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.
That article is seriously flawed. For instance, in the picture above your LED picture it shows this: -

[![enter image description here](https://i.stack.imgur.com/wDGdA.png)](https://i.stack.imgur.com/wDGdA.png)

At first glance is seems OK then you notice the control voltages and they say that 5V turns the motor off. What utter rubbish - the motor will still be on - it needs above 11 volts to turn it off and this is a pretty serious error in my book.

They do redeem themselves with an explanation about this but the article is pretty unprofessional and could lead to someone building that circuit and never being able to turn the motor off.

Then they show what they say is a two input AND gate: -

[![enter image description here](https://i.stack.imgur.com/b624P.png)](https://i.stack.imgur.com/b624P.png)

However, what they fail to recognize is that with the top transistor base turned-off, the emitter of the lower transistor will be at a logic level of 1 when its base is high. Bad stuff really.

Then they show this H bridge (yuk): -

[![enter image description here](https://i.stack.imgur.com/sZjtP.gif)](https://i.stack.imgur.com/sZjtP.gif)

There is no excuse for not showing base resistors given as they made a big deal about it early on in the article.

My advice is don't believe that article.