Question: I was reading this [sparkfun article](https://learn.sparkfun.com/tutorials/transistors/applications-i-switches) to learn more about how transistors work, but I am confused by one of their diagrams. Here it is:
[![enter image description here](https://i.stack.imgur.com/mu6wm.png)](https://i.stack.imgur.com/mu6wm.png)

If there is no current flowing, why is there a 1.3V drop across the LED? I looked at the VI characteristic graph for LEDs and it should be 0v when I=0, which is the case here. I think I am missing something, can someone point out exactly what? 

Thanks!

Answer: 
They are probably trying to show you the voltage the LED would drop when some current is going thru it. I agree that is inconsistent with how they are showing the voltages across other components. In reality, the LED will have very close to 0 V across it when the transistor is off. The only current will be the transistor leakage, which quite small.

There is a case where what they show could actually be true. LEDs also work as photocells in reverse, although rather poorly. With no load on it and in reasonable light, the LED will develop a voltage close to its normal forward operating voltage. However, the impedance of that will be so high that even a ordinary voltmeter can load it. Whoever made that diagram may have probed around the circuit with a voltmeter, and at that illumination and that voltmeter, that's what was reported across the LED. Whatever part of the supply voltage that doesn't appear across the LED would then be across the transistors, since no current is flowing.

Another possibility is that when they probed across the transistor with a voltmeter, the meter caused enough current to flow for the LED to develop 1.3 V across it, so the meter read 3.7 V. They then subtracted 5 V from 3.7 V to say the LED had 1.3 V across it. If the LED was measured directly, it would have less voltage across it.
You are right.

When the current is zero, the voltage will be zero.  

However, when simulating 0mA is not necessarily absolutely zero.

What I speculate their simulation tool to make that image does is also calculate a small leakage current through the transistor. To be honest I find it fairly shoddy they do not explain about that (I'm assuming, since I am too lazy to read all of it), because it is confusing, as you have experienced first hand.

A transistor may still conduct a tiny bit if you do not force a current into the base, due to small bits of unperfectness. When you express that leakage as mA it will say "0mA". But a simulation tool may estimate up to several dozens of μA (which rounds down to 0mA in the view) as leakage, depending on the type of model it uses to simulate.

This means the LED will see a small current, which will put it, apparently, at the 1.3V point of its estimated curve.

Of course, this is speculation, not knowing how they made the picture, but I think it's likely enough the simulator "estimated" all of those leakage numbers at the points you see. You should be aware that a simulation is just a simulation and never absolute truth. I think in the real world you would not see such a significant voltage across the LED, although with old fashion transistors, the first generations basically, it's possible.

If in the real world you see that happening, this may help a little bit:

![schematic](https://i.stack.imgur.com/guDD8.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fguDD8.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

Here, if you open the switch R2 will help pull the transistor closed. With modern transistors (like newly made 2N3904's and such) controlling only a simple LED or Relay these resistors are usually not at all that important, but when you want to be sure there is as little off-leakage as possible, they do still help.