Hi again David,

Upon re-reading your web page, I realize that you want this circuit to
consume as little power as possible,since it will be operating even when
the device is turned off. SO, I have two possible additions/corrections to
make to my previous circuit:

#1) Increase resistor values. The 10K may be able to be made as high as
100K, depending upon how much current your inverter draws. Also, the 4.7K
pot on the input could probably be changed to a 100K pot. As is true of the
original circuit, I have not tested these values.

#2) You COULD use a MOSFET as you originally proposed. I am not sure if you
will find FETs with enough of a "knee" to be fully on at >4v and fully off
(or off enough to limit current drain to a very low level) at 2.5v. Here is
what you can try,though (keep in mind that discrete low-power mosfets are
hard to find,most discrete mosfets are power transistors):



Input
o--------+              +5v
         |1Meg pot       |Source
         R<------------|[           P-MOSFET
         |         Gate  |Drain
        GND              |
                        Inverter
                         |
                        GND

This works again by using a pot to adjust the input voltage (pot may not be
needed). When the input voltage is at 4v, the Gate-source voltage is only
around -1v, not enough to turn on the FET, but,when the input drops to 2.5v
or lower, gate-source voltage increases to about -2.5 or -3v, enough to
turn it on. Even though this is a simpler circuit than the one I previously
gave, it has two problems: #1) discrete mosfets usually come in big
(TO-220) packages and cost as much as several small bipolar transistors.
#2) You may have trouble finding a MOSFET with a small enough (2.5 to 3v)
turn-on threshold.

The basic idea of using any FET as a switch is that the channel (from Drain
to Source) acts like a very low constant-current device when the voltage
between the gate and source is small, but when the voltage form the gate to
source increases above a turn on threshold (usually several volts,but often
10 or more for large power mosfets), the channel then acts like a constant
low resistance. Whether the gate needs to be more positive or more negative
than the drain for conduction depends upon the type of the device.
N-MOSFETs require the gate be more positive and P-MOSFETs require the
opposite.


You might look into JFETs,too. They also have the advantage of very low
gate current AND they often come in small (TO-92) packages, BUT, they very
often won't conduct more than 20 or 30 mA and have a high channel
resistance (100 ohms or so), so it really depends on what your inverter needs.

Sean

|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
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