# How does a NOT gate bypass the output?

I have found the basic tutorials on gates and the way they work.

However, I'm not understanding conceptually how a NOT gate, when activated, bypasses the output.

Circuit as follows:

simulate this circuit – Schematic created using CircuitLab

When switch goes on, LED goes off - SIMPLE & Useful.

But, how does this happen?

Why doesn't ANYTHING go through the LED?
Is this because it's a diode?
.. Is the voltage drop across the transistor too much for the LED to be lit up?

What if it was a speaker?
.. Would it emit a very small/quiet noise even while the switch is on?

I can't find a tutorial, or reference online that explains this to me.

Resources or assistance would be greatly appreciated.

• Edit your question, and click on the schematic button. Commented Feb 15, 2014 at 3:45

Use the schematic tool to draw your circuits.

The output voltage level is taken as shown in the above image. As you can see, when the base of the transistor is low (0V), the NPN device is not conducting and is high-Z (en open circuit). At this point, Rc acts like a pull-up resistor and for a logic low input, the output is logic high.

When the transistor is turned on by an appropriate input voltage at the base, the transistor conducts heavily and essentially shorts to ground. I am not sure on the exact values to expect, but I want to say that Vce(sat) is approximately 0.2V or so which is clearly a logic low.

Also, to answer your other questions, current does flow through the LED (which is a diode like you suggest), however such current is on the order of micro amps or smaller. As suggested in the comments below, voltage seen at the collector of the transistor (VceSat) is far less than the cut-in voltage of the LED.

Consider this plot of the forward and reverse characteristics of a diode.

Despite the forward voltage being a bit higher with an LED, the concept still applies. Significant current conduction is required for the LED to emit light. However, such conduction only occurs when sufficient biasing of the device is realized. Given the design you presented in your post, approximately 200mV will be seen across the LED, which is about a tenth of the necessary voltage to force an LED to conduct current (Vf varies by color and other factors).

• And the final step is that Vce(sat) is lower than the forward voltage of the LED, which means that the current flows through the transistor instead. Commented Feb 15, 2014 at 4:18
• Yes, that would be the final statement to make. I did not realize the OP had the LED configured as shown in the added diagram. Commented Feb 15, 2014 at 4:25
• Ok, I think I got this. Thanks for the explanations there. Switch on - means: Transistor 'on'. Transistor has a low resistance, and high conductance (Relative to the LED). So there may be a current & voltage going to the LED, but it is simply too low. I'll get myself a multimeter and get some measurements going. Commented Feb 15, 2014 at 4:32
• Actually, the transistor will conduct current and force the voltage at its collector to be approximately 200mV. This voltage is less than the cut-in forward voltage of the LED. Thus, the LED does not conduct significant current and does not turn on. Check out the newly added i-v plot of a diode. Broadly speaking, the LED will not conduct enough current to emit light until sufficient biasing voltage is produced at its terminals - which might be approx. 2V or so for a red LED. Commented Feb 15, 2014 at 4:35
• Ideally, yes. However, the transistor device is not ideal. The actual model might look something like this: imgur.com/0Q1Lcnv Commented Feb 15, 2014 at 7:32

The LED does not glow because since the emitter is grounded and the collector to Vcc resistance is high (1kohm), the transistor will happily drop off the entire 6V (or almost) across the 1k resistor. hence the collector terminal is going to be at near zero voltages and since the LED is connected in parallel to the collector-earth junction, the voltage across the LED is going to be the same. Thus anything less than 0.7V, the LED wont glow.