Question

For a p-channel JFET, the current in the output circuit can be controlled by applying

whyyy explan and give me for n channel and what is mean by vds vgs and idss id and who i know the name of region is it cut off and so on

Answer 1

The Junction Field Effect Transistor, or JFET, is a voltage controlled three terminal unipolar semiconductor device available in N-channel and P-channel configurations.The field effect transistor is a three terminal device that is constructed with no PN-junctions within the main current carrying path between the Drain and the Source terminals. These terminals correspond in function to the Collector and the Emitter respectively of the bipolar transistor. The current path between these two terminals is called the “channel” which may be made of either a P-type or an N-type semiconductor material.

A P-Channel JFET is a JFET whose channel is composed primarily of holes as the charge carrier. This means that when the transistor is turned on, it is primarily the movement of holes which constitutes the current flow.This is in contrast to N-Channel JFETs, whose channel is composed primarily of electrons, which constitute the current flow. A P-Channel JFET is composed of a gate, a source and a drain terminal.

It is made with an p-type silicon channel that contains 2 n-type silicon terminals placed on either side. The gate lead is connected to the N-type terminals, while the drain and source leads are connected to either ends of the P-type channel.When no voltage is applied to the gate of a P-Channel JFET, current (holes) flows freely through the central P-channel. This is why JFETs are referred to as "normally on" devices. Even without any voltage, they conduct current across from source to drain.

The characteristics curve of a P Channel JFET transistor shown below is the the graph of the drain current, ID versus the gate-source voltage, VGS.This curve represents the transconductance, or simply the gain, of the transistor.The transconductance of a transistor really means the gain of the transistor.So this transconductance characteristics below shows the gain of the transistor, how much current the transistor outputs based on the voltage input into the gate terminal. Remember that gain is the the output over the input. The input is how much voltage is fed to the gate terminal. The output is how much current the transistor outputs.

The Regions that make this characteristic curve are the following:

Cutoff Region- This is the region where the JFET transistor is off, meaning no drain current, ID flows through the source-drain region.

Ohmic Region- This is the region where the JFET transistor begins to show some resistance to the drain current ID that is beginning to flow through the source-drain region. This is the only region in the curve where the response is linear.

Saturation Region- This is the region where the JFET transistor is fully operational and maximum current is flowing. During this region, the JFET is On and active.

Breakdown Region- This is the region where the voltage that is supplied to the source terminal of the transistor exceeds the necessary maximum. At this point, the JFET loses its ability to resist current because too much voltage is applied across its source-drain terminals. The transistor breaks down and current flows from source to drain.

If you really want to make sense of all the technical details of the graph above, you have to really that a P-channel normally receives positive voltage to the source terminal of the JFET. So the source terminal receives positive voltage and the drain terminal is normally grounded. So the source terminal is positive relative to the drain terminal. Notice that the voltage on the horizontal of the graph represents the voltage, VDS. VDS is the voltage across the drain and the source, in that order. Since we, again, feed positive voltage to the source terminal and ground the drain terminal, the drain terminal is negative with respect to the source terminal. This is why you see negative voltages for VDS. A negative voltage for VDS just means that we're feeding positive voltage to the source terminal. So if you think of it that way, it makes a lot of sense. If you look all the way to the left of the curve at VDS being around 0V, no drain current can flow because the source terminal needs positive voltage. So if we increase positive voltage to the source terminal which means we're making the drain terminal more negative, we increase the output drain current. About +10V to the source is the midpoint of the graph (which is -10V VDS). And as we go above about +20V or so the source terminal, we reach the transistor's breakdown point.