Electronic – Why is this Circuit Measuring 1.8v

ledvoltage

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

Best Answer

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.

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The 2 rails at the top of a breadboard are all connected along the straight line. In the picture, you are connecting both ends of the resistor to the same rail, so it's as if it's not even there (all the current will bypass it and travel through the LED).

It's the equivalent circuit of having the resistor ends connected (same voltage):

          +---R---+      <----- Resistor is useless, current will bypass it
          |       |
 +--------+-------+-----+
 |                      |
(+)                     |
 V                    (LED)
(-)                     |
 |                      |
 +----------------------+

If you are trying to have an LED to let you know when your power is connected, you will either need to use a pin line outside of the voltage rails (and make sure the current passes through the resistor), or else use an LED with a built-in resistor. As your circuit is now though, you run the risk of shorting out your battery, depending on how much current the LED allows through.

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When you press S1, you're storing a charge on the gate which has a small capacitance Cgs. This charge maintains the electric field which keeps the channel between the drain and source. Once you press S2, the charge on the gate is deplected and the channel is turned off

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