Voltage and current don't seem to match up

Question

Between two pins I'm measuring 2.1V but 0.04 mA. What does this mean?

I'm trying to light up an LED on an LPC2148. I set pin 2 to high and connected my LED to it and pin 14. It doesn't light up. When I use a multimeter to measure the voltage between these two pins, I measure enough voltage, but when I switch the multimeter to measure current, I read almost no current.

So what's up?

Answer 1

It means that you have very little idea of what you are doing :-).
BUT keep asking questions! :-) - it's a good way to learn.

Using a "water analogy" is a good way to get a feel for electrical terms.
As George Box said " All models are wrong, but some models are useful". ie no analogy is perfect but even an imperfect analogy can help understanding.

• Voltage is similar to the pressure in a water flow situation.

• Pressure can be caused by "head" such as height of a dam = a battery, or by a pump = an alternator or generator.

  • Water current flow is about the same as electrical current flow.

  • Resistance to water flow such as from a thin pipe, is similar to electrical resistance.

Some things get harder and the models may be even less exact but:

• A Capacitor is like having a water tank with in and out pipes at the bottom on opposite sides and a flexible rubber dividing wall separating the two halves. The size of the tank and the flexibility of the rubber are related to capacitor size.

• An inductor is like having a length of tubing made of flexible rubber so that it will expand as water flows through it under pressure.

SO

Your 2.1V and 0.04 mA are like looking at system and seeing some water pressure (voltage) and water-current flow (current in mA). The size of the voltage and the size of the current will depend on the "circuit" (pipe connections, lengths, sizes etc.). Until you explain quite a lot better how you measured these things and what your circuit is like we cannot explain why they are as they are.

Note that voltage and current ALWAYS match up.
Nature is never wrong :-).
Any time we think it is wrong it means that we do not properly understand it (yet).

Answer 2

Schematic or it didn't happen!

You're not telling us everything, but the most obvious explanation would be in Ohm's Law:

\$ R = \dfrac{V}{I} = \dfrac{2.1V}{0.04mA} = 52500\Omega \$

So you could be measuring the 2.1V drop over a 52.5k\$\Omega\$ resistor.
With the right resistor value any voltage/current ratio can be achieved, so "not matching up" doesn't make sense.