0
\$\begingroup\$

I know that Ohm's Law is V = IR but I am getting some very confusing current readings from multiple multimeters. Here is my very basic breadboard configuration:
V = 6v
R = 200 Ohms
I = Should equal 30 mA but it does not (see multimeter pic)
Breadboard

The multimeter shows 18.8 mA but I do not know why. I have also tried with 3.3v and it registers 6.6 - 7 mA.

I would like to know why this is happening. I have tried different resistors, LEDs, multimeters, breadboards, power supplies but it continues to happen and I am very confused with the readings.

Multimeter

\$\endgroup\$
6
\$\begingroup\$

Diodes do not follow Ohm's law. Instead, when forward biased they have roughly a constant forward voltage. For an LED, this could be on the order of ~2V to 3V though you should check the datasheet.

Assuming your circuit looks like this (you should include a schematic of your circuit to clear up ambiguities):

schematic

simulate this circuit – Schematic created using CircuitLab

Then the voltage across the resistor is \$V1 - V_D \approx 4V\$. Then using Ohm's law:

\begin{gather} I_{R1} = \frac{4V}{R1} = \frac{4V}{200 \Omega} = 20mA \end{gather}

Which is much closer to the reading you were getting. Further discrepancies can come from any number of sources:

  1. I just roughly estimated \$V_D\$. You can measure this with another multimeter to figure out what it is.
  2. Resistors have some tolerance. Again, measure the resistance before inserting the meter.
  3. The multi-meter has a shunt resistor inside to measure current, and also has some tolerance. You can measure the shunt resistance with another meter, though it's possible the resistance is small enough that you'll need to use a 4-wire Kelvin bridge in order to measure it accurately. You can find the multimeter's tolerance rating on the box.

etc.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ By "do not follow Ohm's law" do you not mean that they do not have a linear voltage/current relationship? If current passes through an object and a voltage difference is observed across that object, it is exhibiting a resistance - that is Ohm's law. \$\endgroup\$ – CharlieHanson Jul 3 '15 at 2:10
  • 4
    \$\begingroup\$ Ohm's law requires that the relationship between voltage/current must be linear. Diodes are non-linear, and therefore don't follow Ohm's law. Just having a voltage difference across an object when current is flowing through it doesn't mean it follows Ohm's law: for example, an ideal voltage source doesn't follow Ohm's law. \$\endgroup\$ – helloworld922 Jul 3 '15 at 2:31
3
\$\begingroup\$

When you are calculating the current, are you allowing for the voltage drop in the LED?

A typical red LED will have about 1.8 volts across it when lit, leaving 4.2 volts across the resistor. With your 200 Ohm resistor, this should give a current of 21 mA. Yellow and green LEDs have slightly higher voltage drops, so will result in a slightly lower current.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

The LED drops some of the volts. Because voltage is lost over the LED, then in combination with the resistor, there will be some current going through it. If the LED drops 2V, then the resulting voltage over the resistor would be 4V. 4/200 is 20mA, so that's close enough. The LED probably drops 2.2V not only 2.

Also, if your LED is rated to 30mA, do not try to drive it at 30mA, as the lifespan will be greatly shortened, and designing to a "max rating" in engineering is a terrible idea. Instead, run it more like 15-20mA for safety and reliability. Depends on the use of course, of it's blinking or pulsed then that's different to a constant-on LED.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.