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'[OT] Quick electronics MOSFET design help?'
1999\06\07@214548 by David Knaack

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<DIV><FONT color=#000000 size=2>Howdy,</FONT></DIV>
<DIV><FONT color=#000000 size=2></FONT>&nbsp;</DIV>
<DIV><FONT color=#000000 size=2>I know this is off topic, I'm not sure how
tolerant this list is to a few</FONT></DIV>
<DIV><FONT color=#000000 size=2></FONT><FONT size=2>off topic posts so please
accept my apologies in advance if this is</FONT></DIV>
<DIV><FONT size=2>too far off from PICs for your sensibilities. </FONT><FONT
size=2>I'll try to keep it brief.</FONT></DIV>
<DIV><FONT size=2>As usual, replies can be directly to me to avoid cluttering
the list.</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT color=#000000 size=2>I play with modifying TI calculators in my spare
time, one modification</FONT></DIV>
<DIV><FONT color=#000000 size=2></FONT><FONT size=2>requires a switch.&nbsp;
Instead of using a physical switch I want to control</FONT></DIV>
<DIV><FONT size=2>it with a circuit that senses if the calculator is on or off
and switches</FONT></DIV>
<DIV><FONT size=2>the device appropriately.&nbsp; I have a point on the calc
that is 5v when the</FONT></DIV>
<DIV><FONT size=2>calc is off, 0v or 2.5v when it is on (depending on screen
contrast).</FONT></DIV>
<DIV><FONT size=2>I want the power to the device to be off when the voltage at
this point</FONT></DIV>
<DIV><FONT size=2>is above about 4v (actually Vcc, it varies).</FONT></DIV>
<DIV><FONT size=2></FONT>&nbsp;</DIV>
<DIV><FONT size=2>I am currently using a circuit I designed with some junk box
transistors,</FONT></DIV>
<DIV><FONT size=2>however, not having any formal training, I haven't done it
optimally, and</FONT></DIV>
<DIV><FONT size=2>it is quite sensitive to variations in component values (about
10% of my</FONT></DIV>
<DIV><FONT size=2>transistors work in this design).&nbsp; Also, the transistors
drop too much</FONT></DIV>
<DIV><FONT size=2>voltage, the power to my device (an inverter) is less than
what it could</FONT></DIV>
<DIV><FONT size=2>be.&nbsp; Therefore, I would like to redesign the circuit to
use a MOSFET</FONT></DIV>
<DIV><FONT size=2>or something similar that has very low on resistance.&nbsp;
However, not</FONT></DIV>
<DIV><FONT size=2>having used them before, I have no idea how to proceed.&nbsp;
As usual</FONT></DIV>
<DIV><FONT size=2>with electronics, there is no 'How to use a XXXXX' to be
found, or even</FONT></DIV>
<DIV><FONT size=2>a 'How NOT to use a XXXXX'.</FONT></DIV>
<DIV><FONT size=2></FONT>&nbsp;</DIV>
<DIV><FONT size=2>Details for the device can be found at my web
page:</FONT></DIV>
<DIV><FONT size=2><A
href="http://www.genetech.net/dknaack/autopower.html">www.genetech.net/dknaack/autopower.html</A></FONT></DIV>
<DIV><FONT size=2></FONT>&nbsp;</DIV>
<DIV><FONT color=#000000 size=2>Thanks in advance if you can help me
out.</FONT></DIV>
<DIV><FONT color=#000000 size=2></FONT><FONT size=2>DK</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT color=#000000 size=2>I want you to be fully aware of your role in
Usenet.&nbsp; It goes like this:<BR>Bill Gates is the organ grinder, playing a
song called &quot;Web TV&quot;.&nbsp; You are<BR>the little monkey in a bright
red hat hopping up and down and acting silly<BR>for our amusement.<BR>&nbsp; --
<A href="spam_OUTdgriffiTakeThisOuTspamultrix6.cs.csubak.edu">.....dgriffiKILLspamspam@spam@ultrix6.cs.csubak.edu</A>
to <A
href="MAHKspamKILLspamwebtv.net">.....MAHKKILLspamspam.....webtv.net</A>,<BR>&nbsp;&nbsp;&nbsp;&nbsp; in
alt.pizza.delivery.drivers<BR></FONT></DIV></BODY></HTML>

