# How does a gate voltage determine the state of a transistor?

It is known that in a transistor, like the FET below, it has an 'on' state when current flows from source to drain, and an 'off' state when there is no current flowing between junctions.

This current is mediated by gate voltage, shown as $V_{GS}$ in the diagram below. When $V_{GS}$ is greater than some threshold $T$, current flows and the transistor is on.

My question is this: how does the gate voltage mediate between on and off? That is, if $V_{GS}$ is applied to a transistor, that transistor will be in state $S$ - this means that the $V_{GS}$ must also be in state $S$ (on or off). If this is the case, then the transistor's state tells you nothing more than knowing the gate's state. Is it just a feed-forward from previous transistors?

• Perhaps better on Electrical Engineering. However, the gate voltage in a FET modulates the carrier concentrations under the gate. Go get your pictured device to conduct, you need to make the region under the gate n-type to connect the source to the drain. So, you bias the gate to attract electrons from the substrate to come to the gate, inverting the material from p-type to n-type. – Jon Custer Feb 1 '18 at 13:55
• Right, I do get that. But I’m wondering how the gate knows to modulate its voltage in the first place - how does the gate voltage know when to apply itself and when not to? – Daniel R. Livingston Feb 2 '18 at 2:17

Firstly I didn't get it when you say "I’m wondering how the gate knows to modulate its voltage in the first place - how does the gate voltage know when to apply itself and when not to?"

When you apply a potential difference bigger than $V_T$, after charging up the gate capacitance which is determined by physical properties i.e $A$ (area), $W$ (width) and $t_{ox}$ (thickness of the oxide) and material property like $\epsilon$

"I’m wondering how the gate knows to modulate its voltage in the first place"

Technically speaking, since there is a oxide layer between metal and p substrade gate is actually capacitor. So current leads the voltage meaning that actually $V_G$ happen in second place. The time to build up this $V_{GS}$ (%10 to %90) called rise time of the transistor and it is very important parameter of operating frequency.

However,this capacitance is pretty small (order of $pF$-$fF$) so rise time is in the range nanoseconds.

After building sufficient $V_{GS}$ the rest is straightforward. Due to the electric field electrons create n channel.

• Nail, thanks for your well-written answer. I suppose I'm still missing something quite basic. An analogy might be: imagine flicking on and off a light switch with some logic (a transistor switching between $S_0$ and $S_1$). While the light is turning on and off due to current flowing (a gate voltage being applied), the lightbulb is not mediating its own flow with respect to other lightbulbs - you are controlling the light switch (the gate voltage isn't the 'prime mover' of the transistor, something else with logic is mediating the gate). Who is controlling the 'light switch' on a CPU? – Daniel R. Livingston Jun 29 '18 at 16:18
• @DanielR.Livingston Other separate transistors. I suggest you learn about digital logic and on how to build logic gates from transistors (NMOS is conceptually simplest, I think?). You can try your circuits out on a simple simulator like CircuitJS. A CPU is a really big and complicated logic circuit but it's still just a logic circuit, so start playing with small ones. – user253751 Nov 6 '18 at 5:02
• @immibis , thanks very much. I’ll look into what you’ve written. I’ve wondered about this problem for years! I’m a software guy if you couldn’t tell ;) – Daniel R. Livingston Nov 6 '18 at 5:04
• @DanielR.Livingston I made a 2-bit binary adder just for you, using only transistors and resistors and power, in that simulator: pastebin.com/88ck8i9Q (use File -> Import from Text). High voltage (green) represents 1 and low voltage (grey) represents 0. Click on the H/L letter inputs to switch them. I'll also note that this simulator is not considered very professional, but it is beginner-friendly. – user253751 Nov 6 '18 at 23:43
• Oh this is absolutely fascinating. Please put this in an answer so I can mark as the 'correct' one. Looking at the diagrams, the answer to my source of confusion is clear. Thank you very much! – Daniel R. Livingston Nov 7 '18 at 23:57