On Mar 29 2006, 5:34 pm, "billcalley" email@example.com wrote:
I happened to find this thread and thought I would comment.
To understand the JFET pinch-off one needs to understand the mechanism of current flow. The channel in a JFET is a resistive material with a certain cross-sectional area. As a voltage, Vds, is applied across the resistive material a current flows (ohm's law). The gate is implanted/ diffused into the resistive material using a doping of type opposite to the channel (N in P channel, P in N channel). This PN junction forms a depletion region (low carrier concentration), which restricts the cross-sectional area. Increasing the reverse bias on the gate- channel junction increases the width of the depletion region. For a N- channel JFET, the reverse bias at the source end of the channel is - Vgs. The reverse bias at the drain end is -Vgd. Since the drain voltage, Vd, is larger than the source voltage, Vs, the reverse bias at the drain end is larger than the reverse bias at the source end. Thus, the depletion region at the drain is larger than that at the source and the current is more restricted at the drain end than the source end. As the drain voltage continues to increase (with fixed gate voltage), the channel area continues to narrow. At some point, the decrease in cross-sectional area (due to increased Vd) counteracts the effect of the increased resistive conduction (also due to increased Vd). At this point, the device stops behaving like a resistor and acts more like a current source with the amount of current controlled on the source end. The point at which the constriction counteracts the increased lateral field (Vds), is known as the pinch-off point. The voltage from gate-to-drain at which this happens is Vp.
So the device is pinched off when:
Vd-Vg>-Vp (sign convention on Vp) Vd-Vs+Vs-Vg>-Vp Vds>Vgs-Vp=Vds,sat So for Vds>Vds,sat the output current is roughly constant (with constant Vgs)
This is what Bill refers to with his first definition.
Now, let look at what happens on the source end. Like the drain, when the source voltage increases to a certain point relative to the gate voltage, the depletion region on the source side "pinches-off" the channel. Assuming, Vds>0, then the entire channel is pinched off (if Vds<0, you have reversed the definition of source and drain). There is no current flow. This happens when: Vs-Vg>-Vp
This is what Bill refers to with his second definition.
Art Kalb Amplifier Design Analog Devices, Inc.