You need a voltage comparator.
http://www.allaboutcircuits.com/vol_6/chpt_6/2.html
The circuit shown here will only drive a circuit of
low current when on like around 10 ma's or so. if you
need more, you can use a transistor to boost current.
check back with the guys on that later on also, you can
use an alternate part like a LM311 which can drive a small
relay coil.
http://www.fairchildsemi.com/ds/LM/LM311.pdf
on this, you would need to connect the COM and VEE together
when using a single ended supply. also, the output only sinks
which means it pulls to common and does not source an output
equal or near the power supply voltage.. This isn't a problem because
you can thus pull AUX control line functions to common that have higher
voltages than what this chip uses. For example, you can operate the
chip at 5 volts and pull a load to common up to 40 volts..
This is more favorable when signaling various AUX equipment.
You may want recheck the values on the pots, those sizes sounds a bit
on the low side..
Are you sure there isn't a "K" on the display on your meter?
this would make more sense..
the readings you would be getting would then be 10K ohms and 6 K
ohms.
etc..
http://webpages.charter.net/jamie_5"
On May 29, 7:09 pm, Jamie
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
> lizardbo...@gmail.com wrote:
> > I would like to convert some output (ohm resistance?) from a
[quoted text clipped - 47 lines]
>
> http://webpages.charter.net/jamie_5"
Again, thx for the help. I've already learned more than I knew
before.
I need to add that each pot actually represents an Axis, so there is a
positive switch point and a negative switch point. Don't know if this
makes a difference...I'm sure it does.
In the example you linked to, on the final diagram...the pot I want to
compare (joystick axis) would go in place of the Windmill. Could I
replace the pot in that last diagram with a 6k ohm resistor since this
is the tipping point?
Jamie - 30 May 2008 01:29 GMT
> On May 29, 7:09 pm, Jamie
> <jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
[quoted text clipped - 63 lines]
> replace the pot in that last diagram with a 6k ohm resistor since this
> is the tipping point?
Well, that's actually a generator but I understand what you're talking
about.
The generator produces voltages from 0 up to what ever. that circuit
will switch on when the (+) input exceeds the (-) input.
What you need to do is if you plan on using the (-) for the pot
input as it shows in the image. You need to place a fixed voltage that
sets the window trip point on the (+) input.
Or, you could use the circuit as is and put the wiper of the pot in
the joy stick on the (+) input, one outer connection on the common
(ground), and the other outer connection of the pot on the +Voltage
supply that is operating the comparator. In this case, you would need a
Trim pot on the (-) as you see it in the print to set the trip voltage.
So, if you were to place the joy stick in the middle, the wiper on one
of the pots should be generating half of the supplied voltage, in this
case 9 Volts.
You don't need to use a Trim pot as shown in the print for the (-)
input, you can use a pair of resistors to form a voltage divider how
ever, in your case where the POTS in the stick most likely are not
perfectly aligned, you may want the trim pot on the (-) as shown to
adjust your settings. A value of 10k is ok, you can use lower values also.
Now keep in mind with this simple circuit, the stick is only going
to respond in one direct of the pot you connect it to.
If you're looking for lets say a window effect on one pot where you
move left /right for example which should move the same pot, you would
then need a "Window comparator". The simplest approach is to use the
other OP-AMP that is in the 1458 part that is being used. You would
build an equal circuit as the first one how ever, the inputs are
reversed. For the pot from the stick, you would connect the wiper to the
(-) input and the preset trim pot, you connect to the (+) input..
So what happens is this Op-amp will work in reverse logic as the
other.
If you want the 2 to drive a single common load, you need to use
diodes on the output of each so that when one is off and the other is
on, it won't short on each other..
This is where a LM393 series comes into play, since the output of
this comparator is a open collector which means it sinks only (pulls to
common), you can tie a number of these all together on their outputs
to form a single signal. It also has 2 units in a 8 pin body so this
makes it easy to use as a window comparator and in your case you would
need 3 of these chips to do the window compare for all three pots unless
you don't need a window compare and only a one sided compare.
This should give you something to think about :)
http://www.onsemi.com/pub_link/Collateral/LM393-D.PDF
http://webpages.charter.net/jamie_5"