Measured actual voltage and it dropped?

Hello all,

Today, I setup a simple LED circuit with 220 Ohm resistor in series with an LED. I tested to see the potential difference from Pin 9 with reference to ground with no LED or resistor and found that the output voltage was 4.95 V. I then checked the voltage across the LED circuit and found the total voltage was 4.75 V? What happened to the 0.2 V and is there a way to know how much voltage will be dropped for any circuit using the Arduino Uno?

Hi, welcome to the forum. Arduino pins have some internal resistance. True LED current is (VCC - Vf of the LED) / (external resistor + internal resistor). You can ignore internal resistance in most cases, but I think it can be ~25-40ohm. Leo..

So your saying that I just have to add some extra resistance into my calculations to account for this voltage drop? Is there a way to find out how much internal resistance there is? I'm pretty disappointed that my actual results were not close to my theoretical results.

Is there a way to find out how much internal resistance there is?

You've almost already done it.

The LED current = (Vpin-Vf(led))/220

The voltage drop in the pin voltage under that load is 0.2V

So let's assume the the LED forward voltage is exactly 2.0V. (You can measure it for your calculation.) The current for a Vf of 2.0V = (4.75-2.0)/220 = 12.5mA The pin internal resistance = 0.2V/12.5mA = 16 ohms. That's a pretty good value.

Why do yo think that voltdrop is a problem? Leo..

0.2V; it could be worse :D

The 328 spec states that you may expect as low as 4.2V (@ 85 degrees Celsius; 4.1V @ 105 degrees Celsius) at a load of 20mA.

I see! ok thank you all very much for all the help! I was just really surprised that my actual results were really off and didn't know how to account for this loss.

mac124:
I’m pretty disappointed that my actual results were not close to my theoretical results.

That always happens when your theoretical model of what is happening is simplified from what is actually happening. Most of the time simplified theory is good enough though.

This theoretical series resistance on any output is known by the name “output impedance”.

The voltage loss is from the Rds (Resistance, drain to source) of the MOSFET transistors that connect an output pin to VCC (for a high out) or to Gnd (for a low out). We can calculate what that Rds might be. You saw 4.75V with 220 ohm resistor and an LED. If the Vf of the LED was 2.7V (somewhat typical for Red) then the current flow was (4.75V - 2.7V)/220 ohm = .0093A, or 9.3mA. This is just basic Ohms Law, V=IR, or V/R = I. With no current flow, Vout was 4.95. With a 9.3mA load (current flow), Vout was 4.75V. Then (4.95V - 4.75V)/.0093A = 21.5 ohm.

CrossRoads:
If the Vf of the LED was 2.7V (somewhat typical for Red)

Standard red LEDs don’t usually have that high a forward voltage. Mine are about 2V.

And this indicates that my LEDs aren’t unusual:- LED - Basic Red 5mm

2.2, 2.5, 2.7, whatever. Measure the actual Vf of the LED used and do the math. Only makes a few mA difference and a few ohm difference in the calculated Rds.