🚀 go-pugleaf

I must not understand JFETs yet.

Background...

I am trying to build a circuit that measures current flow very accurately
without inducing a significant voltage drop.  The following circuit is shown
on page 2-25 of the Motorola Linear and Interface Integrated Circuits
databook. ("+" indicates connection of crossing wires)

                   +-----------------------------+
                   |                             |
            +------|------+                      |
            |      |      |                      |
            |      |   |-----\                   |
            |      |   |  V+  \               |------|
  INPUT-----+--R1--+---|-      \              |      |
            |          |        \             | LOAD |
            R2         | LF355   )--+         |      |
            |          |        /   |         |------|
            +----------|+      /    |            |
            |          |  V-  /     |            |
            |          |-----/      |           GND
     2N4092 |--|          |         |
      JFET     |<---------|---------+
            |--|          |
OUTPUT------+            GND
            |
            R3
            |
           GND

OUTPUT = R3*(R1/R2)*(current in load)

Typically, R1 is quite small, about .1 ohms.  R2 is about 100 ohms and
R3 is about 5K.  This gives 5V per amp in load as output.

Now the question...

The way I understand JFETs, N channel types in particular, is that you must
supply a negative gate voltage on the order of -5V to turn them off
(the 2N4092 is a depletion mode JFET).

Since the LF355 is supplied from the INPUT voltage to ground, its output
range must be within this voltage, therefore it cannot turn off the
JFET.

Is this true?  Does this circuit work?

I've also seen this circuit in a National databook, and I have seen it with
a bipolar NPN substituted (but isn't the base drive current an error term?).

Mike Ciholas

email:  mikec@ai.mit.edu
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