LED v-drop calculation: Can you check my equations?

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Folks,

I've pulled two LEDs our of a Cellphone board and I want to use them. When I measured one with my multimeter while it was still on the board I got a 0.672 vdrop. My supply voltage is 3.3v, I'd like to put 20mA through the leds. I worked out the resistance from that with this equation:

R = (3.3v-(2*.672v))/.02A
R = 97.2 ohms

So I designed my circuit with a 100ohm resistor. Not that I have it put together I find the LEDs to be very dim. I measured the current through the two (connected in series) and I get 3.45mA.

Am I doing the math right if I then calculate the voltage drop from this reading and the 100 ohm resistor?

V = (100 ohms) * (0.00345A)
V = 0.345v

Vdrop = Vsource -Vcalcuated
Vdrop = 3.3v - 0.345v
Vdrop = 2.955v

So I'm reading about a 3v drop. I If I calculate the resistance value I want i get 15 ohms

R = (Vsource - Vdrop) / A
R = (3.3v - 3v) / .02A
R = 15 ohms

So a 15 ohm resistor is what I need? Does is this look correct?

Thanks!

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There is no way to tell without a datasheet on the LED, the current the LED will draw will change nonlinearly with the voltage. The best to do would be try out a few different resistances in series with the LED and don't worry about the current, just worry about it being the appropriate brightness. I would guess your 15 ohms would be somewhat close to producing 20mA but it might produce less.

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That "voltage drop" is very suspicious. It is way to low for any LED. It is almost like it was being controlled by PWM and the meter was averaging. You would expect at least something in the vicinity of 2V, minimum (OK, maybe 1.8 in a few cases). But you should expect an even higher voltage at 20ma.

For a 3.3V supply, stacking two LEDs in series in probably not a useful way to go.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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ka7ehk wrote:
That "voltage drop" is very suspicious. It is way to low for any LED. It is almost like it was being controlled by PWM and the meter was averaging.

Either that, or he was using the "diode forward voltage check" on his meter. Because of the IV curve, they'll show very low Vf for the LEDs.

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ka7ehk wrote:
That "voltage drop" is very suspicious. It is way to low for any LED. It is almost like it was being controlled by PWM and the meter was averaging. You would expect at least something in the vicinity of 2V, minimum (OK, maybe 1.8 in a few cases). But you should expect an even higher voltage at 20ma.

For a 3.3V supply, stacking two LEDs in series in probably not a useful way to go.

Jim

You will notice that when he actually tested out the VDrop of the LED he got 3V, with a 3.45mA current draw. It is likely that the rated VDrop for the LED is higher than 3V considering the low amount of current he was moving through it.

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I was using the diode forward voltage check, you are correct.

As for diodes in series.... hmmm. I think that I am actually in parallel. Here's how I have things hooked up:

Vtg----/\/\/\/-------------
        Resistor    |     |
                    |     |
                   ---   ---
              LED1 \ /   \ / LED2
                    |     |
GND------------------------

I did try out several different resistors and noticed that results are not linear:

Resistor:    Current:
15 Ohm       12 mA
10 Ohm       15.3 mA
2.2 Ohm      ~29mA
1 Ohm        >32mA

Looks like 10 ohms is my best bet.

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The first measured 0,7 V drop are to low for a LED even at very low currents. Having 3 volts for two LEDs in series is possible for 2 red LEDs, but still quite low (or less likely at high temperatures).

So there may not be a simple way to run the two LED in series from a 3.3 V supply. A simple resistor will not do the trick anymore: for a reasonable current (e.g. 15 mA) you would need about an extra 50 mV per LED. So you would have only 200 mV for the resistor. That is very close to compensate for temperature-dependencies.
It somewhat depends on the application, but it may be better to get one high brightness LED instead of the two dim one. The higher efficiency LEDs tend to have a slightly higher voltage (e.g. 1.6V) than the very cheap and dim ones.

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Kleinstein wrote:
The first measured 0,7 V drop are to low for a LED even at very low currents.

It's not far off from times that I've measured some LEDs with my diode check. A few weeks ago, I checked the polarity on some Optek 1W red and yellow LEDs, and they lit up *just barely*, and measured around 1V.

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I once measured some red SMD leds which did not light up on a PCB. It turned out the notch was at the anode end, not at the cathode end.

Anyway, the LEDs did light up when examined with a multimeter in diode mode, showing around 1.6V. They were quite nanoscopic though, 0406 and 0805, so very low current ones.

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Hi Barney,

You may do better to start with a 5 or 6 or 9 V DC supply, and a bigger resistor, say 470 ohms, and ONE LED.

Get a nice brightness level by adjusting your resistor value, then measure the forward voltage drop across the LED, and measure the voltage drop across the resistor to calculate your current.

With that information, measured separately for each LED, you will be in a better position to determine how to power it from a 3+ voltage supply.

Also, it is reasonable to put a string of LEDs in series, so all of them get the same current, when you have a high enough driving voltage, (Which you probably do not have, in this case).

It would generally not be a great idea to put them in parallel, as they will have slightly different forward voltages, will draw different amounts of current, and will be of unequal intensity. (That said, it is occassionally done...). The "better" approach would be to use a separate resistor for each LED.

JC

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LED brightness is better controlled by means of PWM,
if you have a uc pin available for it.