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1999\06\07@222242 by Sean Breheny

face picon face
<x-rich>Hi David,


I don't think that there is any need to switch to mosfets here, although,
you could if you wanted. Here is what I suggest (I'm sure the gurus here
will have an even better idea):



                          +5V                      +5V

                           |                        |

Input                       10K                    Inverter

                           |                       /

o-----+                     +----------------------|    Any small NPN
(Q2)

     | 4.7K pot           /                        >

     R<<-------->|--------|    Any small NPN         \

     |       RED LED      >   (Q1)                GND

    GND                    \

                           GND


Here's how it works: You set the pot to divide the input down so that the
threshold voltage (which you gave as usually 4v) gets divided down to
about 2.6v. 2.6v is about equal to the base-emitter turn-on voltage plus
the LED's voltage drop. When Q1 transistor is OFF (the input voltage is
LESS than 4v), current can flow thru the 10K resistor and fully turn on
Q2. When Q1 is ON, it will clamp the base of Q2 to GND and turn Q2 off.


Sean




At 08:46 PM 6/7/99 -0500, you wrote:

>>>>

<excerpt><smaller>Howdy,

</smaller>

<smaller>I know this is off topic, I'm not sure how tolerant this list is
to a few

off topic posts so please accept my apologies in advance if this is

too far off from PICs for your sensibilities. I'll try to keep it
brief.

As usual, replies can be directly to me to avoid cluttering the list.

</smaller>

<smaller>I play with modifying TI calculators in my spare time, one
modification

requires a switch.  Instead of using a physical switch I want to
control

it with a circuit that senses if the calculator is on or off and
switches

the device appropriately.  I have a point on the calc that is 5v when
the

calc is off, 0v or 2.5v when it is on (depending on screen contrast).

I want the power to the device to be off when the voltage at this point

is above about 4v (actually Vcc, it varies).

</smaller>

<smaller>I am currently using a circuit I designed with some junk box
transistors,

however, not having any formal training, I haven't done it optimally,
and

it is quite sensitive to variations in component values (about 10% of
my

transistors work in this design).  Also, the transistors drop too much

voltage, the power to my device (an inverter) is less than what it
could

be.  Therefore, I would like to redesign the circuit to use a MOSFET

or something similar that has very low on resistance.  However, not

having used them before, I have no idea how to proceed.  As usual

with electronics, there is no 'How to use a XXXXX' to be found, or even

a 'How NOT to use a XXXXX'.

</smaller>

<smaller>Details for the device can be found at my web page:

<<http://www.genetech.net/dknaack/autopower.html>http://www.genetech.net/dknaack
/autopower.html

</smaller>

<smaller>Thanks in advance if you can help me out.

DK

</smaller>

<smaller>I want you to be fully aware of your role in Usenet.  It goes
like this:

Bill Gates is the organ grinder, playing a song called "Web TV".  You
are

the little monkey in a bright red hat hopping up and down and acting
silly

for our amusement.

 -- <<EraseMEdgriffispam_OUTspamTakeThisOuTultrix6.cs.csubak.edu>dgriffispamspam_OUTultrix6.cs.csubak.edu
to <<@spam@MAHKKILLspamspamwebtv.net>KILLspamMAHKKILLspamspamwebtv.net,

    in alt.pizza.delivery.drivers

</smaller>

</excerpt><<<<<<<<




|

| Sean Breheny

| Amateur Radio Callsign: KA3YXM

| Electrical Engineering Student

\--------------=----------------

Save lives, please look at http://www.all.org

Personal page: http://www.people.cornell.edu/pages/shb7

RemoveMEshb7TakeThisOuTspamcornell.edu ICQ #: 3329174
________________________________________________________
NetZero - We believe in a FREE Internet.  Shouldn't you?
Get your FREE Internet Access and Email at
http://www.netzero.net/download/index.html

