So I made a small 5v Power Supply (Regulated) using one of Sparkfuns tutorials.

enter image description here

I used a Slightly Different LED (a 2.6v with 20mA) and a 270 ohm resistor ( have no idea why they use a 330 Ohm one...270 was even overkill for mine.

Anyways Im using my Multimeter and measuring the voltages (Marked in purple, - for Ground probe, and + for my + probe).

Im getting a correct 5v reading after the LM7805 (well like 4.95 volts.....close enough), but when I measure after the LED I get 1.8v....

This doesn't make any sense.......

Here are some pictures of the circuit itself, I have drawn arrows to where I measured the 1.8v with my Probe (Positive red probe), the Resistor is a 270ohm, and I was incorrect about the Forward Voltage, it's actually 2.25 (MAX was 2.6......thats what got me confused). Sorry for the Long pictures, but wide-ways they are kinda hard to see. Capacitor values are same as schematic.

edit: Corrected Picture. edit2: Added picture of circuit

  • \$\begingroup\$ I don't understand what you are measuring: if your probe is across the resistor, it's pretty normal what you get. But if the resistor is overkill or not you calculate from the current that you want. \$\endgroup\$
    – clabacchio
    Apr 24, 2012 at 12:56
  • 2
    \$\begingroup\$ I'm guessing you somehow managed to measure the voltage accross the LED. This is more likely if it's red since those have lower voltage than other colors and 1.8 V is about right. \$\endgroup\$ Apr 24, 2012 at 13:12
  • \$\begingroup\$ I just reuploaded new picture, my apologies. \$\endgroup\$
    – user3073
    Apr 24, 2012 at 13:15
  • \$\begingroup\$ Voltage is ALWAYS relative to something. If you measure +1.8V it is relaive to some other point in the circuit. You need to specifiy both points. \$\endgroup\$
    – Russell McMahon
    Apr 24, 2012 at 14:05
  • \$\begingroup\$ See addition to my answer re LED voltage and colour. \$\endgroup\$
    – Russell McMahon
    Apr 24, 2012 at 14:21

4 Answers 4


If I'm not mistaken, your schematic and breadboard circuits do not match.

In the schematic you have your LED before your resistor. On the breadboard you have your resistor before your LED.

Since you are measuring the anode of the LED in the physical circuit, what you are measuring is the voltage drop after the resistor.

  • \$\begingroup\$ I thought it didn't matter where the Resistor went in regards to the LED....and I thought Resistors only reduced current not voltage? \$\endgroup\$
    – user3073
    Apr 24, 2012 at 16:59
  • 1
    \$\begingroup\$ @Sauron, simple Ohms law. V = IR. If a current flows through a resistor, it will have a voltage drop across that resistor according to that formula. \$\endgroup\$
    – Jon L
    Apr 24, 2012 at 17:00
  • \$\begingroup\$ Oh.....I see, well that makes sense. I guess I just thought it didn't matter in relation to the LED....but I guess if my resistor is before it......ya.....well that makes sense. God I feel so dumb lol \$\endgroup\$
    – user3073
    Apr 24, 2012 at 17:05
  • \$\begingroup\$ @Sauron, gotta start somewhere! and no better place than ohm's law :D \$\endgroup\$
    – Jon L
    Apr 24, 2012 at 17:07
  • 4
    \$\begingroup\$ @Sauron It does NOT matter where the LED goes as far as being able to function. However, it does mater for what voltages you will see. You were measuring the voltage across the LED to be 1.8v. The 3v you are seeing now roughly makes sense as well. I would expect to see 5v-1.8v=3.2v. \$\endgroup\$
    – Kellenjb
    Apr 24, 2012 at 17:43

Your question does not make sense as stated, but one can guess at what is happening.

If you measure the LED Anode = + side you are measuring the identical location to the +5 output As you report two different voltages for the same point there is something wrong with what you are reporting.

You did not report LED colour, and this makes a difference.

