In a NOT gate circuit, does charge not flow through both the transistor and output line when the transistor is “on”

digital-logictransistors

I've tried to recreate a circuit from what I think would be a NOT gate, based on some other schematics and breadboard examples that I've seen.

schematic

simulate this circuit – Schematic created using CircuitLab

For this to be practical, I think SW1 would go away, and instead be substituted for some other voltage source as an input, but I've inserted SW1 and attached it to the same voltage source as everything else just to simplify the diagram…

Anyway, my assumption is that when there is charge going into the base of the transistor, the charge flows through the collector, out the emitter, and straight to ground (the negative end of the voltage source). So, in this case, does the charge not flow through R2 and through the LED, somehow? It seems to me that this conclusion must be correct, or else the LED would light up, right? But apparently, electricity does not take the "path of least resistance", so to speak.

Conversely, if the base of the transistor does not have any current, I assume the current stops at the collector of the transistor, but since the circuit must be complete, pressure moves the charge through R2 and the LED, then to ground.

I feel that I am missing some fundamental principle involved in this. Any thoughts?

Best Answer

It is conventional to draw circuits so signals flow from left to right, like so:

schematic

simulate this circuit – Schematic created using CircuitLab

(You need a base resistor, R3, to limit base current, and a resistor from base to ground to ensure that the transistor does turn off when the switch is open)

With the switch open, the base will be held at ground by R4, so no current will flow through the transistor, so the collector will rise to a voltage determined by R1, R2, and the 2 volts or so voltage drop in the LED - this will allow current to flow through the LED and light it.

With the switch closed, the base will be pulled up, allowing current to flow through the transistor. If the resistors are selected correctly the transistor will be saturated, pulling the collector down to about 0.2 volts. As the LED requires about 2 volts across it, it will not pass any current.

I really dislike the sentence "Current always takes the path of least resistance", as many beginners seem to read it as "Current takes only the path of least resistance". In face, an electric current takes all possible paths, withthe lower resistance paths passing higher currents than the higher resistance paths.