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

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. 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?

• I assume a resistor is missing between the switch and the base of the transistor? Otherwise you'll force VBE = 9 V in the transistor where usually this should be 0.7 V or so. The circuit as is should let out the magic smoke. Jul 9, 2014 at 1:25
• @swineone, yeah, you're right, thanks for pointing that out. Just a "typo"; edited my question to reflect your correction. Jul 9, 2014 at 1:38

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

• Thanks for the answer. What do you mean about needing a resistor from base to ground to ensure the transistor doesn't turn off when the switch is open? I thought you would want the transistor to be off (i.e. not allowing current to pass from collector to emitter) when the swith is open. Jul 9, 2014 at 11:13
• Also, what do you mean by "pulling" the collector down to 0.2v? Why does this make it so the LED cannot be forward biased? Jul 9, 2014 at 11:21
• @JoshBeam he says a resistor is needed to ensure the transistor DOES turn off. If you just open the switch the base is left floating, and I'd expected some accumulated charge there might not lead to an immediate turn-off (that should certainly be the case with a MOSFET rather than a BJT). As for your second question, see my answer below about the issue of biasing the LED. Jul 9, 2014 at 12:03
• Accepted this answer because of this: "an electric current takes all possible paths, withthe lower resistance paths passing higher currents than the higher resistance paths." I think that was the key to my whole misunderstanding. Jul 12, 2014 at 20:56

You are correct for the most part, but what you have here is essentially two switches that are working in inverted states.

If you drive the base of the transistor then you will draw about 29mA collector current from V1 and force, assuming saturation, the collector to be about 200mV (Vce sat). Current would flow through the diode if the voltage at that node was sufficient to forward bias the PN junction of the device, but that is not the case. So, what happens here is that when the transistor is ON, the diode switch is reverse biased (off), and when the transistor is OFF, it provides a high-resistance path (an open) and current will flow through the two series resistances and drive the diode. simulate this circuit – Schematic created using CircuitLab

• Definitely suggest the pull-down base resistor as shown below in Peter's schematic. Jul 9, 2014 at 1:57
• Thanks for answering. When the transistor base has charge pushed through it, why is the voltage at the LED not sufficient to forward bias it? What is the threshold the voltage has to pass in order to forward bias, say, a red LED, and why can the voltage not pass that threshold only when the transistor has charge flowing through it? Jul 9, 2014 at 11:18
• @JoshBeam, That may be fit for an entirely new question. For information as to why the LED needs certain biasing conditions to conduct see here and the section just below it as well. Perhaps for a bit on insight as to why the collector voltage (which is driving your LED in this case) drops, see here. So, in short, the construction of the diode requires an E field to be applied across the device such that the inherent field is negated thereby allowing current flow. Jul 9, 2014 at 12:39
• .. This negation closes what is known as the depletion region between the P and N doped regions. When the diode is reverse biased, this region acts like a very high resistance, and when forward biased the region has some on-state resistance (perhaps a few ohms). In the case of your schematic, if the transistor saturates and Vce(sat) = 200mV, then you are not allowing sufficient biasing voltage for the LED and consequentially it does not conduct (i.e. depletion region is open and as such the path is very high resistance). Jul 9, 2014 at 12:42
• thanks. I think what I'm not understanding is, why does the saturation level of the transistor have anything whatsoever to do with enough voltage going to the LED (thus, disallowing a forward bias in the LED)? Jul 9, 2014 at 13:12

It's simple, really. When the transistor conducts, the current has to choose between going through the collector of Q1 to the emitter and then ground, or going through R2 and D1. The diode only allows current to flow when biased, which in practice (for the range of currents in which the LED is simultaneously visible yet not damaged) means a constant voltage drop in the LED, perhaps 1.5V to 2.5V depending on LED color and other properties. There's also an IR voltage drop in R2; say you must apply at least 1 mA for the LED to be visible, then you're looking at a 0.3 V voltage drop in R2 ($300 \Omega \times 1 \textrm{mA}$).

So, unless you apply somewhere between 1.8 V and 2.8 V in the branch of the circuit containing the LED, it'll basically look like an open circuit (very high resistance). Now the transistor conducts current much more easily, clamping the voltage at a Vce(sat) = 200 mV as mentioned by the other answer. At that voltage, the LED is not biased at all, so it does look like an open circuit, while the transistor looks almost like a perfect short circuit (save for the voltage drop of 200 mV). So indeed the current chooses the past of least resistance, which is the transistor in this case.

When the transistor is not conducting, then no amount of voltage applied to the collector will allow current to pass. On the other hand, when applying 1.8 to 2.8 V to the branch containing the LED, it will bias the LED and let current through. So again the path of least resistance is being chosen: the resistance through a transistor with no current flowing into the base is effectively infinite.