Very elementary question, I think...

I need to measure the Vf for some LED displays I received that have no datasheet so that I can size the resistor appropriately using the equation (Vsupply - Vf )/I. So I apply 12v to the Common Anode and run a resistor to ground for one of the segments. I measure the voltage across the LED (and before the resistor) and get 7v. So here's my question:

Is the value I plug into the equation Vf = 7? or is it equal to the drop of 5V?

Thanks!

 

Well if you measure across the diode and get 7V, then Vf is 7V. Be sure to check that the current used is in the correct range to get good results.

 

I measure the voltage across the LED (and before the resistor) and get 7v.

Be sure you are truly measuring "across the LED" and nothing else in series with it. A "normal" LED will be more like 3V depending on the color.

Note that Vf is a weak function of current, varying maybe 0.5V over the useful range, and a weaker still function of temperature.

 

7 volts sounds more like a reverse voltage breakdown then a forward voltage.

Try turning the thing around and measure it again.

What value resistor are you using? I would suggest something between 10K and 1K for an unknown LED: these values limit the current to below 1.2 and 12 mA, which should be safe for any LED to handle.

 

How many LED's per segment?

 

hmmm, not sure how many LEDs per segment. These are 2.3" displays. I'm sure I'm measuring the voltage correctly, and the 1k resistor I'm using results in a dim segment.

So, I think kubeek answered my question. I use 7V for the Vf. Many thanks!

 

Vary the resistor until you get reasonable brightness. Then use that value resistor.
You can't go wrong there.

 

Sure, that works, except I just want to make sure that what I think is a reasonable brightness isn't pumping 50 mA through the segments and therefore shortening the life of the displays.

In any case, if I am calculating correctly, ~250 ohms should yield 20 mA per segment.

 

Then measure the voltage across the resistor and use Ohm's Law I = V/R to determine the current.

 

(12v - 7v)/ 20 mA = 250 ohms. So I think I'm doing this right. A 250 ohm resistor per segment yields 20 mA. I just wanted to be sure that when I measure 7V across the LED that that is the value I plug into the equation (as opposed to 5V). Many thanks!

 

Incorrect.

The voltage to use for R = V/I is the voltage across the resistor, not the LED forward voltage.

 

Incorrect.

The voltage to use for R = V/I is the voltage across the resistor, not the LED forward voltage.

Thanks Mr. Chips, but I am clearly confused. Isn't the equation for calculating the appropriate resistor for an LED:

Vsupply-Vforward/I= R?

Apparently not, but it seems like I see that everywhere. Thanks again.

 

Your equation is almost correct.


(Vsupply - Vforward) / I = R

 

Yes, thanks, that is what I meant, mea culpa.

So, with that equation, my supply voltage is 12. How do I measure Vf? I did it by measuring across the LED with a 1 k resistor attached and got 7 volts. Common anode displays, one probe at the common anode, one probe at the segment pin and a current-limiting resistor connected to ground AFTER the probe. Not correct?

 

That means 5mA. Try next with a 220R and see what the Vf will be.

 

Yes thanks, sorry I am being unclear, my aim is simply to calculate the resistor properly. To do that I need to first measure Vf properly. I think I now know how to do that. That was my original question, how to measure Vf properly. Thanks to all.

 

Do you realize that Vf changes with current? You might get 7V at 5mA , but for example 7.5V at 20mA.

 

Oh ok, good to know, thanks!

 

...
So, with that equation, my supply voltage is 12. How do I measure Vf? I did it by measuring across the LED with a 1 k resistor attached and got 7 volts.
...

That is good information.
LED Vf = 7v
resistor voltage = 5v

actual current used when testing (we calculate, using the resistor voltage);
I = E/R
I = 5v / 1000 ohms
I = 5mA

So you now know the LED Vf is 7v when passing 5mA through the LED.

Please be aware on these large digit displays; the decimal points usually have less LED chips inside, and have a much lower Vf than the digit segments!

 

RB, many thanks, very clear explanation.