A red LED will drop about 2V - could be up to about 2.3V and may be under 2V. The 1.8V you report MAY be the voltage across the LED. As YOU are placing the test probes YOU MUST tell us where they are when you measure. Otherwise your questions are random and the answer is "The sound of one dog barking".

The LED series resistor drops the voltage not dropped across the LED. IF LD voltage =2V then resistor voltage = 5-2 = 3V. The current in the resistor and in the LED then is R = V/I = 3/330 ~= 9 mA. Changing the resistor changes the current.

The chart below is out of date and LED voltages change as technology improves, but this gives an indication how both colour and current effect a LED's forward voltage.
Draw a line horizontally at say 20 mA and see how the voltage varies with LED colour.

For example, using the LEDs below, and noting that the same colour LED in a newerdesign and/or different technology may give a different result, consider a red and HE green LED from the chart below.
At 20 mA the red LED drops (= "has a forward voltage drop of") about 1.7V (sounds close to your 1.8V) and the HE green drops about 2.2V.

Much more related information here


3 Bright Red
4 Ultra Red
5 Red
6 HE Red
7 Orange
8 Yellow
9 HE Green

enter image description here

  • \$\begingroup\$ Why would the LED Color matter? I stated the Dropout Voltage (2.6v) and mA (20). It's a LED Red for reference however. i'll reupload the image however with where I placed the probes. \$\endgroup\$
    – user3073
    Apr 24, 2012 at 13:14
  • \$\begingroup\$ @Sauron because for physic reasons the drop of the voltage is dependent on the color, and usually for a red one is 1.8 V \$\endgroup\$
    – clabacchio
    Apr 24, 2012 at 13:16
  • \$\begingroup\$ Then I must have misunderstood, Why does the package for the LED I used say 2.6v Drop?.......is that something else entirely. \$\endgroup\$
    – user3073
    Apr 24, 2012 at 13:44
  • \$\begingroup\$ @Sauron - show us yur circuit with ALL voltages shown relative to ground and we will be able to answer all your questions (probably :-) ). So far we have no idea where you are measuring voltages between. \$\endgroup\$
    – Russell McMahon
    Apr 24, 2012 at 14:20
  • \$\begingroup\$ Well the 1.8v is the only one I measured, and I put the Probes exactly where the purple Dots are (+/-) on the picture. \$\endgroup\$
    – user3073
    Apr 24, 2012 at 16:20

The circuit in your photograph is not the same as the circuit in your schematic.

The positions of R1 and D2 are reversed.

So when you measure the voltage you are just measuring the forward voltage of the LED, which is about 1.8 V as expected.

In the schematic, you showed measuring voltage across the series combination of the LED and resistor, which would have read as 5 V.


For an LED you want to set the current flowing through the LED by choosing the current limiting resistor that is appropriate for the forward voltage of your LED.

Based on your LED datasheet (which you haven't linked to as far as I can tell, by the way) the Forward Voltage Drop (I don't know what Dropout Voltage is in the context of LEDs) is 2.6V.

Based on your circuit measurements, the actual Forward Voltage Drop you are experiencing is about 3.15V. The current your LED will have pulled through it is exactly the voltage you measured past the LED (1.8V) divided by the the current limiting resistance (270 Ohm). That is only 6-7mA. A bigger resistance will only make that current smaller, and you really want your current to be around the knee of your IV curve for the LED which is often around 20mA.

Assuming your actual Forward Voltage Drop is 3.15V, you would need to set your current limiting resistance to something not much bigger than 1.8V / 20mA = 90 Ohms. That being said, I've only really seen forward voltage drops that high for Green and Blue LEDs, so that sounds fishy.

  • \$\begingroup\$ It's an older Radio Shack LED, it just says 2.6 on the package. I think i got it like maybe 2-3 years ago, so maybe it's inefficient. Regardless even if it was 2.2v or w/e the 1.8v reading still seems strange. \$\endgroup\$
    – user3073
    Apr 24, 2012 at 16:19

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