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Sci.Electronics.Basics -> Re: JFETs and "pinch-off" - two meanings?
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Author: billcalleyDate: 00:08 09-04-08
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On Mar 29 2006, 5:34 pm, "billcalley" <billcal...@yahoo.com> wrote:
> Hi All,
>
> When reading descriptions on the operation of JFETs, I
repeatedly
> see mention of "pinch-off". Some books on JFETs describepinch-off
> as the point that the drain-to-source voltage (Vds) is just high
enough
> so as not to affect the drain current if Vds happens to increase;
while
> other books describepinch-offas the point that the JFET's gate bias
> is negative enough to shut down, or "pinch-off", all the drain
> current. Which is correct?
>
> Thanks!
>
> -Bill
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
-Vgs>-Vp
Vgs<Vp
This is what Bill refers to with his second definition.
Regards,
Art Kalb
Amplifier Design
Analog Devices, Inc.
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