How current is steady after pinch off voltage? - Page 1
Forum section: electronics / beginners

author: usmansa1
author_rank: Newbie
timestamp: May 18, 2019, 04:03:29 am
content: I was studying the JFET. In the books and other material it is mentioned that when the voltage between the drain and source is increased then drain current is also increased. After a certain point (Pinch Off) the drain current becomes constant and will remain constant even if the drain voltage is increased. If we see the physical structure of the device the depletion region will start to increase when the voltage between the drain and source is increased. After the same certain point (Pinch off Point)depletion region touches and then current cannot move forward.My question here is that then how the current becomes steady or current will pass when the depletion regions are closed. Doesn't the current become zero ?If the depletion region is closed by Voltage between drain and source then what is the role of voltage between gate and source. Although I can see the steady current value is changed at different Voltage gate to source and a logic of faster build of depletion region (Through Voltage to Source and Drain to source) also comes in mind but what is the main logic behind it ?Before replying please consider me as basic learner. Thanks

author: rfeecs
author_rank: Frequent Contributor
timestamp: May 18, 2019, 07:38:17 pm
content: Quote from: usmansa1 on May 18, 2019, 04:03:29 amNo. You may be thinking that current cannot flow through the depletion region. This is not true.For an n-channel JFET for example: at the drain end, although the channel is depleted, there is a strong electric field in the direction from drain to source. This will sweep the electrons right through the depletion region.Further increase in the drain voltage just widens the depletion region towards the drain, with very little effect on the current.

author: Wimberleytech
author_rank: Super Contributor
timestamp: May 19, 2019, 08:40:32 am
content: For a MOSFET:Source at ground, drain voltage is Vd, and gate voltage is Vg.With Vd=0, as Vg increases passed the threshold voltage VT, an inversion layer charge is developed under the gate and connects the drain and source with a resistive region that conducts proportional to the inversion charge. It acts like a resistor in the first order.As Vd increases, current increases because the channel is conducting like a resistor. It is nononlinear, of course, because as Vd increases, the voltage along the channel is increasing and there is less voltage from gate to channel and thus the charge in the channel is decreasing (from source to drain). That is why, the IV curve is not a straight line, but bows down as Vd increases.At some point, Vd is high enough that the voltage from gate to drain is less than VT, thus the inversion layer is not supported at the drain. This is the "pinch off" point. The voltage at that point (the drain end of the channel) can no longer increase. The transistor is operating in what is called the "saturation" region. Vdoltage at the pinch off point no longer changes, so the current Vpinch/R (where R is inversely proportional to the charge in the channel) is fixed.In reality, there is a lot more going on beyond this simple explanation. For example, physical location of the pinch off point is modulated by Vd so that is Vd increases, the inversion region (the channel) gets shorter and thus resistance gets lower and the current goes up. This is called "channel length modulation."