</x-rich>

1999\06\07@231814 by l.allen

picon face
{Quote hidden}

To ensure accurate switching some sort of reference (e.g bandgap) ,
comparator and output is needed otherwise you are at the mercy of
component variations as to the switch on voltage.
I would use something like a MAX 883 voltage regulator with the IN
terminal connector to your inverter VCC and the on/off signal to the
low battery detect pin (see Maxim data sheet) via 2 resistors with
the LBO (pull down) connected to say a 1Meg resistor to VCC and the
gate of a P Channel Mosfet. If VCC  is not around 12 volts (I think
you're web page indicates 6 volts?) then a logic level mosfet will be
needed. Source goes to VCC and Drain to your inverter.   This circuit
should consume a few microamps and is very reliable with a relatively
small number of components. Of course the voltage regulator output is
not used as it is the low battery detect circuit we want.

Lance Allen

1999\06\07@232033 by Sean Breheny

face picon face
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
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
spamBeGoneshb7spamBeGonespamcornell.edu ICQ #: 3329174
________________________________________________________
NetZero - We believe in a FREE Internet.  Shouldn't you?
Get your FREE Internet Access and Email at
http://www.netzero.net/download/index.html

1999\06\08@022320 by Maris

picon face
David --

The following circuit should work reliably using generic Radio Shack
transistors and won't draw any current when off.


                       +Vcc      +Vcc   +Vcc (+6V)
                         |         |      |
                    100K R         |      |
                    R2   R         |      |
                         |         e      |
          D1     R1      |       |/ Q1    |
    in-----|<-----RRR----o-------|  PNP   |
         1N4148   51K    |       |\       | +
       or 1N914          |         c    -----
                    2meg R         |    |   |
                     R3  R    2.2k R    |   |inverter
                         |    R4   R    -----
                         |         |      | -
                         -----------------o
                                   |      |
                                   |      c
                                   |    |/    Q2
                                   o----|    2N2222
                                   |    |\   or any NPN
                              100k R      e
                              R5   R      |
                                   |      |
                                Ground  Ground

Explanation:

With the input held at Vcc, Q1 is off and no current flows in R4, keeping
Q2 off. D1 along with the voltage divider formed by R1 and R2 gives a
turn-on voltage of about 1.5 volts below Vcc, or about +4.5 volts with
respect to GND. With the input high, no current flows in D1, R1, R2, R3,
R4, R5, Q1 or Q2.

When the input is brought down to +2.5 volts, Q1 turns on, allowing current
to flow into the base of Q2 through R4, turning Q2 on. R4 limits the base
current into Q2 to approximately 2.5mA. The base current is amplified by Q2
and provides approximately 100mA to the inverter. If more current is needed
for the inverter, reduce R4 to 1K. The voltage drop across Q2 will be on
the order of .1 to .5 volts depending on the type of NPN transistor, so the
inverter will see around 5.9 to 5.5 volts with a 6 volt supply.

R3 serves to provide a small amount of "snap action" to the circuit by
providing extra current to the base of Q1 when Q2 turns on.

Don't make a connection where the wire from R3 crosses the wire from R4.
Another diode can be placed in series with D1 if needed to increase the
threshold. R2, R3 and R5 can probably be omitted from the circuit to save
space; it should work anyway.

**********************************************************

David Knaack wrote:

I play with modifying TI calculators in my spare time, one modification
requires a switch.  Instead of using a physical switch I want to control
it with a circuit that senses if the calculator is on or off and switches
the device appropriately.  I have a point on the calc that is 5v when the
calc is off, 0v or 2.5v when it is on (depending on screen contrast).
I want the power to the device to be off when the voltage at this point
is above about 4v (actually Vcc, it varies).

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