About JFET (Junction FET)
If you don't understand the BJT outlined above, you won't be able to understand the description of FET this time.
JFET is a little different from the MOSFET (Metal Oxide Semiconductor Field Effect Transistor) that will be summarized later.
I want to tell you first..
It is literally a junction field effect transistor, which is a JFET.
Earlier in BJT, both electrons and holes were involved in the operation of the device, so it was called Bipolar.
These FETs are called Unipolar because only one of the electrons or holes contributes to the operation!
The JFET, which we will learn first, is not used very much these days because it is pushed by the MOSFET. However, you need to have parents to have children and...
Just as ver1.0 is required to have ver2.0, you need to know JFET first to know the MOSFET.
Let’s stop talking and explain now~
The figure below shows the schematic symbols of N-channel JFET and P-channel JFET, respectively.
In the picture above, there are 3 terminals each, and G, D, S representing Gate, Drain, and Source are drawn~
Also, in the direction of the arrow on the Gate terminal, it is distinguished whether it is N or P channel.
Here, a guy called Channel suddenly popped up...
This channel refers to the path through which the actual electrons or holes move and current flows.
I'll explain it more accurately through the picture below~
As shown in the picture above, P-type and N-type semiconductors, which are of the opposite type to the N-type or P-type doped semiconductor, are injected
It will make a gate terminal.
However, since the path through which electrons and holes actually travel is the original basic substrate (N-type semiconductor in N-JFET), this becomes a channel.
So what does the Gate do?
First, I'll explain with only the N channel on the left side of the picture.
Since there is an N-type semiconductor (or N-channel) as a substrate, and a PN junction occurs as P-type semiconductors are located on both sides, diffusion due to the difference in concentration will result in a depletion layer. I organized it at BJT!)
So let's first think that this depletion layer is a space where no current can flow because it has no carriers.
In the picture, is Vds voltage applied to the drain and source? The electron will move to Source -> Drain.
Since the movement direction of electrons and the direction of current are opposite.. Current flows from Drain -> Source.
Also, there is a reverse bias between the gate and the source, so the depletion layer will be much larger than the original Vgs = 0.
In other words, as the reverse voltage increases, the depletion layer will gradually expand, and since the area of the depletion layer is increasing on both sides, as Vgs << 0 (-voltage increases), the two depletion regions will someday meet each other.
This situation where two depletion regions meet is called Pinch-off.
However, in reality, the expansion of the depletion layer is enlarged in a shape that is oriented to one side as shown in the picture above.
For example, in an N-channel JFET, the gate is P-type and the substrate is N-type, but when a voltage of the + pole is applied to the drain, the P-type semiconductor of the gate and the N-type semiconductor of the drain are in reverse state. Because it is biased
(Since the source has a negative voltage, it is a forward bias!) The depletion layer is formed as if it is directed toward the drain.
Therefore, the pinch-off phenomenon also occurs in a shape that is tilted from one side as shown in the picture above.
But at this time, since the depletion layer blocks the channel area as much as L in the picture above, it interferes with the flow of current more.
(It just became a resistance component~)
The above graph is an I-V characteristic graph showing the amount of change in drain current Id according to Vds of N-channel JFET.
(The shape of the graph is taken from MMBF5460LT1 of On semiconductor.)
Importantly, as Vgs increases (as the reverse voltage increases) the depletion layer becomes thicker, so the maximum value of the drain current will decrease.
Since Vgs(off)=-1.2V is shown above, the drain current is almost zero when -1.2V is applied to the gate.
But there is something weird...
Even if Vds increases, Id becomes constant when it reaches some extent.
If you think about it carefully, it is true that the drain current decreases as the depletion layer increases as the voltage increases.
If the gate voltage is kept constant and the Vds voltage is increased, Id must continue to increase in the first place, but later, it becomes saturation.
When I think about it again, even if I want to increase it, the depletion layer between the gate and the drain increases even if Vds increases.. Later, it becomes Pinch-off.
uh? "But when the pinch-off occurs, the depletion layers on both sides are attached, so is the drain current Id to be reduced because the channel is blocked by the depletion layer?"
I will try to summarize what I am curious about.
Since Vds increased, Id should continue to increase..
It seems that Id should decrease because Vds increases and Pinch-off occurs someday, and the depletion layer blocks the channel.
But the result is that Id becomes saturation.. Why doesn't it decrease or increase?
There are usually two reasons for this..
First, even if there is a depletion layer, it is not an area that carriers cannot penetrate the depletion layer.
The force of the electric field across the drain and source. In other words, it does not mean that the current cannot flow because the carrier can move sufficiently by the electric field.
As if there is a lot of soil in a flowerpot, water flows out from the bottom of the flowerpot if you water it.
Secondly, the number of electrons entering the drain from the source terminal is the same anyway, so you can think that the current does not decrease.
Usually, a toll gate is often used as an example...
When vehicles enter the toll gate at an incredibly fast speed, the amount of time it takes for the person receiving the ticket at the toll gate to pass a vehicle
It is similar. That said, some of the vehicles aren't just letting go or not letting go.. Someday they will all pass.
When I think about it, if the vehicle is called a carrier... and the person who receives money at the toll gate is the deprived person...
The number of vehicles passing in a certain time... In other words, the number of carriers can be said to be constant...
The movement of carriers per unit time.. This can be called the movement of electrons per unit time.. This means current!
Ah~That's why the current doesn't decrease.. Anyway, all the carriers supplied will pass through~