Question on LED Voltage. 7.4v too high?

Thread Starter


Joined May 5, 2009
Hello, I am working on a device and I have chosen 2 2.3" 7 segment hi-red LED displays to go in it.

as you can see, it says
Electro-Optical Data (@ 20 mA)
[SIZE=-1]Forward Voltage - Typical: 7.40 V[/SIZE]

I plugged that information into LED wizard and it told me the typical 7.4 was very high. Then after researching around..everyone says a red led is usually around 1.7-2v.

is this just because of how big the display is? Or is the 7.4 for all 7 segments, and I need to divde that number by 7 to get my typical forward voltage?


Joined May 6, 2009
If you look a little further down you will see a schematic showing that each of the segments has 4 LEDs in series. This means that the voltage (at 20 mA) will be 4 times the voltage of one LED. This would be 7.4/4 or about 1.85V per individual LED.

Does this sound about right?

The decimal point only has 2 LEDs. I would expect this to have about 1/2 the drop as the other segments. Unless you are powering the segments with a constant current driver, you may want to have a different value for the series resistor for this string.


Joined Apr 5, 2008

When you look at the schematic on the bottom of the page, you will see that each segment consists of 4 leds and the DP of 2 leds.

What powersupply are you using to drive the display?


Thread Starter


Joined May 5, 2009
Ah that explains it a bit. I am still getting the design down, and to be honest i am new to electronics and really lost on this project. I am getting ready to try to build the board, and don't really know what power supply to use. I modified this schematic:

to try and use a large wheel to count up or down when it is spun CW or CCW. To:

But I have another thread for that in the projects section. ---> Here

Is it ok just to have 1 resistor per segment in this case? Even though it has 4 led's each segment? I currently have 330 ohn resistors to compensate for the 7.4v, Is that ok because the resistor will be set up for those 4 LED's in the segment?

This just have to be battery powered, not sure yet how to power it.

EDIT: Kind of answered my own question in the led wizard. It said an array of 4 led's at 1.85v each, would still use the 330ohm resistor.

In the other thread, I was told to use a 12v batter. this sound about right?
Last edited:


Joined Jul 17, 2007
A 12v sealed lead-acid battery measures around 12.7-12.9v when fully charged.
12.8v - 7.4v = 5.4v remaining.
Now you need a resistor that will allow 20mA or less to pass across that voltage drop.
R=E/I (Resistance = Voltage / Current)
R = 5.4v/20mA = 5.4 / 0.02 = 270 Ohms. This is a standard value of resistance.

If you use 330 Ohms, you would get I=E/R; or I = 5.4/330 = 16.4mA
That would still make the LED segments pretty bright, and would extend their service life.

For the decimal point:
12.8v - 3.7v = 9.1v
R = 9.1v/20mA = 9.1/0.02 = 455 Ohms. 470 Ohms is the closest standard value.

If you want to have the 16.4mA current instead:
R = 9.1 / 16.4mA = 9.1/0.0164 = 555 Ohms. 560 Ohms is the closest standard value.

Table of standard resistance values: