How does this transistor NOT gate work?

Question

I only have a basic knowledge of electronics so I apologise if this is a stupid question.

I've created a NOT gate using a 2N2222 NPN transistor shown below:

When the switch is open, the LED is on and when it is closed the LED is off.

I understand that in parallel circuits all branches receive the same voltage, regardless of what is happening in other branches.

My issue is that the transistor and the LED + resistor are in parallel to each other. Hence, according to the parallel circuits rule, the LED should always receive the same voltage (and, as resistance is constant, the same current), and therefore be on, regardless of whether the transistor is conducting or not.

If that was the case the LED would be on whether or not the switch was closed or open and the NOT gate would not work.

Could you please explain why this is not the case?

I hope my question makes sense and thanks in advance.

Answer 1

Excuse the crude edit:

Assuming an ideal transistor, this is what your circuit becomes when the switch is closed.

Both sides of the section of the circuit with the resistor and LED are lowered to the same electric potential. There is no voltage across and therefore no current through the LED.

Answer 2

The graph below shows voltage between Base and Emitter of a P2N2222, for Collector currents up to 500mA:-

When the switch is OFF the Base is open-circuit so the transistor will be off and all the current will go to the LED. Depending on the LED type its voltage drop could be between ~1.1V (infrared LED) to ~3.3V (white LED), and current through the 470Ω resistor will be somewhere between ~9mA and ~5mA.

When the switch is ON the Base-Emitter junction will draw current, dropping the voltage to ~0.8V and turning the transistor on. The transistor has current gain so more current will flow through the Collector-Emitter junction which will pull the voltage even lower, but that also reduces Base voltage which would turn the transistor off, so it will settle at ~0.7V with Collector current at ~11mA (corresponding to voltage drop of 6V-0.7V = 5.3V across the 470Ω resistor).

At this lower voltage the LED draws very little current so it is effectively 'off'.

Answer 3

When the switch is closed, the transistor turns on, or saturates. That means the voltage between its collector C and its emitter E is very low, e.g. 0.3 V and its impedance is also low. The CE then acts as a very low voltage battery (replacing the 6 V battery), which does not have enough voltage to light the LED.