$\begingroup\$ Circuit diagram ? It is Power source -> LED -> Resistor ->GND \$\endgroup\$ – Anton Stafeyev Oct 29 '18 at 18:08
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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 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

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  • \$\begingroup\$ Thanks for the diagram. is there any mathematicval formula to calculate voltage drops at different curerrent ? \$\endgroup\$ – Anton Stafeyev Oct 29 '18 at 19:08
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    \$\begingroup\$ No. The way it goes you go to the datasheet select a luminous intensity, get the current for that intensity, then go to the IV curve to the the Vf at that current, then use the Vf and current to calculate the resistor value. . \$\endgroup\$ – Misunderstood Oct 29 '18 at 19:23
  • \$\begingroup\$ in my case i need to know LED's Vfd very precisly to display RGB value. ranginf from 0-255. so it is very important to know exact forward voltage so i can see the luminosity as exact as possible. how would one do it ? because hand plotting it wouldn't be as precise as mathematical conversions \$\endgroup\$ – Anton Stafeyev Oct 29 '18 at 20:01
  • \$\begingroup\$ See addition to my answer. \$\endgroup\$ – Misunderstood Oct 29 '18 at 20:04
  • \$\begingroup\$ here is the formula i came up with. 255RGBV = I*max * (Vs - Vfd)/R \$\endgroup\$ – Anton Stafeyev Oct 29 '18 at 20:05
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enter image description here

Figure 1. Typical LED I-V curves. Source: LEDnique.

The forward voltage drop, VF, across an LED is not a constant but varies with the current. If we take the blue LED in Figure 1 we can see that at 40 mA it will have a VF of 3.0 V. If we reduce the current to 20 mA the VF will reduce to about 2.6 V.

... and according to Kirchhoff law it would affect my current bacause I am connecting this LED before the resistor.

The order of components in a series circuit does not matter. The same current flows through all of them.

There is also a video on the linked page which shows how to draw the I-V graph.

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  • \$\begingroup\$ in my case i need to know LED's Vfd very precisly to display RGB value. ranginf from 0-255. so it is very important to know exact forward voltage so i can se the luminosity as exact as possible. how would one do it ? \$\endgroup\$ – Anton Stafeyev Oct 29 '18 at 20:01
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    \$\begingroup\$ @Anton you need a constant current source for that, not a simple voltage source with a resistor. Besides, the I-V characteristic may vary greatly with temperature, so your current design can't allow the accuracy you seem to require. \$\endgroup\$ – dim Oct 29 '18 at 20:36
  • \$\begingroup\$ what if i used transistor to just controll the current that flows via LED without a resistor \$\endgroup\$ – Anton Stafeyev Oct 29 '18 at 21:26
  • \$\begingroup\$ It would be difficult to control. See constant current driver for details on how it is done with a current sensing resistor. In any case, you need to be controlling LED current, not voltage. You can't rely on the \$ V_F \$ being consistent from LED to LED. See Vf and binning for more on this. \$\endgroup\$ – Transistor Oct 29 '18 at 22:36
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Here's what happens. Voltage generator and series resistance have the following V-I characteristic - note that by changing the value of the voltage, the characteristic shifts. On the V axis the characteristic reaches the open circuit voltage, while on the I axis you have the short circuit current. This is known as the 'load line'.

enter image description here

The diode has the following characteristic (this is a simplified Shockley relation, but you can derive yours from the datasheet)

enter image description here

When you put the two parts together, the circuit will work so as to have the same voltage and the same current in both parts. The intersection of the generator and diode characteristic is the pair Vd Id you are looking for:

enter image description here

As you can see, if you raise the voltage, you move the point of intersection and you end up with a higher voltage drop across the diode. Note, however, that the diode characteristic hasn't changed.

The quiescent point (Vd, Id) is the solution of the system of two equations that describe the characteristic. The nonlinearity of the diode characteristic is such that you cannot solve it analytically (but have a look at this answer by jonk). You can either solve it graphically (by drawing the load line on the curve supplied by the diode manufacturer) or numerically (by hand with an iterative method, or by computer firing up Matlab or Mathematica - you might need to scan the characteristic to produce a model) or, if you are given a Spice model, by using Spice.

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  • \$\begingroup\$ i solved it by replacing resistor with transistor so i just limit the current via NPN transistor instead of complicating things with resistor \$\endgroup\$ – Anton Stafeyev Oct 29 '18 at 23:45

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