I have a 12v unregulated power supply. It reads as 15.6v. I need to get the voltage down to 3v.
I have done the calculations and I apparently need 630 ohm resistor. I have a 680 ohm resistor. When I test the voltage across this resistor, it goes from 15.6v down to 9v.
I am VEREY confused. Why is it 9v and not the expected 3v? 
Show us your calculation.
The voltage drop across a resistor depends on the current and the resistance, have a look at Ohm's Law
If this is for power you can't use a resistor, you need a buck converter, there are plenty for sale on your favourite e commerce web site.
PerryBebbington:
Show us your calculation.The voltage drop across a resistor depends on the current and the resistance, have a look at Ohm's Law
If this is for power you can't use a resistor, you need a buck converter, there are plenty for sale on your favourite e commerce web site.
Here are the numbers I am working with:
Start Voltage = 15.6v
End Voltage = 3v
0603 LED = .02 Amps
The I am even more confused. Please provide a schematic. If you have the LED in series with the resistor then the voltage is defined by the Vf of the LED, not the resistor. What is the Vf of a 0603 LED?
PerryBebbington:
The I am even more confused. Please provide a schematic. If you have the LED in series with the resistor then the voltage is defined by the Vf of the LED, not the resistor. What is the Vf of a 0603 LED?
I found this on the 0603 SMD LED. Is this what you are looking for?
Voltage - Forward (Vf) (Typ) = 2V
Current - Test = 20mA
So you want to make 5v out of your 12 volt source, sure use one of these https://smile.amazon.com/gp/product/B07Q5W1BG3/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1, the add in a MCP1700 to get the rest 5V down to 3.3V.
OK, now draw a schematic showing how you have wired it up and where you are measuring the voltages. You said 3V in your original post but what you have just said suggests 2V. Without knowing how you have it wired I am guessing.
You can draw on paper and photograph it.
A photo of what you have wired up helps a lot too.
Thanks,
EDIT
Measure the voltage across every component and across the power supply and show those on your schematic.
Is this the only items you will be powering ?
12v——[Resistor]——[Anode LED Cathode]——Ground
larryd:
Is this the only items you will be powering ?12v——[Resistor]——[Anode LED Cathode]——Ground
Yes.
(12v - LED(voltage) ) ÷ 20mA = resistance needed
12v - 3v(LED) = 9v across the resistor
9v ÷ .02A = 450 Ω resistance needed for 20mA of current flow
W(resistor) = Vr * Ir = 9v * .02A = .180 Watts use 1/2 watt.
However, many modern LEDs are plenty bright at 5mA 
larryd:
(12v - LED(voltage) ) ÷ 20mA = resistance needed12v - 3v = 9v
9v ÷ .02A = 450 Ω
This is the math I used. Here is the calculation with the correct input voltage:
15.6v - 3v = 12.6v
12.6v /.02A = 630 ohm
I setup a circuit with just a 680 ohm resistor. No LED. Shouldn't I be reading 3v or less (used 680 instead of 630) across the resistor? I am reading 9v.
Is the LED properly connected, i.e. the anode goes towards the positive 12v terminal.
It might be possible you have damaged the LED, try a new one.
Always show us a good image of the actual wiring.
Thank you all for your suggestions on how to fix my circuit. The problem is that I do not have a circuit problem.
I am just trying to understand the fundamentals. In this case, resistors. I thought I should get 3v or less across my resistor and I am getting 9v. I am trying to understand why 9v in place of the expected 3v. There is obviously some information I am missing. I guess I will go back to YouTube university and search for the missing information. LOL
The voltage across the resistor is equal to the supply voltage minus the LED voltage.
If the supply voltage is 12v and the LED voltage is 3v the resistor voltage is 12v - 3v = 9v
Note:
The sum of the voltage drops in your circuit must add up to the supply voltage.
V(led) + V(r) = V(supply) 3v + 9v = 12v
someboredguy:
I setup a circuit with just a 680 ohm resistor. No LED. Shouldn't I be reading 3v or less (used 680 instead of 630) across the resistor? I am reading 9v.
Without an LED the circuit is not complete so there is no point trying to measure anything.
But anyway the point of your calculation is that the 3V should be across the LED. The rest of the 15.6V is what is dropped across the resistor.
Steve
slipstick:
Without an LED the circuit is not complete so there is no point trying to measure anything.But anyway the point of your calculation is that the 3V should be across the LED. The rest of the 15.6V is what is dropped across the resistor.
Steve
I think this helps a lot!
I thought the purpose of a resistor was to drop it to the desired input voltage before the LED. Therefore, I was expecting the voltage across just the resistor to be the 3v needed by the LED.
I am just trying to understand the fundamentals. In this case, resistors. I thought I should get 3v or less across my resistor and I am getting 9v.
The "issue" is, LEDs are highly-non-linear (like all diodes). Their resistance changes with voltage. When you increase the voltage slightly, the resistance goes down by a LOT.
But if we make some calculations and choose the correct resistor, the resistor controls/limits the current and the LED voltage "magically falls into place".
In case you don't know this, voltage divides in series circuits, proportionally to the resistance (Voltage Divider) so the voltage across the resistor plus the voltage across the LED must equal the power supply voltage.
And, the same current flows through series components so you have the same current (~20mA) through the resistor and LED. So if we know the current through the resistor we know the current through the LED.
Since your power supply is unregulated it may not be exactly 15.6V with a load connected. Ohm's Law (from which the voltage divider concept is derived) is a law of nature (with man-made units of measure) so it's always true and the voltages MUST sum-up!
It is clear now that I am missing fundamental knowledge.
I have been a software developer for 25+ years. I have always wanted to experiment with circuit building. I finally decided to play around with the Arduino using the ELEGOO Uno. I am enjoying it quite a lot! I am discovering that there is quite a bit of fundamentals I will need to learn to get the most out of this new passion.
Thank you all for your help.
someboredguy:
It is clear now that I am missing fundamental knowledge.
I have been a software developer for 25+ years.
My nephew is a software developer, when I mention hardware he replies "I leave that to the hardware guys :o )
I leave that to the hardware guys and gals.